wrong_submission_id
stringlengths 10
10
| problem_id
stringlengths 6
6
| user_id
stringlengths 10
10
| time_limit
float64 1k
8k
| memory_limit
float64 131k
1.05M
| wrong_status
stringclasses 2
values | wrong_cpu_time
float64 10
40k
| wrong_memory
float64 2.94k
3.37M
| wrong_code_size
int64 1
15.5k
| problem_description
stringlengths 1
4.75k
| wrong_code
stringlengths 1
6.92k
| acc_submission_id
stringlengths 10
10
| acc_status
stringclasses 1
value | acc_cpu_time
float64 10
27.8k
| acc_memory
float64 2.94k
960k
| acc_code_size
int64 19
14.9k
| acc_code
stringlengths 19
14.9k
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s329520151
|
p03556
|
u411278350
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 56
|
Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer.
|
N = int(input())
N_s = (N ** (1/2)) // 1
print(N_s ** 2)
|
s343515257
|
Accepted
| 38
| 2,940
| 131
|
N = int(input())
ans = 1
for i in range(1, N):
if ans <= i ** 2 <= N:
ans = i ** 2
else:
break
print(ans)
|
s195794975
|
p03693
|
u373047809
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 56
|
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
print("YENOS"[max(int(input().replace(" ",""))%4,1)::2])
|
s678321021
|
Accepted
| 17
| 2,940
| 40
|
print("YNEOS"[int(input()[::2])%4>0::2])
|
s754060924
|
p03777
|
u329058683
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 69
|
Two deer, AtCoDeer and TopCoDeer, are playing a game called _Honest or Dishonest_. In this game, an honest player always tells the truth, and an dishonest player always tell lies. You are given two characters a and b as the input. Each of them is either `H` or `D`, and carries the following information: If a=`H`, AtCoDeer is honest; if a=`D`, AtCoDeer is dishonest. If b=`H`, AtCoDeer is saying that TopCoDeer is honest; if b=`D`, AtCoDeer is saying that TopCoDeer is dishonest. Given this information, determine whether TopCoDeer is honest.
|
L=list(input().split())
if set(L)==2:
print("H")
else:
print("D")
|
s883220956
|
Accepted
| 17
| 2,940
| 60
|
A,B=input().split()
if A==B:
print("H")
else:
print("D")
|
s266196609
|
p00049
|
u821624310
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,328
| 345
|
ある学級の生徒の出席番号と ABO 血液型を保存したデータを読み込んで、おのおのの血液型の人数を出力するプログラムを作成してください。なお、ABO 血液型には、A 型、B 型、AB 型、O 型の４種類の血液型があります。
|
import sys
A = 0
B = 0
AB = 0
O = 0
for line in sys.stdin:
if line == "\n":
break
No, b_type = line.rstrip().split(",")
print(No, b_type)
if b_type == "A":
A += 1
elif b_type == "B":
B += 1
elif b_type == "AB":
AB += 1
else:
O += 1
print(A)
print(B)
print(AB)
print(O)
|
s942819132
|
Accepted
| 20
| 7,300
| 238
|
import sys
blood = {"A": 0, "B": 0, "AB": 0, "O": 0}
for line in sys.stdin:
if line == "\n":
break
no, b = line.rstrip().split(",")
blood[b] += 1
print(blood["A"])
print(blood["B"])
print(blood["AB"])
print(blood["O"])
|
s808004082
|
p04043
|
u506086925
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 124
|
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
a, b, c = map(int,input().split())
sn = [a,b,c]
if sn.count(7)==1 and sn.count(5)==2:
print("Yes")
else:
print("No")
|
s903411580
|
Accepted
| 17
| 2,940
| 124
|
a, b, c = map(int,input().split())
sn = [a,b,c]
if sn.count(7)==1 and sn.count(5)==2:
print("YES")
else:
print("NO")
|
s707524327
|
p03698
|
u640603056
| 2,000
| 262,144
|
Wrong Answer
| 21
| 3,316
| 142
|
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
from collections import Counter
lst = list(input())
c = sorted(Counter(lst).values(), reverse=True)
if c[0]==0: print("yes")
else: print("no")
|
s348450953
|
Accepted
| 20
| 3,316
| 142
|
from collections import Counter
lst = list(input())
c = sorted(Counter(lst).values(), reverse=True)
if c[0]==1: print("yes")
else: print("no")
|
s676078138
|
p03998
|
u827141374
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 428
|
Alice, Bob and Charlie are playing _Card Game for Three_ , as below: * At first, each of the three players has a deck consisting of some number of cards. Each card has a letter `a`, `b` or `c` written on it. The orders of the cards in the decks cannot be rearranged. * The players take turns. Alice goes first. * If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says `a`, Alice takes the next turn.) * If the current player's deck is empty, the game ends and the current player wins the game. You are given the initial decks of the players. More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way. Determine the winner of the game.
|
A = [x for x in input()]
B = [x for x in input()]
C = [x for x in input()]
next = 'a'
while True:
if next == 'a':
next = A.pop(0)
if len(A) == 0:
print('A')
break
elif next == 'b':
next = B.pop(0)
if len(B) == 0:
print('B')
break
elif next == 'c':
next = C.pop(0)
if len(C) == 0:
print('C')
break
|
s997759109
|
Accepted
| 17
| 3,064
| 428
|
A = [x for x in input()]
B = [x for x in input()]
C = [x for x in input()]
next = 'a'
while True:
if next == 'a':
if len(A) == 0:
print('A')
break
next = A.pop(0)
elif next == 'b':
if len(B) == 0:
print('B')
break
next = B.pop(0)
elif next == 'c':
if len(C) == 0:
print('C')
break
next = C.pop(0)
|
s631602602
|
p03150
|
u580316619
| 2,000
| 1,048,576
|
Wrong Answer
| 29
| 9,084
| 142
|
A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once. Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string.
|
s=input()
ans='No'
for i in range(1,len(s)):
for j in range(i,len(s)):
if s[:i]+s[j:]=='keyence':
ans='Yes'
break
print(ans)
|
s209220544
|
Accepted
| 29
| 8,996
| 142
|
s=input()
ans='NO'
for i in range(1,len(s)):
for j in range(i,len(s)):
if s[:i]+s[j:]=='keyence':
ans='YES'
break
print(ans)
|
s382831582
|
p03494
|
u259861571
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,104
| 2,940
| 293
|
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
n = input()
a = list(map(int, input().split()))
count = 0
odd = ""
while True:
for i, n in enumerate(a):
if n % 2 == 0:
a[i] == n / 2
count += 1
else:
odd = "finish"
break
if odd == "finish":
break
print(count)
|
s177787318
|
Accepted
| 20
| 2,940
| 198
|
n = input()
a = list(map(int, input().split()))
ans = a[0]
for i in range(int(n)):
count = 0
while(a[i] % 2 == 0):
a[i] /= 2
count += 1
ans = min(ans, count)
print(ans)
|
s684333289
|
p03494
|
u814986259
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 264
|
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
N = int(input())
A = list(map(int, input().split()))
flag = True
ans = 0
while(1):
for i in range(N):
if A[i] % 2 != 0:
flag = False
break
if flag:
A = list(map(lambda x:x/2, A))
else:
break
print(ans)
|
s997863089
|
Accepted
| 19
| 3,060
| 281
|
N = int(input())
A = list(map(int, input().split()))
flag = True
ans = 0
while(1):
for i in range(N):
if A[i] % 2 != 0:
flag = False
break
if flag:
A = list(map(lambda x:x/2, A))
ans += 1
else:
break
print(ans)
|
s920746617
|
p02259
|
u821624310
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,580
| 219
|
Write a program of the Bubble Sort algorithm which sorts a sequence _A_ in ascending order. The algorithm should be based on the following pseudocode: BubbleSort(A) 1 for i = 0 to A.length-1 2 for j = A.length-1 downto i+1 3 if A[j] < A[j-1] 4 swap A[j] and A[j-1] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 4 of the pseudocode.
|
N = int(input())
A = list(map(int, input().split()))
n = N
while n > 0:
for i in range(N-1, 0, -1):
if A[i] < A[i-1]:
t = A[i]
A[i] = A[i-1]
A[i-1] = t
n -= 1
print(A)
|
s644846010
|
Accepted
| 20
| 7,732
| 370
|
N = int(input())
A = [int(n) for n in input().split()]
flg = 1
cnt = 0
while flg:
flg = 0
for i in range(N-1, 0, -1):
if A[i] < A[i-1]:
t = A[i]
A[i] = A[i-1]
A[i-1] = t
flg = 1
cnt += 1
for i in range(N):
if i == N - 1:
print(A[i])
else:
print(A[i], end=" ")
print(cnt)
|
s156860191
|
p03737
|
u027675217
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 47
|
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
|
a,b,c = input().split()
print(a[0]+b[0]+c[0])
|
s518881801
|
Accepted
| 17
| 2,940
| 109
|
a,b,c = input().split()
a_d=a.capitalize()
b_d=b.capitalize()
c_d=c.capitalize()
print(a_d[0]+b_d[0]+c_d[0])
|
s760909041
|
p03067
|
u366269356
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 98
|
There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise.
|
a,b,c = map(int,input().split())
if a < c < b or b < c < a:
print('yes')
else:
print('no')
|
s796402402
|
Accepted
| 17
| 2,940
| 133
|
A,B,C = map(int, input().split())
if A < C and C < B:
print('Yes')
elif A > C and C > B:
print('Yes')
else:
print('No')
|
s521190140
|
p03598
|
u773865844
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 245
|
There are N balls in the xy-plane. The coordinates of the i-th of them is (x_i, i). Thus, we have one ball on each of the N lines y = 1, y = 2, ..., y = N. In order to collect these balls, Snuke prepared 2N robots, N of type A and N of type B. Then, he placed the i-th type-A robot at coordinates (0, i), and the i-th type-B robot at coordinates (K, i). Thus, now we have one type-A robot and one type-B robot on each of the N lines y = 1, y = 2, ..., y = N. When activated, each type of robot will operate as follows. * When a type-A robot is activated at coordinates (0, a), it will move to the position of the ball on the line y = a, collect the ball, move back to its original position (0, a) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. * When a type-B robot is activated at coordinates (K, b), it will move to the position of the ball on the line y = b, collect the ball, move back to its original position (K, b) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. Snuke will activate some of the 2N robots to collect all of the balls. Find the minimum possible total distance covered by robots.
|
import sys
lines = sys.stdin.readlines()
N = int(lines[0])
K = int(lines[1])
X = lines[2].split(' ')
x =[]
y = 0
for i in range(N):
x.append(int(X[i]))
if 2*x[i] >= K:
dist = K-x[i]
else:
dist = x[i]
y = y + dist
print(y)
|
s144140647
|
Accepted
| 18
| 3,064
| 251
|
import sys
lines = sys.stdin.readlines()
N = int(lines[0])
K = int(lines[1])
X = lines[2].split(' ')
x =[]
y = 0
for i in range(N):
x.append(int(X[i]))
if 2*x[i] >= K:
dist = 2*(K-x[i])
else:
dist = 2*x[i]
y = y + dist
print(y)
|
s744068991
|
p03698
|
u509739538
| 2,000
| 262,144
|
Wrong Answer
| 22
| 3,444
| 2,587
|
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
import math
from collections import deque
from collections import defaultdict
def readInt():
return int(input())
def readInts():
return list(map(int, input().split()))
def readChar():
return input()
def readChars():
return input().split()
def factorization(n):
res = []
if n%2==0:
res.append(2)
for i in range(3,math.floor(n//2)+1,2):
if n%i==0:
c = 0
for j in res:
if i%j==0:
c=1
if c==0:
res.append(i)
return res
def fact2(n):
p = factorization(n)
res = []
for i in p:
c=0
z=n
while 1:
if z%i==0:
c+=1
z/=i
else:
break
res.append([i,c])
return res
def fact(n):
ans = 1
m=n
for _i in range(n-1):
ans*=m
m-=1
return ans
def comb(n,r):
if n<r:
return 0
l = min(r,n-r)
m=n
u=1
for _i in range(l):
u*=m
m-=1
return u//fact(l)
def combmod(n,r,mod):
return (fact(n)/fact(n-r)*pow(fact(r),mod-2,mod))%mod
def printQueue(q):
r=copyQueue(q)
ans=[0]*r.qsize()
for i in range(r.qsize()-1,-1,-1):
ans[i] = r.get()
print(ans)
class UnionFind():
def __init__(self, n):
self.n = n
self.parents = [-1]*n
def find(self, x): # root
if self.parents[x]<0:
return x
else:
self.parents[x] = self.find(self.parents[x])
return self.parents[x]
def union(self,x,y):
x = self.find(x)
y = self.find(y)
if x==y:
return
if self.parents[x]>self.parents[y]:
x,y = y,x
self.parents[x]+=self.parents[y]
self.parents[y]=x
def size(self,x):
return -1*self.parents[self.find(x)]
def same(self,x,y):
return self.find(x)==self.find(y)
def members(self,x): # much time
root = self.find(x)
return [i for i in range(self.n) if self.find(i)==root]
def roots(self):
return [i for i,x in enumerate(self.parents) if x<0]
def group_count(self):
return len(self.roots())
def all_group_members(self):
return {r: self.members(r) for r in self.roots()} # 1~n
def bitArr(n):
x = 1
zero = "0"*n
ans = []
ans.append([0]*n)
for i in range(2**n-1):
ans.append(list(map(lambda x:int(x),list((zero+bin(x)[2:])[-1*n:]))))
x+=1
return ans;
def arrsSum(a1,a2):
for i in range(len(a1)):
a1[i]+=a2[i]
return a1
def maxValue(a,b,v):
v2 = v
for i in range(v2,-1,-1):
for j in range(v2//a+1):
k = i-a*j
if k%b==0:
return i
return -1
def copyQueue(q):
nq = queue.Queue()
n = q.qsize()
for i in range(n):
x = q.get()
q.put(x)
nq.put(x)
return nq
s = readChar()
d = defaultdict(int)
for i in s:
d[s]+=1
for i in d.values():
if i!=1:
print("no")
exit()
print("yes")
|
s416648989
|
Accepted
| 23
| 3,444
| 2,587
|
import math
from collections import deque
from collections import defaultdict
def readInt():
return int(input())
def readInts():
return list(map(int, input().split()))
def readChar():
return input()
def readChars():
return input().split()
def factorization(n):
res = []
if n%2==0:
res.append(2)
for i in range(3,math.floor(n//2)+1,2):
if n%i==0:
c = 0
for j in res:
if i%j==0:
c=1
if c==0:
res.append(i)
return res
def fact2(n):
p = factorization(n)
res = []
for i in p:
c=0
z=n
while 1:
if z%i==0:
c+=1
z/=i
else:
break
res.append([i,c])
return res
def fact(n):
ans = 1
m=n
for _i in range(n-1):
ans*=m
m-=1
return ans
def comb(n,r):
if n<r:
return 0
l = min(r,n-r)
m=n
u=1
for _i in range(l):
u*=m
m-=1
return u//fact(l)
def combmod(n,r,mod):
return (fact(n)/fact(n-r)*pow(fact(r),mod-2,mod))%mod
def printQueue(q):
r=copyQueue(q)
ans=[0]*r.qsize()
for i in range(r.qsize()-1,-1,-1):
ans[i] = r.get()
print(ans)
class UnionFind():
def __init__(self, n):
self.n = n
self.parents = [-1]*n
def find(self, x): # root
if self.parents[x]<0:
return x
else:
self.parents[x] = self.find(self.parents[x])
return self.parents[x]
def union(self,x,y):
x = self.find(x)
y = self.find(y)
if x==y:
return
if self.parents[x]>self.parents[y]:
x,y = y,x
self.parents[x]+=self.parents[y]
self.parents[y]=x
def size(self,x):
return -1*self.parents[self.find(x)]
def same(self,x,y):
return self.find(x)==self.find(y)
def members(self,x): # much time
root = self.find(x)
return [i for i in range(self.n) if self.find(i)==root]
def roots(self):
return [i for i,x in enumerate(self.parents) if x<0]
def group_count(self):
return len(self.roots())
def all_group_members(self):
return {r: self.members(r) for r in self.roots()} # 1~n
def bitArr(n):
x = 1
zero = "0"*n
ans = []
ans.append([0]*n)
for i in range(2**n-1):
ans.append(list(map(lambda x:int(x),list((zero+bin(x)[2:])[-1*n:]))))
x+=1
return ans;
def arrsSum(a1,a2):
for i in range(len(a1)):
a1[i]+=a2[i]
return a1
def maxValue(a,b,v):
v2 = v
for i in range(v2,-1,-1):
for j in range(v2//a+1):
k = i-a*j
if k%b==0:
return i
return -1
def copyQueue(q):
nq = queue.Queue()
n = q.qsize()
for i in range(n):
x = q.get()
q.put(x)
nq.put(x)
return nq
s = readChar()
d = defaultdict(int)
for i in s:
d[i]+=1
for i in d.values():
if i!=1:
print("no")
exit()
print("yes")
|
s558048428
|
p02270
|
u022407960
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,720
| 1,428
|
In the first line, two integers $n$ and $k$ are given separated by a space character. In the following $n$ lines, $w_i$ are given respectively.
|
# encoding: utf-8
import sys
import math
class Solution:
def __init__(self):
"""
init input array
"""
self.__input = sys.stdin.readlines()
@property
def solution(self):
product_num = int(self.__input[0].split()[0])
truck_num = int(self.__input[0].split()[1])
weight_list = list(map(int, self.__input[1:]))
assert product_num == len(weight_list)
print(product_num, truck_num, weight_list)
truck_capacity = max(math.ceil(sum(weight_list) / truck_num), max(weight_list))
return str(truck_capacity)
# for each in array_2:
# if self.allocation(key=each, array=weight_list, array_length=max(weight_list)):
# truck_capacity += 1
#
# return str(truck_capacity)
#
# @staticmethod
# def allocation(key, array, array_length):
# # print(len(set(array_1).intersection(set(array_2))))
# left, right = max(array), array_length
# while left < right:
# mid = (left + right) // 2
# if key == array[mid]:
# return True
# elif key < array[mid]:
# right = mid
# else:
# left = mid + 1
# return False
if __name__ == '__main__':
case = Solution()
print(case.solution)
|
s603226808
|
Accepted
| 260
| 18,596
| 1,376
|
# encoding: utf-8
import sys
class Solution:
def __init__(self):
"""
init input array
"""
self.__input = sys.stdin.readlines()
@property
def solution(self):
product_num = int(self.__input[0].split()[0])
truck_num = int(self.__input[0].split()[1])
weight_list = list(map(int, self.__input[1:]))
# binary search
left, right, mid = max(weight_list), sum(weight_list), 0
while left < right:
mid = (left + right) // 2
# check return True : p could be smaller
if self.check(max_load_cursor=mid, weight_list=weight_list, truck_num=truck_num):
right = mid
else:
mid += 1
left = mid
return str(mid)
@staticmethod
def check(max_load_cursor, weight_list, truck_num):
track_count = 1
current_load = 0
for item_weight in weight_list:
current_load += item_weight
# change to next truck
if current_load > max_load_cursor:
current_load = item_weight
track_count += 1
if track_count > truck_num:
return False
return True
if __name__ == '__main__':
case = Solution()
print(case.solution)
|
s778898068
|
p03680
|
u494748969
| 2,000
| 262,144
|
Wrong Answer
| 162
| 13,460
| 219
|
Takahashi wants to gain muscle, and decides to work out at AtCoder Gym. The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up. These buttons are numbered 1 through N. When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i. When Button i is not lighten up, nothing will happen by pressing it. Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up. Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons.
|
n = int(input())
a = [0] * n
for i in range(n):
a.append(int(input()))
count = 0
num = 0
for i in range(n):
num = a[num]
count += 1
if num == 2:
break
if num != 2:
count = -1
print(count)
|
s295551036
|
Accepted
| 165
| 13,044
| 217
|
n = int(input())
a = []
for i in range(n):
a.append(int(input()))
count = 0
num = 1
for i in range(n):
num = a[num - 1]
count += 1
if num == 2:
break
if num != 2:
count = -1
print(count)
|
s590375864
|
p03795
|
u353919145
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 261
|
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
n=int(input())
can=n*800
p=n
des=200
if n>=15 and n<30:
total=can-200
elif n>=30 and n<45:
total=can-400
elif n>=45 and n<60:
total=can-600
elif n>=60 and n<75:
total=can-800
elif n>=75 and n<90:
total=can-1000
elif n>=90 and n<=100:
total=can-1200
|
s811025024
|
Accepted
| 23
| 9,016
| 186
|
from sys import stdin, stdout
def main() -> None:
line = int(stdin.readline())
x = line * 800
tmp = line / 15
y = int(tmp) * 200
rta = x - y
print(rta)
main()
|
s816128702
|
p00014
|
u032662562
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,580
| 315
|
Write a program which computes the area of a shape represented by the following three lines: $y = x^2$ $y = 0$ $x = 600$ It is clear that the area is $72000000$, if you use an integral you learn in high school. On the other hand, we can obtain an approximative area of the shape by adding up areas of many rectangles in the shape as shown in the following figure: $f(x) = x^2$ The approximative area $s$ where the width of the rectangles is $d$ is: area of rectangle where its width is $d$ and height is $f(d)$ $+$ area of rectangle where its width is $d$ and height is $f(2d)$ $+$ area of rectangle where its width is $d$ and height is $f(3d)$ $+$ ... area of rectangle where its width is $d$ and height is $f(600 - d)$ The more we decrease $d$, the higer-precision value which is close to $72000000$ we could obtain. Your program should read the integer $d$ which is a divisor of $600$, and print the area $s$.
|
def integral(d):
print("d=%d" % d)
n = 600//d
print("n=%d" % n)
s = 0.0
for i in range(n):
s += (d*i)**2
return(s*d)
if __name__ == '__main__':
while True:
try:
d = int(input())
print(int(integral(d)))
except:
break
|
s522806639
|
Accepted
| 30
| 7,656
| 266
|
def integral(d):
n = 600//d
s = 0.0
for i in range(n):
s += (d*i)**2
return(s*d)
if __name__ == '__main__':
while True:
try:
d = int(input())
print(int(integral(d)))
except:
break
|
s823592240
|
p03550
|
u843135954
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,188
| 275
|
We have a deck consisting of N cards. Each card has an integer written on it. The integer on the i-th card from the top is a_i. Two people X and Y will play a game using this deck. Initially, X has a card with Z written on it in his hand, and Y has a card with W written on it in his hand. Then, starting from X, they will alternately perform the following action: * Draw some number of cards from the top of the deck. Then, discard the card in his hand and keep the last drawn card instead. Here, at least one card must be drawn. The game ends when there is no more card in the deck. The score of the game is the absolute difference of the integers written on the cards in the two players' hand. X will play the game so that the score will be maximized, and Y will play the game so that the score will be minimized. What will be the score of the game?
|
import sys
stdin = sys.stdin
sys.setrecursionlimit(10**6)
ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
nn = lambda: list(stdin.readline().split())
ns = lambda: stdin.readline().rstrip()
n,z,w = na()
a = na()
print(max(abs(a[-1]-w),abs(z-w)))
|
s460286295
|
Accepted
| 20
| 3,188
| 330
|
import sys
stdin = sys.stdin
sys.setrecursionlimit(10**6)
ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
nn = lambda: list(stdin.readline().split())
ns = lambda: stdin.readline().rstrip()
n,z,w = na()
a = na()
if n == 1:
print(abs(a[-1]-w))
exit()
print(max(abs(a[-1]-w),abs(a[-2]-a[-1])))
|
s733682162
|
p02669
|
u194472175
| 2,000
| 1,048,576
|
Wrong Answer
| 194
| 12,552
| 1,150
|
You start with the number 0 and you want to reach the number N. You can change the number, paying a certain amount of coins, with the following operations: * Multiply the number by 2, paying A coins. * Multiply the number by 3, paying B coins. * Multiply the number by 5, paying C coins. * Increase or decrease the number by 1, paying D coins. You can perform these operations in arbitrary order and an arbitrary number of times. What is the minimum number of coins you need to reach N? **You have to solve T testcases.**
|
import sys
import heapq
from collections import deque
T=int(input())
for t in range(T):
N,A,B,C,D= map(int,input().split())
q = [(0,N)]
heapq.heapify(q)
seen=set()
while q:
cost,n = heapq.heappop(q)
if n in seen:
continue
seen.add(n)
if n==0:
print(cost)
break
heapq.heappush(q,(cost+n*D,0))
if n%2==0:
heapq.heappush(q,(cost+A,n//2))
else:
heapq.heappush(q,(cost+D+A,(n+1)//2))
heapq.heappush(q,(cost+D+A,(n-1)//2))
if n%3==0:
heapq.heappush(q,(cost+C,n//3))
else:
rest=n % 3
heapq.heappush(q,(cost+rest*D+B,(n-rest+3)//3))
heapq.heappush(q,(cost+(3-rest)*D+B,(n-rest+3)//3))
if n % 5 ==0:
heapq.heappush(q,(cost+C,n//5))
else:
rest= n%5
heapq.heappush(q,(cost+rest*D+C,(n-rest)//5))
heapq.heappush(q,(cost+(5-rest)*D+C,(n-rest+5)//5))
|
s680384983
|
Accepted
| 174
| 12,816
| 998
|
import sys
import heapq
T = int(input())
from collections import deque
for t in range(T):
N,A,B,C,D = map(int,input().split())
q = [(0, N)]
heapq.heapify(q)
seen = set()
while q:
cost,n = heapq.heappop(q)
if n in seen:
continue
seen.add(n)
if n == 0:
print(cost)
break
heapq.heappush(q, (cost + n * D, 0))
if n % 2 == 0:
heapq.heappush(q, (cost + A, n // 2))
else:
heapq.heappush(q, (cost + D + A, (n + 1) // 2))
heapq.heappush(q, (cost + D + A, (n - 1) // 2))
if n % 3 == 0:
heapq.heappush(q, (cost + B, n // 3))
else:
rest = n % 3
heapq.heappush(q, (cost + rest * D + B, (n - rest) // 3))
heapq.heappush(q, (cost + (3 - rest) * D + B, (n - rest + 3) // 3))
if n % 5 == 0:
heapq.heappush(q, (cost + C, n // 5))
else:
rest = n % 5
heapq.heappush(q, (cost + rest * D + C, (n - rest) // 5))
heapq.heappush(q, (cost + (5 - rest) * D + C, (n - rest + 5) // 5))
|
s498478248
|
p03352
|
u049354454
| 2,000
| 1,048,576
|
Time Limit Exceeded
| 2,104
| 3,064
| 453
|
You are given a positive integer X. Find the largest _perfect power_ that is at most X. Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2.
|
# coding: utf-8
# Your code here!
INPUT=int(input())
b=1
p=2
END=0
while True:
while True:
now_num = b**p
if b == 1:
end_num=now_num
break
elif p==2 and INPUT<now_num:
END = 1
break
elif INPUT < now_num:
end_num = b**(p-1)
break
else:
p+=1
if END == 1:
ans=end_num
break
b+=1
print(ans)
|
s972279505
|
Accepted
| 18
| 3,060
| 314
|
from math import sqrt
X=int(input())
sqrt_X=int(sqrt(X))
ans=1
for b in range(1,sqrt_X + 1):
for p in range(2,sqrt_X+1):
now_num = b**p
if now_num <= X and ans <= now_num:
ans =now_num
print(ans)
|
s321071283
|
p02972
|
u672220554
| 2,000
| 1,048,576
|
Wrong Answer
| 875
| 16,820
| 477
|
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.
|
n = int(input())
a = list(map(int,input().split()))
res = [0 for i in range(n)]
for i in range(n,n//2,-1):
res[i-1] = a[i-1]
for i in range(n//2,0,-1):
l = []
flag = 0
t = i*2
while t <= n:
flag = flag ^ res[t-1]
t = t+i
if a[i-1] != flag:
res[i-1] = 1
if res == [0 for i in range(n)]:
print(0)
else:
lres = []
for i in range(n):
if res[i] == 1:
lres.append(str(i+1))
print(" ".join(lres))
|
s518409560
|
Accepted
| 450
| 17,212
| 422
|
def main():
n = int(input())
a = list(map(int,input().split()))
res = [0 for i in range(n)]
lres = []
for i in range(n,0,-1):
flag = 0
t = i*2
while t <= n:
flag = flag ^ res[t-1]
t = t+i
if a[i-1] != flag:
res[i-1] = 1
lres.append(str(i))
print(len(lres))
if len(lres) != 0:
print(" ".join(lres))
main()
|
s870342598
|
p03545
|
u876616721
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 579
|
Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.
|
x = input()
#print(x)
y = [int(c) for c in x]
#print(y)
a = y[0]
b = y[1]
c = y[2]
d = y[3]
if a+b+c+d == 7:
print("a+b+c+d=",a+b+c+d,sep='')
exit()
if a+b+c-d == 7:
print("a+b+c-d=",a+b+c-d,sep='')
exit()
if a+b-c+d == 7:
print("a+b-c+d=",a+b-c+d,sep='')
exit()
if a-b+c+d == 7:
print("a-b+c+d=",a-b+c+d,sep='')
exit()
if a-b-c+d == 7:
print("a-b-c+d=",a-b-c+d,sep='')
exit()
if a-b+c-d == 7:
print("a-b+c-d=",a-b+c-d,sep='')
exit()
if a-b-c-d == 7:
print("a-b-c-d=",a-b-c-d,sep='')
exit()
if a+b-c-d == 7:
print("a+b-c-d=",a+b-c-d,sep='')
exit()
|
s424105742
|
Accepted
| 17
| 3,064
| 691
|
x = input()
#print(x)
y = [int(c) for c in x]
#print(y)
a = y[0]
b = y[1]
c = y[2]
d = y[3]
if a+b+c+d == 7:
print(a,"+",b,"+",c,"+",d,"=",a+b+c+d, sep='')
exit()
if a+b+c-d == 7:
print(a,"+",b,"+",c,"-",d,"=",a+b+c-d, sep='')
exit()
if a+b-c+d == 7:
print(a,"+",b,"-",c,"+",d,"=",a+b-c+d, sep='')
exit()
if a-b+c+d == 7:
print(a,"-",b,"+",c,"+",d,"=",a-b+c+d, sep='')
exit()
if a-b-c+d == 7:
print(a,"-",b,"-",c,"+",d,"=",a-b-c+d, sep='')
exit()
if a-b+c-d == 7:
print(a,"-",b,"+",c,"-",d,"=",a-b+c-d, sep='')
exit()
if a-b-c-d == 7:
print(a,"-",b,"-",c,"-",d,"=",a-b-c-d, sep='')
exit()
if a+b-c-d == 7:
print(a,"+",b,"-",c,"-",d,"=",a+b-c-d, sep='')
exit()
|
s872629750
|
p03495
|
u268516119
| 2,000
| 262,144
|
Wrong Answer
| 190
| 39,292
| 239
|
Takahashi has N balls. Initially, an integer A_i is written on the i-th ball. He would like to rewrite the integer on some balls so that there are at most K different integers written on the N balls. Find the minimum number of balls that Takahashi needs to rewrite the integers on them.
|
from collections import Counter as ct
N,K=map(int,input().split())
A=[int(i) for i in input().split()]
count=ct(A)
div=len(count)
if div<=K:print(0)
else:print(sum([i[0] for i in sorted(count.items(),key= lambda x:x[1])[:div-K]]))
|
s900188916
|
Accepted
| 188
| 39,288
| 239
|
from collections import Counter as ct
N,K=map(int,input().split())
A=[int(i) for i in input().split()]
count=ct(A)
div=len(count)
if div<=K:print(0)
else:print(sum([i[1] for i in sorted(count.items(),key= lambda x:x[1])[:div-K]]))
|
s687409363
|
p03386
|
u614181788
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 177
|
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
a,b,k = map(int,input().split())
aa = list(range(a,min(a+k,b)))
bb = list(range(max(b-k+1,a),b+1))
aa.extend(bb)
c = set(aa)
d = list(c)
for i in range(len(d)):
print(d[i])
|
s530129514
|
Accepted
| 17
| 3,064
| 191
|
a,b,k = map(int,input().split())
aa = list(range(a,min(a+k,b)))
bb = list(range(max(b-k+1,a),b+1))
aa.extend(bb)
c = set(aa)
d = list(c)
d = sorted(d)
for i in range(len(d)):
print(d[i])
|
s550431972
|
p03025
|
u201234972
| 2,000
| 1,048,576
|
Wrong Answer
| 299
| 18,804
| 644
|
Takahashi and Aoki will play a game. They will repeatedly play it until one of them have N wins in total. When they play the game once, Takahashi wins with probability A %, Aoki wins with probability B %, and the game ends in a draw (that is, nobody wins) with probability C %. Find the expected number of games that will be played, and print it as follows. We can represent the expected value as P/Q with coprime integers P and Q. Print the integer R between 0 and 10^9+6 (inclusive) such that R \times Q \equiv P\pmod {10^9+7}. (Such an integer R always uniquely exists under the constraints of this problem.)
|
def getInv(N):
inv = [0] * (N + 1)
inv[0] = 1
inv[1] = 1
for i in range(2, N + 1):
inv[i] = (-(Q // i) * inv[Q%i]) % Q
return inv
Q = 10**9+7
N, A, B, C = map( int, input().split())
modinv = getInv( max(2*N-1, 100))
a = A*modinv[100]%Q
b = B*modinv[100]%Q
Powa = [1]*(N+1)
Powb = [1]*(N+1)
for i in range(N):
Powa[i+1] = (Powa[i]*a)%Q
Powb[i+1] = (Powb[i]*b)%Q
t = 100*modinv[100-C]%Q
ans = 0
nowc = 1
ans = Powa[N]*N + Powb[N]*N
ans %= Q
for k in range(N+1, N*2):
nowc *= (k-1)*modinv[k-N]
nowc %= Q
ans += nowc*( Powa[N]*Powb[k-N] + Powb[N]*Powa[k-N])%Q*k%Q
ans %= Q
print(ans*t%Q)
|
s044024904
|
Accepted
| 296
| 18,804
| 644
|
def getInv(N):
inv = [0] * (N + 1)
inv[0] = 1
inv[1] = 1
for i in range(2, N + 1):
inv[i] = (-(Q // i) * inv[Q%i]) % Q
return inv
Q = 10**9+7
N, A, B, C = map( int, input().split())
modinv = getInv( max(2*N-1, 100))
a = A*modinv[A+B]%Q
b = B*modinv[A+B]%Q
Powa = [1]*(N+1)
Powb = [1]*(N+1)
for i in range(N):
Powa[i+1] = (Powa[i]*a)%Q
Powb[i+1] = (Powb[i]*b)%Q
t = 100*modinv[100-C]%Q
ans = 0
nowc = 1
ans = Powa[N]*N + Powb[N]*N
ans %= Q
for k in range(N+1, N*2):
nowc *= (k-1)*modinv[k-N]
nowc %= Q
ans += nowc*( Powa[N]*Powb[k-N] + Powb[N]*Powa[k-N])%Q*k%Q
ans %= Q
print(ans*t%Q)
|
s241955345
|
p03861
|
u368882459
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 52
|
You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x?
|
a, b, x = map(int, input().split())
print(b//x-a//x)
|
s589895380
|
Accepted
| 17
| 2,940
| 56
|
a, b, x = map(int, input().split())
print(b//x-(a-1)//x)
|
s262190493
|
p04011
|
u209619667
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 150
|
There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee.
|
N = int(input())
K = int(input())
X = int(input())
Y = int(input())
s = 0
for n in range(N):
if n <= K:
s = s + X
else:
s = s + Y
print(s)
|
s676883791
|
Accepted
| 19
| 2,940
| 154
|
N = int(input())
K = int(input())
X = int(input())
Y = int(input())
s = 0
for n in range(1,N+1):
if n <= K:
s = s + X
else:
s = s + Y
print(s)
|
s682862262
|
p03359
|
u993461026
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 68
|
In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year. For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year. In this country, a date is called _Takahashi_ when the month and the day are equal as numbers. For example, 5-5 is Takahashi. How many days from 2018-1-1 through 2018-a-b are Takahashi?
|
a, b = map(int, input().rstrip().split())
print(a if a < b else a-1)
|
s175613052
|
Accepted
| 17
| 2,940
| 69
|
a, b = map(int, input().rstrip().split())
print(a if a <= b else a-1)
|
s829483684
|
p03657
|
u373593739
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 157
|
Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.
|
def biscoitos():
A = input()
B = input()
soma = int(A)+int(B)
if soma%3 == 0:
print('Possible')
else:
print('Impossible')
|
s824039851
|
Accepted
| 17
| 3,060
| 185
|
A, B = map(int,input().split())
soma = A+B
if A%3 == 0:
print('Possible')
elif B%3 == 0:
print ('Possible')
elif soma%3 == 0:
print('Possible')
else:
print('Impossible')
|
s594939669
|
p00001
|
u422087503
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,376
| 98
|
There is a data which provides heights (in meter) of mountains. The data is only for ten mountains. Write a program which prints heights of the top three mountains in descending order.
|
h = []
for i in range(10):
h.append(input())
h.sort(reverse=True)
for i in range(3):
print(h[i])
|
s927324921
|
Accepted
| 20
| 7,684
| 103
|
h = []
for i in range(10):
h.append(int(input()))
h.sort(reverse=True)
for i in range(3):
print(h[i])
|
s864520673
|
p03860
|
u782654209
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 30
|
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
print('A'+str(input())[0]+'C')
|
s410490992
|
Accepted
| 17
| 2,940
| 54
|
print('A'+list(map(str,input().split(' ')))[1][0]+'C')
|
s668315648
|
p03644
|
u819208902
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 100
|
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
if __name__ == '__main__':
n = int()
i = 1
while i * 2 < n:
i *= 2
print(i)
|
s995517123
|
Accepted
| 18
| 3,064
| 108
|
if __name__ == '__main__':
n = int(input())
i = 1
while i * 2 <= n:
i *= 2
print(i)
|
s453040796
|
p03719
|
u797550216
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 95
|
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
a,b,c = map(int,input().split())
if a >= c and c <= b:
print('Yes')
else:
print('No')
|
s590340481
|
Accepted
| 17
| 2,940
| 95
|
a,b,c = map(int,input().split())
if c >= a and c <= b:
print('Yes')
else:
print('No')
|
s199403725
|
p03407
|
u969190727
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 63
|
An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.
|
a,b,c=map(int,input().split())
print("Yes" if a+b<=c else "No")
|
s067086011
|
Accepted
| 17
| 2,940
| 63
|
a,b,c=map(int,input().split())
print("Yes" if a+b>=c else "No")
|
s898998826
|
p02608
|
u664652017
| 2,000
| 1,048,576
|
Wrong Answer
| 2,205
| 9,132
| 410
|
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
def fx(n):
cnt = 0
for i in range(1,10000):
if (i+1+1)**2 - i*1 - 1*1 - 1*i > n: break
for j in range(1, 10000):
if (1+j+1)**2 - 1*j - j*1 - 1*1 > n : break
for k in range(1, 10000):
formula = (i+j+k)**2 - i*j - j*k - k*i
if formula == n:
cnt+=1
if formula >= n:
break
return cnt
n = int(input())
for i in range(1,n):
print(fx(i))
|
s382780078
|
Accepted
| 223
| 9,208
| 495
|
def fx(n):
s = [0]*(n+1)
for i in range(1,10000):
if (i+1+1)**2 - i*1 - 1*1 - 1*i > n: break
for j in range(1, 10000):
if (1+j+1)**2 - 1*j - j*1 - 1*1 > n : break
for k in range(1, 10000):
if (1+1+k)**2 - 1*1 - 1*k - k*1 > n: break
formula = (i+j+k)**2 - i*j - j*k - k*i
if formula <= n and formula > 0:
s[formula] += 1
if formula >= n:
break
return s
n = int(input())
s = fx(n)
for i in range(1,n+1):
print(s[i])
|
s570396831
|
p03698
|
u969211566
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 92
|
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
s = input()
n = len(s)
l = set(s)
m = len(l)
if n == m:
print("Yes")
else:
print("No")
|
s678590240
|
Accepted
| 17
| 2,940
| 92
|
s = input()
n = len(s)
l = set(s)
m = len(l)
if n == m:
print("yes")
else:
print("no")
|
s249648802
|
p02261
|
u357267874
| 1,000
| 131,072
|
Wrong Answer
| 30
| 6,352
| 1,968
|
Let's arrange a deck of cards. There are totally 36 cards of 4 suits(S, H, C, D) and 9 values (1, 2, ... 9). For example, 'eight of heart' is represented by H8 and 'one of diamonds' is represented by D1. Your task is to write a program which sorts a given set of cards in ascending order by their values using the Bubble Sort algorithms and the Selection Sort algorithm respectively. These algorithms should be based on the following pseudocode: BubbleSort(C) 1 for i = 0 to C.length-1 2 for j = C.length-1 downto i+1 3 if C[j].value < C[j-1].value 4 swap C[j] and C[j-1] SelectionSort(C) 1 for i = 0 to C.length-1 2 mini = i 3 for j = i to C.length-1 4 if C[j].value < C[mini].value 5 mini = j 6 swap C[i] and C[mini] Note that, indices for array elements are based on 0-origin. For each algorithm, report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).
|
import copy
class Card:
def __init__(self, suit, number):
self.suit = suit
self.number = number
self.view = self.suit + str(self.number)
def bubble_sort(A):
count = 0
while True:
swapped = False
for i in range(len(A)-1):
if A[i+1].number < A[i].number:
A[i+1], A[i] = A[i], A[i+1]
count += 1
swapped = True
if not swapped:
return count
def selection_sort(A):
count = 0
for i in range(len(A)):
min_value = A[i].number
min_value_index = i
# print('- i:', i, 'A[i]', A[i], '-')
for j in range(i, len(A)):
if A[j].number < min_value:
min_value = A[j].number
min_value_index = j
if i != min_value_index:
count += 1
A[i], A[min_value_index] = A[min_value_index], A[i]
# print('swap!', A)
return count
n = int(input())
A = []
for row in input().split():
suit, number = list(row)
A.append(Card(suit, int(number)))
def is_stable(A, B):
N = len(A)
for i_A in range(N-1):
for j_A in range(i_A+1, N):
for i_B in range(N-1):
for j_B in range(i_B+1, N):
if A[i_A].number == A[j_A].number and A[i_A].view == B[j_B].view and A[j_A].view == B[i_B].view:
return False
return B
bubble_sort_A = copy.deepcopy(A)
selection_sort_A = copy.deepcopy(A)
bubble_sort(bubble_sort_A)
print(*[elem.view for elem in bubble_sort_A])
if is_stable(A, bubble_sort_A):
print('Stable')
else:
print('Not Stable')
selection_sort(selection_sort_A)
print(*[elem.view for elem in selection_sort_A])
if is_stable(A, selection_sort_A):
print('Stable')
else:
print('Not Stable')
|
s325942064
|
Accepted
| 90
| 6,360
| 1,968
|
import copy
class Card:
def __init__(self, suit, number):
self.suit = suit
self.number = number
self.view = self.suit + str(self.number)
def bubble_sort(A):
count = 0
while True:
swapped = False
for i in range(len(A)-1):
if A[i+1].number < A[i].number:
A[i+1], A[i] = A[i], A[i+1]
count += 1
swapped = True
if not swapped:
return count
def selection_sort(A):
count = 0
for i in range(len(A)):
min_value = A[i].number
min_value_index = i
# print('- i:', i, 'A[i]', A[i], '-')
for j in range(i, len(A)):
if A[j].number < min_value:
min_value = A[j].number
min_value_index = j
if i != min_value_index:
count += 1
A[i], A[min_value_index] = A[min_value_index], A[i]
# print('swap!', A)
return count
n = int(input())
A = []
for row in input().split():
suit, number = list(row)
A.append(Card(suit, int(number)))
def is_stable(A, B):
N = len(A)
for i_A in range(N-1):
for j_A in range(i_A+1, N):
for i_B in range(N-1):
for j_B in range(i_B+1, N):
if A[i_A].number == A[j_A].number and A[i_A].view == B[j_B].view and A[j_A].view == B[i_B].view:
return False
return B
bubble_sort_A = copy.deepcopy(A)
selection_sort_A = copy.deepcopy(A)
bubble_sort(bubble_sort_A)
print(*[elem.view for elem in bubble_sort_A])
if is_stable(A, bubble_sort_A):
print('Stable')
else:
print('Not stable')
selection_sort(selection_sort_A)
print(*[elem.view for elem in selection_sort_A])
if is_stable(A, selection_sort_A):
print('Stable')
else:
print('Not stable')
|
s695027765
|
p04031
|
u870297120
| 2,000
| 262,144
|
Wrong Answer
| 20
| 3,060
| 197
|
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal **integers** by transforming some of them. He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2). Find the minimum total cost to achieve his objective.
|
N = int(input())
nums = list(map(int, input().split(' ')))
tmp = 0
max = 10**100
for i in nums:
for j in nums:
tmp += (j-i)**2
if max > tmp:
max = tmp
tmp = 0
print(max)
|
s125379835
|
Accepted
| 25
| 3,060
| 200
|
N = int(input())
nums = list(map(int, input().split(' ')))
temp = 0
cost = []
for i in range(-100, 101):
for j in nums:
temp += (j-i)**2
cost.append(temp)
temp = 0
print(min(cost))
|
s695840682
|
p02866
|
u038854253
| 2,000
| 1,048,576
|
Wrong Answer
| 85
| 14,036
| 457
|
Given is an integer sequence D_1,...,D_N of N elements. Find the number, modulo 998244353, of trees with N vertices numbered 1 to N that satisfy the following condition: * For every integer i from 1 to N, the distance between Vertex 1 and Vertex i is D_i.
|
divider = 998244353
num = int(input())
depths = [int(x) for x in input().split()]
max_depth = max(depths)
if depths[0] != 0:
print(0)
exit(0)
if max_depth == 0:
print(0)
exit(0)
nodes = [0] * (max_depth+1)
for depth in depths:
nodes[depth] += 1
if nodes[0] != 1 or min(nodes) == 0:
print(0)
exit(0)
arr = [(nodes[i] - 1) + nodes[i+1] for i in range(0, max_depth)]
ans = 1
for a in arr:
ans = (ans * a) % divider
print(ans)
exit(0)
|
s404585390
|
Accepted
| 100
| 13,892
| 450
|
divider = 998244353
num = int(input())
depths = [int(x) for x in input().split()]
max_depth = max(depths)
if depths[0] != 0:
print(0)
exit(0)
if max_depth == 0:
print(0)
exit(0)
nodes = [0] * (max_depth+1)
for depth in depths:
nodes[depth] += 1
if nodes[0] != 1 or min(nodes) == 0:
print(0)
exit(0)
arr = [nodes[i] ** nodes[i+1] for i in range(max_depth)]
ans = 1
for a in arr:
ans = (ans * a) % divider
print(ans)
exit(0)
|
s105466560
|
p03433
|
u545334404
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 104
|
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
n = int(input())
a = int(input())
n_1 = n % 500
if a >= n_1:
print("YES")
else:
print("NO")
|
s179056024
|
Accepted
| 17
| 2,940
| 84
|
N=int(input())
A=int(input())
if N % 500 <= A:
print("Yes")
else:
print("No")
|
s151204364
|
p03386
|
u609814378
| 2,000
| 262,144
|
Wrong Answer
| 2,151
| 794,648
| 236
|
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
A,B,K = map(int, input().split())
ans = []
ans2 = []
ans3 = []
for i in range(A,B+1):
ans.append(i)
for i in ans[0:K]:
ans2.append(i)
for j in ans[-K:]:
ans3.append(j)
ans4 = ans2 + ans3
for i in ans4:
print(ans4)
|
s623340452
|
Accepted
| 17
| 3,060
| 214
|
a, b, k = map(int, input().split())
t = []
for i in range(k):
if i+a <= b:
t.append(i+a)
for i in range(k):
if b-i >= a:
t.append(b-i)
y = list(set(t))
for x in sorted(y):
print(x)
|
s315968122
|
p03853
|
u088974156
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 78
|
There is an image with a height of H pixels and a width of W pixels. Each of the pixels is represented by either `.` or `*`. The character representing the pixel at the i-th row from the top and the j-th column from the left, is denoted by C_{i,j}. Extend this image vertically so that its height is doubled. That is, print a image with a height of 2H pixels and a width of W pixels where the pixel at the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division is rounded down).
|
h,w=map(int,input().split())
for i in range(h):
s=input()
print(s+"¥n"+s)
|
s777706953
|
Accepted
| 18
| 3,060
| 77
|
h,w=map(int,input().split())
for i in range(h):
s=input()
print(s+"\n"+s)
|
s203747109
|
p03024
|
u384261199
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 160
|
Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility.
|
S = input()
o_, x_ = 0, 0
for s in S:
if s == "o":
o_ += 1
else:
x_ += 1
if o_ > (15-len(S)) + 7:
print("YES")
else:
print("NO")
|
s121077881
|
Accepted
| 17
| 2,940
| 160
|
S = input()
o_, x_ = 0, 0
for s in S:
if s == "o":
o_ += 1
else:
x_ += 1
if o_ > 7 - (15-len(S)):
print("YES")
else:
print("NO")
|
s322221495
|
p03556
|
u555356625
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 29
|
Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer.
|
print(int(int(input())**0.5))
|
s189238064
|
Accepted
| 17
| 3,060
| 34
|
print((int(int(input())**0.5))**2)
|
s362670868
|
p03448
|
u278670845
| 2,000
| 262,144
|
Wrong Answer
| 50
| 3,064
| 434
|
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
import sys
a = int(input())
b = int(input())
c = int(input())
n = int(input())
count = 0
if n<100:
if c > 0:
count = 1
print(count)
sys.exit()
elif n<500:
for i in range(b+1):
for j in range(c+1):
if 100*b+50*c==n:
count += 1
print(count)
sys.exit()
else:
for i in range(a+1):
for j in range(b+1):
for k in range(c+1):
if 500*a+100*b+50*c==n:
cout += 1
print(count)
|
s829594700
|
Accepted
| 50
| 3,060
| 221
|
import sys
a = int(input())
b = int(input())
c = int(input())
n = int(input())
count = 0
for i in range(a+1):
for j in range(b+1):
for k in range(c+1):
if 500*i+100*j+50*k==n:
count += 1
print(count)
|
s300098117
|
p03369
|
u731665172
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 33
|
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
|
s=input()
print(700+s.count('o'))
|
s503414224
|
Accepted
| 19
| 3,060
| 37
|
s=input()
print(700+100*s.count('o'))
|
s639830425
|
p03400
|
u623349537
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 186
|
Some number of chocolate pieces were prepared for a training camp. The camp had N participants and lasted for D days. The i-th participant (1 \leq i \leq N) ate one chocolate piece on each of the following days in the camp: the 1-st day, the (A_i + 1)-th day, the (2A_i + 1)-th day, and so on. As a result, there were X chocolate pieces remaining at the end of the camp. During the camp, nobody except the participants ate chocolate pieces. Find the number of chocolate pieces prepared at the beginning of the camp.
|
N = int(input())
D, X = map(int, input().split())
A = [0 for i in range(N)]
for i in range(N):
A[i] = int(input())
ans = X
for a in A:
ans += ((D - 1)// A[i] + 1)
print(ans)
|
s629999605
|
Accepted
| 18
| 3,060
| 183
|
N = int(input())
D, X = map(int, input().split())
A = [0 for i in range(N)]
for i in range(N):
A[i] = int(input())
ans = X
for a in A:
ans += ((D - 1)// a + 1)
print(ans)
|
s223376903
|
p00016
|
u747479790
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,912
| 218
|
When a boy was cleaning up after his grand father passing, he found an old paper: In addition, other side of the paper says that "go ahead a number of steps equivalent to the first integer, and turn clockwise by degrees equivalent to the second integer". His grand mother says that Sanbonmatsu was standing at the center of town. However, now buildings are crammed side by side and people can not walk along exactly what the paper says in. Your task is to write a program which hunts for the treature on the paper. For simplicity, 1 step is equivalent to 1 meter. Input consists of several pairs of two integers d (the first integer) and t (the second integer) separated by a comma. Input ends with "0, 0". Your program should print the coordinate (x, y) of the end point. There is the treature where x meters to the east and y meters to the north from the center of town. You can assume that d ≤ 100 and -180 ≤ t ≤ 180\.
|
import math as m
x,y,d=0,0,90
while True:
r,t=map(int,input().split(","))
if (r,t)==(0,0):
break
x+=r*m.cos(m.radians(d))
y+=r*m.sin(m.radians(d))
d-=t
print(int(m.sqrt(x**2+y**2)))
print(d)
|
s918769557
|
Accepted
| 20
| 7,916
| 214
|
# 0016
import math as m
x,y,d=0,0,90
while True:
r,t=map(int,input().split(","))
if (r,t)==(0,0):
break
x+=r*m.cos(m.radians(d))
y+=r*m.sin(m.radians(d))
d-=t
print(int(x))
print(int(y))
|
s951355384
|
p03674
|
u338824669
| 2,000
| 262,144
|
Wrong Answer
| 2,114
| 141,816
| 751
|
You are given an integer sequence of length n+1, a_1,a_2,...,a_{n+1}, which consists of the n integers 1,...,n. It is known that each of the n integers 1,...,n appears at least once in this sequence. For each integer k=1,...,n+1, find the number of the different subsequences (not necessarily contiguous) of the given sequence with length k, modulo 10^9+7.
|
def cmb(n, r, p):
if (r<0) or (n<r):
return 0
r=min(r, n-r)
return fact[n]*factinv[r]*factinv[n-r]%p
p=10**9+7
N=10**6
fact=[1,1]
factinv=[1,1]
inv=[0,1]
for i in range(2, N+1):
fact.append((fact[-1]*i)%p)
inv.append((-inv[p%i]*(p//i)%p))
factinv.append((factinv[-1]*inv[-1])%p)
N=int(input())
A=list(map(int,input().split()))
index=[[] for _ in range(N+1)]
for i in range(N+1):
index[A[i]].append(i)
for i in range(1,N+1):
if len(index[i])==2:
dup=i
mod=10**9+7
n=index[dup][0]
m=N-index[dup][1]
l=n+m
print(n,m,l)
for k in range(1,N+2):
if k==1:
print(N)
continue
ans=cmb(N+1,k,mod)
if l>=k-1:
for i in range(k):
ans-=cmb(n,i,mod)*cmb(m,k-1-i,mod)
print(ans)
|
s017222608
|
Accepted
| 1,621
| 132,708
| 647
|
def cmb(n, r, p):
if (r<0) or (n<r):
return 0
r=min(r, n-r)
return fact[n]*factinv[r]*factinv[n-r]%p
p=10**9+7
N=10**6
fact=[1,1]
factinv=[1,1]
inv=[0,1]
for i in range(2, N+1):
fact.append((fact[-1]*i)%p)
inv.append((-inv[p%i]*(p//i)%p))
factinv.append((factinv[-1]*inv[-1])%p)
N=int(input())
A=list(map(int,input().split()))
index=[-1]*N
for i,a in enumerate(A):
if index[a-1]!=-1:
dup_first,dup_second=index[a-1],i
break
index[a-1]=i
mod=10**9+7
n=dup_first
m=N-dup_second
l=n+m
for k in range(1,N+2):
if k==1:
print(N)
continue
ans=(cmb(N+1,k,mod)-cmb(l,k-1,mod))%mod
print(ans)
|
s471836506
|
p03433
|
u360038884
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 85
|
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
n = int(input())
a = int(input())
if n % 500 < a:
print('No')
else:
print('Yes')
|
s763935233
|
Accepted
| 17
| 2,940
| 86
|
n = int(input())
a = int(input())
if n % 500 <= a:
print('Yes')
else:
print('No')
|
s894489376
|
p03599
|
u106778233
| 3,000
| 262,144
|
Wrong Answer
| 3,309
| 27,824
| 504
|
Snuke is making sugar water in a beaker. Initially, the beaker is empty. Snuke can perform the following four types of operations any number of times. He may choose not to perform some types of operations. * Operation 1: Pour 100A grams of water into the beaker. * Operation 2: Pour 100B grams of water into the beaker. * Operation 3: Put C grams of sugar into the beaker. * Operation 4: Put D grams of sugar into the beaker. In our experimental environment, E grams of sugar can dissolve into 100 grams of water. Snuke will make sugar water with the highest possible density. The beaker can contain at most F grams of substances (water and sugar combined), and there must not be any undissolved sugar in the beaker. Find the mass of the sugar water Snuke will make, and the mass of sugar dissolved in it. If there is more than one candidate, any of them will be accepted. We remind you that the sugar water that contains a grams of water and b grams of sugar is \frac{100b}{a + b} percent. Also, in this problem, pure water that does not contain any sugar is regarded as 0 percent density sugar water.
|
water = []
a,b,c,d,e,f = map(int, input().split())
for i in range(f//a+1):
for j in range(f//b+1):
if 0<a*i+b*j<=f:
water.append(a*i+b*j)
sugar = []
for i in range(f//c+1):
for j in range(f//d+1):
if 0<c*i+d*j<=f:
sugar.append(c*i+d*j)
ans1 =0
ans2 =0
ans3 =0
for i in water:
for j in sugar:
if i+j<=f and j/i*100<=e:
if j/(i+j) > ans3:
ans1 = i+j
ans2 = j
print(ans1,ans2)
|
s648111610
|
Accepted
| 29
| 9,136
| 347
|
A, B, C, D, E, F = map(int, input().split())
ans1 = A
ans2 = 0
for i in range(A, F // 100 + 1):
if any((i - A * j) % B == 0 for j in range(i // A + 1)):
x = min(E * i, F - i * 100)
e = max([C * j + (x - C * j) // D * D for j in range(x // C + 1)])
if e * ans1 > ans2 * i:
ans1 = i
ans2 = e
print(ans1 * 100 + ans2, ans2)
|
s607466588
|
p04043
|
u897302879
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 78
|
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
print("YES" if sorted(list(map(int, input().split()))) == [3, 3, 5] else "NO")
|
s670375142
|
Accepted
| 17
| 2,940
| 79
|
print("YES" if sorted(list(map(int, input().split()))) == [5, 5, 7] else "NO")
|
s639828226
|
p04011
|
u434208140
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 94
|
There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee.
|
n=int(input())
k=int(input())
x=int(input())
y=int(input())
print(k*x+y*(n-x) if n>k else n*x)
|
s981000940
|
Accepted
| 17
| 2,940
| 94
|
n=int(input())
k=int(input())
x=int(input())
y=int(input())
print(k*x+y*(n-k) if n>k else n*x)
|
s260487663
|
p03455
|
u668924588
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 78
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = map(int, input().split())
answer = 'Even' if a * b % 2 == 0 else 'Odd'
|
s270875907
|
Accepted
| 18
| 2,940
| 94
|
a, b = map(int, input().split())
answer = 'Even' if a * b % 2 == 0 else 'Odd'
print(answer)
|
s408980753
|
p02603
|
u682271925
| 2,000
| 1,048,576
|
Wrong Answer
| 29
| 9,192
| 276
|
To become a millionaire, M-kun has decided to make money by trading in the next N days. Currently, he has 1000 yen and no stocks - only one kind of stock is issued in the country where he lives. He is famous across the country for his ability to foresee the future. He already knows that the price of one stock in the next N days will be as follows: * A_1 yen on the 1-st day, A_2 yen on the 2-nd day, ..., A_N yen on the N-th day. In the i-th day, M-kun can make the following trade **any number of times** (possibly zero), **within the amount of money and stocks that he has at the time**. * Buy stock: Pay A_i yen and receive one stock. * Sell stock: Sell one stock for A_i yen. What is the maximum possible amount of money that M-kun can have in the end by trading optimally?
|
n = int(input())
li = list(map(int,input().split()))
money = 1000
stock = 0
for i in range(n-1):
if stock > 0:
money += stock * li[i]
if li[i] < li[i+1]:
stock = money // li[i]
money -= stock * li[i]
if stock > 0:
money += stock * li[n-1]
print(money)
|
s395054083
|
Accepted
| 30
| 9,180
| 199
|
n = int(input())
li = list(map(int,input().split()))
money = 1000
stock = 0
for i in range(n-1):
if li[i] < li[i+1]:
stock = money // li[i]
money += stock * (li[i+1]-li[i])
print(money)
|
s027283359
|
p03605
|
u617203831
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 40
|
It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?
|
print("Yes" if input() in "9" else "No")
|
s138814003
|
Accepted
| 17
| 2,940
| 40
|
print("Yes" if "9" in input() else "No")
|
s463836014
|
p03862
|
u064408584
| 2,000
| 262,144
|
Wrong Answer
| 136
| 14,132
| 301
|
There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies. Snuke can perform the following operation any number of times: * Choose a box containing at least one candy, and eat one of the candies in the chosen box. His objective is as follows: * Any two neighboring boxes contain at most x candies in total. Find the minimum number of operations required to achieve the objective.
|
n,x=map(int, input().split())
a=list(map(int, input().split()))
count=0
for i in a:
if i>x:count+=i-x
print(count)
a=[min(x,i) for i in a]
print(a)
a2=[a[0]]
for i in range(1,n):
if a2[-1]+a[i]-x>0:
count+=a2[-1]+a[i]-x
a2.append(x-a2[-1])
else:a2.append(a[i])
print(count)
|
s953316893
|
Accepted
| 126
| 14,252
| 279
|
n,x=map(int, input().split())
a=list(map(int, input().split()))
count=0
for i in a:
if i>x:count+=i-x
a=[min(x,i) for i in a]
a2=[a[0]]
for i in range(1,n):
if a2[-1]+a[i]-x>0:
count+=a2[-1]+a[i]-x
a2.append(x-a2[-1])
else:a2.append(a[i])
print(count)
|
s554813227
|
p02390
|
u592365052
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,600
| 123
|
Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively.
|
S = int(input())
a = S / 60 / 60
h = int(a)
b = (a - h) * 60
m = int(b)
s = (b - m) * 60
print("{}:{}:{}".format(h, m, s))
|
s783102504
|
Accepted
| 20
| 5,584
| 95
|
S = int(input())
h = S // 3600
m = S % 3600 // 60
s = S % 60
print("{}:{}:{}".format(h, m, s))
|
s195557114
|
p02747
|
u887199039
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 73
|
A Hitachi string is a concatenation of one or more copies of the string `hi`. For example, `hi` and `hihi` are Hitachi strings, while `ha` and `hii` are not. Given a string S, determine whether S is a Hitachi string.
|
hitachi = input()
if hitachi in 'hi':
print('Yes')
else:
print('No')
|
s141920579
|
Accepted
| 17
| 2,940
| 187
|
S = input()
hitachi = 'hi'
i = 1
while True:
corect = hitachi * i
if S == corect:
print('Yes')
break
i += 1
if i == 20:
print('No')
break
|
s330773079
|
p03593
|
u238510421
| 2,000
| 262,144
|
Wrong Answer
| 1,783
| 21,404
| 699
|
We have an H-by-W matrix. Let a_{ij} be the element at the i-th row from the top and j-th column from the left. In this matrix, each a_{ij} is a lowercase English letter. Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A. Here, he wants to satisfy the following condition: * Every row and column in A' can be read as a palindrome. Determine whether he can create a matrix satisfying the condition.
|
from collections import Counter
import numpy as np
hw = input().split()
h = int(hw[0])
w = int(hw[1])
a_all = ""
for i in range(h):
a_all += input()
count = Counter(a_all)
four_num = w//2 * h//2
two_num = w%2 * h//2 + h%2 * w//2
one_num = w%2 * h*2
nums = list(count.values())
nums_array = np.array(nums)
check=0
for i in range(four_num):
if len(nums_array[nums_array>=4])>=1:
nums_array[np.where(nums_array>=4)[0][0]] -= 4
else:
check = 1
break
for i in range(two_num):
if len(nums_array[nums_array>=2])>=1:
nums_array[np.where(nums_array>=2)[0][0]] -= 2
else:
check = 1
if check == 1:
print("No")
else:
print("Yes")
|
s160533536
|
Accepted
| 1,900
| 21,528
| 860
|
from collections import Counter
import numpy as np
hw = input().split()
h = int(hw[0])
w = int(hw[1])
# print(h,w)
a_all = ""
for i in range(h):
a_all += input()
count = Counter(a_all)
four_num = (w//2) * (h//2)
two_num = (w%2) * (h//2) + (h%2) * (w//2)
one_num = (w%2) * (h%2)
# print("test:",four_num,two_num,one_num)
nums = list(count.values())
nums_array = np.array(nums)
check=0
# print(nums_array)
for i in range(four_num):
if len(nums_array[nums_array>=4])>=1:
nums_array[np.where(nums_array>=4)[0][0]] -= 4
# print(nums_array)
else:
check = 1
break
for i in range(two_num):
if len(nums_array[nums_array>=2])>=1:
nums_array[np.where(nums_array>=2)[0][0]] -= 2
# print(nums_array)
else:
check = 1
break
if check == 1:
print("No")
else:
print("Yes")
|
s409830232
|
p03378
|
u375616706
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 251
|
There are N + 1 squares arranged in a row, numbered 0, 1, ..., N from left to right. Initially, you are in Square X. You can freely travel between adjacent squares. Your goal is to reach Square 0 or Square N. However, for each i = 1, 2, ..., M, there is a toll gate in Square A_i, and traveling to Square A_i incurs a cost of 1. It is guaranteed that there is no toll gate in Square 0, Square X and Square N. Find the minimum cost incurred before reaching the goal.
|
# python template for atcoder1
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline
N, M, X = map(int, input().split())
a = list(map(int, input().split()))
ans = 0
for i in reversed(range(X)):
if i in a:
ans += 1
print(ans)
|
s250641177
|
Accepted
| 18
| 3,060
| 253
|
# python template for atcoder1
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline
N, M, X = map(int, input().split())
a = list(map(int, input().split()))
l = [0]*(N+1)
for i in a:
l[i] += 1
print(min(sum(l[0:X]), sum(l[X+1: N+1])))
|
s778250151
|
p03456
|
u372670441
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 137
|
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.
|
import math
a,b=map(str,input().split())
c=int(str(a)+str(b))
A=math.sqrt(c)
if int(A)==float(A):
print("yes")
else:
print("no")
|
s061237325
|
Accepted
| 18
| 2,940
| 137
|
import math
a,b=map(str,input().split())
c=int(str(a)+str(b))
A=math.sqrt(c)
if int(A)==float(A):
print("Yes")
else:
print("No")
|
s632210175
|
p02608
|
u927740808
| 2,000
| 1,048,576
|
Wrong Answer
| 123
| 9,364
| 298
|
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
n = int(input())
c = [0]*10001
for x in range(1,101):
if(x > n):
break
for y in range(x,101):
for z in range(y,101):
p = int(((x+y)*(x+y) + (y+z)*(y+z) + (z+x)*(z+x))/2)
if(n >= p):
c[p] += 1
for i in range(1,n+1):
print(c[i])
|
s526901211
|
Accepted
| 533
| 9,432
| 297
|
n = int(input())
c = [0]*10001
for x in range(1,101):
if(x > n):
break
for y in range(1,101):
for z in range(1,101):
p = int(((x+y)*(x+y) + (y+z)*(y+z) + (z+x)*(z+x))/2)
if(n >= p):
c[p] += 1
for i in range(1,n+1):
print(c[i])
|
s764666349
|
p03493
|
u213401801
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 77
|
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.
|
s=list(input())
result=0
for i in s:
if i=='0':
result+=1
print(result)
|
s874545892
|
Accepted
| 17
| 2,940
| 83
|
s=list(input())
result=0
for i in s:
if i=='1':
result=result+1
print(result)
|
s345828750
|
p03693
|
u882620594
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 99
|
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
a=list(map(int,input().split()))
if a[1]*10 + a[2] % 4 == 0:
print("YES")
else:
print("NO")
|
s920758331
|
Accepted
| 18
| 2,940
| 101
|
a=list(map(int,input().split()))
if (a[1]*10 + a[2]) % 4 == 0:
print("YES")
else:
print("NO")
|
s874724040
|
p03544
|
u440161695
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 66
|
It is November 18 now in Japan. By the way, 11 and 18 are adjacent Lucas numbers. You are given an integer N. Find the N-th Lucas number. Here, the i-th Lucas number L_i is defined as follows: * L_0=2 * L_1=1 * L_i=L_{i-1}+L_{i-2} (i≥2)
|
n=int(input())
a,b=2,1
for i in range(n-1):
a,b=b,(a+b)
print(a)
|
s702694983
|
Accepted
| 17
| 2,940
| 66
|
n=int(input())
a,b=2,1
for i in range(n-1):
a,b=b,(a+b)
print(b)
|
s151541161
|
p03860
|
u788856752
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 43
|
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
l = list(input())
print("A" + l[0] + "C")
|
s064756466
|
Accepted
| 17
| 2,940
| 42
|
l = list(input())
print("A" + l[8] + "C")
|
s906731252
|
p03472
|
u062147869
| 2,000
| 262,144
|
Wrong Answer
| 349
| 11,316
| 506
|
You are going out for a walk, when you suddenly encounter a monster. Fortunately, you have N katana (swords), Katana 1, Katana 2, …, Katana N, and can perform the following two kinds of attacks in any order: * Wield one of the katana you have. When you wield Katana i (1 ≤ i ≤ N), the monster receives a_i points of damage. The same katana can be wielded any number of times. * Throw one of the katana you have. When you throw Katana i (1 ≤ i ≤ N) at the monster, it receives b_i points of damage, and you lose the katana. That is, you can no longer wield or throw that katana. The monster will vanish when the total damage it has received is H points or more. At least how many attacks do you need in order to vanish it in total?
|
from bisect import bisect_right,bisect_left
import sys
N,H = map(int,input().split())
A = []
B = []
for i in range(N):
a,b = map(int,input().split())
A.append(a)
B.append(b)
B.sort()
ama=max(A)
x = bisect_right(B,ama)
ans = 0
if x ==N:
ans = H//ama +1
print(ans)
sys.exit()
for i in range(x,N):
if H<=0:
print(ans)
sys.exit()
if H>0:
H-=B[i]
ans +=1
if H>0:
if H%ama ==0:
ans += H//ama
else:
ans += H//ama
print(ans)
|
s428486890
|
Accepted
| 353
| 12,084
| 446
|
from bisect import bisect_right,bisect_left
import sys
N,H = map(int,input().split())
A = []
B = []
for i in range(N):
a,b = map(int,input().split())
A.append(a)
B.append(b)
B =sorted(B,reverse=True)
ama=max(A)
ans =0
for i in range(N):
if H<=0:
print(ans)
sys.exit()
elif B[i]>=ama:
H-=B[i]
ans +=1
else:
break
if H>0:
ans += H//ama
if H%ama!=0:
ans +=1
print(ans)
|
s428885561
|
p02645
|
u185806788
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 9,020
| 37
|
When you asked some guy in your class his name, he called himself S, where S is a string of length between 3 and 20 (inclusive) consisting of lowercase English letters. You have decided to choose some three consecutive characters from S and make it his nickname. Print a string that is a valid nickname for him.
|
S=input()
print("S[0]"+"S[1]"+"S[2]")
|
s072200283
|
Accepted
| 23
| 9,080
| 22
|
S=input()
print(S[:3])
|
s058560542
|
p02396
|
u313600138
| 1,000
| 131,072
|
Wrong Answer
| 120
| 5,592
| 63
|
In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem.
|
while True:
x=int(input())
if x==0:
break
print(x)
|
s386702072
|
Accepted
| 160
| 5,608
| 103
|
i=0
while True:
x=int(input())
i=i+1
if x == 0:
break
print('Case',' ',i,':',' ',x,sep='')
|
s838358960
|
p03485
|
u604655161
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 121
|
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
def ABC_82_A():
a,b = map(int, input().split())
print(int((a+b)/2)+1)
if __name__ == '__main__':
ABC_82_A()
|
s506819292
|
Accepted
| 17
| 2,940
| 121
|
def ABC_82_A():
a,b = map(int, input().split())
print(int((a+b+1)/2))
if __name__ == '__main__':
ABC_82_A()
|
s252661962
|
p02420
|
u897625141
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,656
| 468
|
Your task is to shuffle a deck of n cards, each of which is marked by a alphabetical letter. A single shuffle action takes out h cards from the bottom of the deck and moves them to the top of the deck. The deck of cards is represented by a string as follows. abcdeefab The first character and the last character correspond to the card located at the bottom of the deck and the card on the top of the deck respectively. For example, a shuffle with h = 4 to the above deck, moves the first 4 characters "abcd" to the end of the remaining characters "eefab", and generates the following deck: eefababcd You can repeat such shuffle operations. Write a program which reads a deck (a string) and a sequence of h, and prints the final state (a string).
|
import re
array = []
lines = []
count = 0
while True:
n = input()
if n == "-":
break
array.append(n)
for i in range(len(array)):
if re.compile("[a-z]").search(array[i]):
if count > 0:
lines.append(stt)
count += 1
stt = array[i]
else:
stt = stt[int(array[i]):len(stt)]+stt[0:int(array[i])]
if i == len(array)-1:
lines.append(stt)
for i in range(len(lines)):
print(lines[i])
|
s119935485
|
Accepted
| 40
| 7,848
| 530
|
import re
array = []
lines = []
count = 0
while True:
n = input()
if n == "-":
break
array.append(n)
for i in range(len(array)):
if re.compile("[a-z]").search(array[i]):
fuck = i+1
if count > 0:
lines.append(stt)
count += 1
stt = array[i]
else:
if i == fuck:
continue
stt = stt[int(array[i]):len(stt)]+stt[0:int(array[i])]
if i == len(array)-1:
lines.append(stt)
for i in range(len(lines)):
print(lines[i])
|
s136697929
|
p02401
|
u131984977
| 1,000
| 131,072
|
Wrong Answer
| 30
| 6,736
| 306
|
Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part.
|
from sys import stdin
while True:
(a, op, b) = stdin.readline().split(' ')
a = int(a)
b = int(b)
if op == "?":
break;
elif op == "+":
print(a + b)
elif op == "-":
print(a - b)
elif op == "*":
print(a * b)
elif op == "/":
print(a / b)
|
s950642966
|
Accepted
| 40
| 7,636
| 291
|
while True:
a, op, b = input().split()
a = int(a)
b = int(b)
if op == '?':
break
if op == '+':
print(a + b)
if op == '-':
print(a - b)
if op == '*':
print(a * b)
if op == '/':
print(a // b)
|
s805742784
|
p04043
|
u137228327
| 2,000
| 262,144
|
Wrong Answer
| 27
| 8,948
| 233
|
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
lst = list(map(int,input().split()))
#print(lst)
count5 = 0
count7 = 0
for i in range(len(lst)):
if i == 5:
count5+=1
elif i == 7:
count7+=1
if count7==1 and count5 ==2:
print('YES')
else:
print('NO')
|
s755623975
|
Accepted
| 23
| 9,032
| 257
|
# coding: utf-8
# Your code here!
lst = list(map(int,input().split()))
#print(lst)
count5 = 0
count7 = 0
for i in lst:
if i == 5:
count5+=1
elif i == 7:
count7+=1
if count7==1 and count5 ==2:
print('YES')
else:
print('NO')
|
s682472624
|
p02613
|
u644516473
| 2,000
| 1,048,576
|
Wrong Answer
| 147
| 9,476
| 200
|
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
from collections import defaultdict
N = int(input())
d = defaultdict(int)
for _ in range(N):
d[input()] += 1
for k in ['AC', 'WA', 'TLE', 'RE']:
v = d[k]
if v:
print(f'{k} x {v}')
|
s328612577
|
Accepted
| 146
| 9,092
| 186
|
from collections import defaultdict
N = int(input())
d = defaultdict(int)
for _ in range(N):
d[input()] += 1
for k in ['AC', 'WA', 'TLE', 'RE']:
v = d[k]
print(f'{k} x {v}')
|
s968863212
|
p02615
|
u466143662
| 2,000
| 1,048,576
|
Wrong Answer
| 182
| 31,440
| 210
|
Quickly after finishing the tutorial of the online game _ATChat_ , you have decided to visit a particular place with N-1 players who happen to be there. These N players, including you, are numbered 1 through N, and the **friendliness** of Player i is A_i. The N players will arrive at the place one by one in some order. To make sure nobody gets lost, you have set the following rule: players who have already arrived there should form a circle, and a player who has just arrived there should cut into the circle somewhere. When each player, except the first one to arrive, arrives at the place, the player gets **comfort** equal to the smaller of the friendliness of the clockwise adjacent player and that of the counter-clockwise adjacent player. The first player to arrive there gets the comfort of 0. What is the maximum total comfort the N players can get by optimally choosing the order of arrivals and the positions in the circle to cut into?
|
n = int(input())
A = sorted(list(map(int, input().split())), reverse=True)
score_list = [0]
for i in range(1, len(A)):
score_list.append(A[i])
score_list.append(A[i])
print(sum(score_list[:n-1]))
|
s994265737
|
Accepted
| 174
| 31,676
| 214
|
n = int(input())
A = sorted(list(map(int, input().split())), reverse=True)
score_list = [0, A[0]]
for i in range(1, len(A)):
score_list.append(A[i])
score_list.append(A[i])
print(sum(score_list[:n]))
|
s254549239
|
p02262
|
u092047183
| 6,000
| 131,072
|
Wrong Answer
| 30
| 7,836
| 677
|
Shell Sort is a generalization of [Insertion Sort](http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_1_A) to arrange a list of $n$ elements $A$. 1 insertionSort(A, n, g) 2 for i = g to n-1 3 v = A[i] 4 j = i - g 5 while j >= 0 && A[j] > v 6 A[j+g] = A[j] 7 j = j - g 8 cnt++ 9 A[j+g] = v 10 11 shellSort(A, n) 12 cnt = 0 13 m = ? 14 G[] = {?, ?,..., ?} 15 for i = 0 to m-1 16 insertionSort(A, n, G[i]) A function shellSort(A, n) performs a function insertionSort(A, n, g), which considers every $g$-th elements. Beginning with large values of $g$, it repeats the insertion sort with smaller $g$. Your task is to complete the above program by filling ?. Write a program which reads an integer $n$ and a sequence $A$, and prints $m$, $G_i (i = 0, 1, ..., m − 1)$ in the pseudo code and the sequence $A$ in ascending order. The output of your program must meet the following requirements: * $1 \leq m \leq 100$ * $0 \leq G_i \leq n$ * cnt does not exceed $\lceil n^{1.5}\rceil$
|
# coding: utf-8
import math
cnt = 0
m = 0
G = []
def insertionSort(A, n, g):
global cnt
for i in range(g, n):
v = A[i]
j = i - g
while j >= 0 and A[j] > v:
A[j+g] = A[j]
j = j - g
cnt += 1
A[j+g] = v
return A
def shellSort(A, n):
global m, G
m = int(math.log(n, 2))
G = [2**x for x in range(m - 1, -1, -1)]
for i in range(m):
A = insertionSort(A, n, G[i])
return A
n = int(input())
A = []
for i in range(n):
A.append(int(input()))
A = shellSort(A, n)
print(m)
print(" ".join(map(str, G)))
print(cnt)
for i in range(n):
print(A[i])
|
s728849356
|
Accepted
| 21,910
| 118,672
| 566
|
# coding: utf-8
import math
import sys
def insertionSort(a, n, g):
ct = 0
for i in range(g, n):
v = a[i]
j = i - g
while j >= 0 and a[j] > v:
a[j+g] = a[j]
j = j - g
ct += 1
a[j+g] = v
return ct
n = int(input())
a = list(map(int, sys.stdin.readlines()))
cnt = 0
m = 0
h = 0
g = []
while True:
h = 3*h + 1
if h > n:
break
g.insert(0, math.ceil(h))
for i in g:
cnt += insertionSort(a, n, i)
print(len(g))
print(*g, sep=" ")
print(cnt)
print(*a, sep="\n")
|
s230526527
|
p03564
|
u763177133
| 2,000
| 262,144
|
Wrong Answer
| 24
| 9,112
| 135
|
Square1001 has seen an electric bulletin board displaying the integer 1. He can perform the following operations A and B to change this value: * Operation A: The displayed value is doubled. * Operation B: The displayed value increases by K. Square1001 needs to perform these operations N times in total. Find the minimum possible value displayed in the board after N operations.
|
n = int(input())
k = int(input())
num = 1
count = 0
while num < k:
num *= 2
count += 1
num + (k * (n-count))
print(num)
|
s223041292
|
Accepted
| 27
| 9,124
| 133
|
n = int(input())
k = int(input())
num = 1
count = 0
while count < n:
num = min(num * 2, num + k)
count += 1
print(num)
|
s942489677
|
p03730
|
u511457539
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 127
|
We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`.
|
A, B, C = map(int, input().split())
for i in range(1000):
if A*i%B == C:
print("Yes")
exit()
print("No")
|
s701510292
|
Accepted
| 17
| 2,940
| 127
|
A, B, C = map(int, input().split())
for i in range(1000):
if A*i%B == C:
print("YES")
exit()
print("NO")
|
s809883617
|
p03493
|
u404240078
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 54
|
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.
|
trout = input().rstrip()
count_zero = trout.count('0')
|
s189779579
|
Accepted
| 19
| 3,060
| 134
|
trout = input().rstrip().lstrip()
num_list = []
for i in range(len(trout)):
num_list.append(int(trout[i]))
print(num_list.count(1))
|
s681028613
|
p03474
|
u679089074
| 2,000
| 262,144
|
Wrong Answer
| 27
| 9,196
| 308
|
The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.
|
#33
import sys
A,B = map(int,input().split())
S = input()
Ans = "No"
if S[A] != "-":
print(Ans)
sys.exit()
for i in range(A):
if S[i] == "-":
print(Ans)
sys.exit()
for i in range(A+1,A+B+1):
if S[i] == "-":
print(Ans)
sys.exit()
Ans = "yes"
print(Ans)
|
s977879707
|
Accepted
| 26
| 9,196
| 308
|
#33
import sys
A,B = map(int,input().split())
S = input()
Ans = "No"
if S[A] != "-":
print(Ans)
sys.exit()
for i in range(A):
if S[i] == "-":
print(Ans)
sys.exit()
for i in range(A+1,A+B+1):
if S[i] == "-":
print(Ans)
sys.exit()
Ans = "Yes"
print(Ans)
|
s171902939
|
p02743
|
u190905976
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 92
|
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
|
a,b,c = (int(i) for i in input().split())
if a+b < c:
print("Yes")
else:
print("No")
|
s719801994
|
Accepted
| 17
| 3,060
| 140
|
a,b,c = (int(i) for i in input().split())
if c-a-b <=0:
print("No")
elif 4*a*b < (c-a-b)*(c-a-b):
print("Yes")
else:
print("No")
|
s378315001
|
p03555
|
u266768906
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 213
|
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
s1 = input()
s2 = input()
if(s1[0] == s2[2]):
if(s1[1] == s2[1]):
if(s1[2] == s2[0]):
print("Yes")
else:
print("No")
else:
print("No")
else:
print("No")
|
s866753339
|
Accepted
| 18
| 3,060
| 213
|
s1 = input()
s2 = input()
if(s1[0] == s2[2]):
if(s1[1] == s2[1]):
if(s1[2] == s2[0]):
print("YES")
else:
print("NO")
else:
print("NO")
else:
print("NO")
|
s841277879
|
p03957
|
u970308980
| 1,000
| 262,144
|
Wrong Answer
| 19
| 3,188
| 67
|
This contest is `CODEFESTIVAL`, which can be shortened to the string `CF` by deleting some characters. Mr. Takahashi, full of curiosity, wondered if he could obtain `CF` from other strings in the same way. You are given a string s consisting of uppercase English letters. Determine whether the string `CF` can be obtained from the string s by deleting some characters.
|
import re
s = input()
print('Yes' if re.search('CF', s) else 'No')
|
s667717692
|
Accepted
| 19
| 3,188
| 96
|
import re
s = input()
s = re.sub(r'[^CF]', '', s)
print('Yes' if re.search(r'CF', s) else 'No')
|
s879918591
|
p03155
|
u749770850
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 88
|
It has been decided that a programming contest sponsored by company A will be held, so we will post the notice on a bulletin board. The bulletin board is in the form of a grid with N rows and N columns, and the notice will occupy a rectangular region with H rows and W columns. How many ways are there to choose where to put the notice so that it completely covers exactly HW squares?
|
n = int(input())
h = int(input())
w = int(input())
print((n - (w + 1)) * (n - (h + 1)))
|
s104306912
|
Accepted
| 17
| 2,940
| 84
|
n = int(input())
h = int(input())
w = int(input())
print((n - w + 1) * (n - h + 1))
|
s248496634
|
p03997
|
u486773779
| 2,000
| 262,144
|
Wrong Answer
| 29
| 9,060
| 62
|
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a=int(input())
b=int(input())
h=int(input())
print((a+b)*h/2)
|
s224142164
|
Accepted
| 27
| 8,984
| 63
|
a=int(input())
b=int(input())
h=int(input())
print((a+b)*h//2)
|
s076505387
|
p03997
|
u178432859
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 67
|
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a = int(input())
b = int(input())
h = int(input())
print((a+b)*h/2)
|
s204123989
|
Accepted
| 17
| 2,940
| 72
|
a = int(input())
b = int(input())
h = int(input())
print(int((a+b)*h/2))
|
s450086208
|
p00001
|
u105694406
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,524
| 149
|
There is a data which provides heights (in meter) of mountains. The data is only for ten mountains. Write a program which prints heights of the top three mountains in descending order.
|
mountain = []
for _ in range(10):
mountain.append(int(input()))
mountain.sort()
mountain.reverse()
print(mountain[0], mountain[1], mountain[2])
|
s432765575
|
Accepted
| 20
| 7,600
| 161
|
mountain = []
for _ in range(10):
mountain.append(int(input()))
mountain.sort()
mountain.reverse()
print(mountain[0])
print(mountain[1])
print(mountain[2])
|
s138256906
|
p03385
|
u207799478
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 243
|
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
def readints():
return list(map(int, input().split()))
s = str(input())
t = ['a', 'b', 'c']
print(sorted(s))
# print(t)
sum = 0
for i in range(3):
if s[i] == t[i]:
sum += 1
if sum == 3:
print("Yes")
else:
print("No")
|
s640298636
|
Accepted
| 17
| 2,940
| 156
|
def readints():
return list(map(int, input().split()))
s = str(input())
if 'a' in s and 'b' in s and 'c' in s:
print("Yes")
else:
print("No")
|
s866921686
|
p02420
|
u519227872
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,608
| 213
|
Your task is to shuffle a deck of n cards, each of which is marked by a alphabetical letter. A single shuffle action takes out h cards from the bottom of the deck and moves them to the top of the deck. The deck of cards is represented by a string as follows. abcdeefab The first character and the last character correspond to the card located at the bottom of the deck and the card on the top of the deck respectively. For example, a shuffle with h = 4 to the above deck, moves the first 4 characters "abcd" to the end of the remaining characters "eefab", and generates the following deck: eefababcd You can repeat such shuffle operations. Write a program which reads a deck (a string) and a sequence of h, and prints the final state (a string).
|
import sys
cards = ""
for line in sys.stdin:
line = line.strip()
try:
#print(line)
h = int(line)
cards = cards[h:] + cards[:h]
except:
print (cards)
cards = line
|
s938319454
|
Accepted
| 20
| 7,756
| 468
|
import sys
lines = [line.strip() for line in sys.stdin]
cards = lines[0]
m = int(lines[1])
lines_index = 2
while m > 0:
line = lines[lines_index]
if line == '-':
break
h = int(line)
cards = cards[h:] + cards[:h]
m -= 1
if m == 0:
print (cards)
lines_index += 1
cards = lines[lines_index]
if cards == '-':
break
lines_index += 1
m = int(lines[lines_index])
lines_index += 1
|
s989025606
|
p02259
|
u462831976
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,688
| 464
|
Write a program of the Bubble Sort algorithm which sorts a sequence _A_ in ascending order. The algorithm should be based on the following pseudocode: BubbleSort(A) 1 for i = 0 to A.length-1 2 for j = A.length-1 downto i+1 3 if A[j] < A[j-1] 4 swap A[j] and A[j-1] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 4 of the pseudocode.
|
# -*- coding: utf-8 -*-
import sys
import os
import math
N = int(input())
A = list(map(int, input().split()))
def bubble_sort(A):
swap_num = 0
while True:
swapped = False
for i in range(N - 1):
if A[i] > A[i + 1]:
A[i], A[i + 1] = A[i + 1], A[i]
swap_num += 1
swapped = True
if not swapped:
return A, swap_num
A, num = bubble_sort(A)
print(num)
print(*A)
|
s051966584
|
Accepted
| 20
| 7,812
| 464
|
# -*- coding: utf-8 -*-
import sys
import os
import math
N = int(input())
A = list(map(int, input().split()))
def bubble_sort(A):
swap_num = 0
while True:
swapped = False
for i in range(N - 1):
if A[i] > A[i + 1]:
A[i], A[i + 1] = A[i + 1], A[i]
swap_num += 1
swapped = True
if not swapped:
return A, swap_num
A, num = bubble_sort(A)
print(*A)
print(num)
|
s001185682
|
p03408
|
u442877951
| 2,000
| 262,144
|
Wrong Answer
| 21
| 3,316
| 480
|
Takahashi has N blue cards and M red cards. A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i. Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen. Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces `atcoder`, he will not earn money even if there are blue cards with `atcoderr`, `atcode`, `btcoder`, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.) At most how much can he earn on balance? Note that the same string may be written on multiple cards.
|
s = []
t = []
N = int(input())
for _ in range(N):
s.append(str(input()))
M = int(input())
for _ in range(M):
t.append(str(input()))
from collections import Counter
sc = Counter(s)
tc = Counter(t)
s_keys, s_values = zip(*sc.most_common())
t_keys, t_values = zip(*tc.most_common())
ans = 0
for i in range(len(s_values)):
for j in range(len(t_values)):
minus = 0
if s_keys[i] == t_keys[j]:
minus = t_values[j]
ans = max(ans, s_values[i] - minus)
print(ans)
|
s384471691
|
Accepted
| 21
| 3,316
| 546
|
s = []
t = []
N = int(input())
for _ in range(N):
s.append(str(input()))
M = int(input())
for _ in range(M):
t.append(str(input()))
from collections import Counter
sc = Counter(s)
tc = Counter(t)
s_keys, s_values = zip(*sc.most_common())
t_keys, t_values = zip(*tc.most_common())
ans = 0
for i in range(len(s_values)):
for j in range(len(t_values)):
minus = 0
if s_keys[i] == t_keys[j]:
minus = t_values[j]
ans = max(ans, s_values[i] - minus)
if s_keys[i] not in t_keys:
ans = max(ans, s_values[i])
print(ans)
|
s054983326
|
p03607
|
u346308892
| 2,000
| 262,144
|
Wrong Answer
| 231
| 17,160
| 253
|
You are playing the following game with Joisino. * Initially, you have a blank sheet of paper. * Joisino announces a number. If that number is written on the sheet, erase the number from the sheet; if not, write the number on the sheet. This process is repeated N times. * Then, you are asked a question: How many numbers are written on the sheet now? The numbers announced by Joisino are given as A_1, ... ,A_N in the order she announces them. How many numbers will be written on the sheet at the end of the game?
|
N=int(input())
a=[]
for i in range(N):
a.append(int(input()))
chk={}
for sa in a:
if sa in chk:
if chk[sa]==1:
chk[sa]=0
else:
chk[sa]=1
else:
chk[sa]=1
print(chk)
print(sum(chk.values()))
|
s452350054
|
Accepted
| 219
| 16,272
| 244
|
N=int(input())
a=[]
for i in range(N):
a.append(int(input()))
chk={}
for sa in a:
if sa in chk:
if chk[sa]==1:
chk[sa]=0
else:
chk[sa]=1
else:
chk[sa]=1
print(sum(chk.values()))
|
s237403977
|
p00071
|
u150984829
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,656
| 341
|
縦 8、横 8 のマスからなる図1 のような平面があります。その平面上に、いくつかの爆弾が置かれています。図2 にその例を示します（● = 爆弾）。 | □| □| □| □| □| □| □| □ ---|---|---|---|---|---|---|--- □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ | □| □| □| ●| □| □| ●| □ ---|---|---|---|---|---|---|--- □| □| □| □| □| ●| □| □ ●| □| □| □| ●| □| □| ● □| □| ●| □| □| □| ●| □ □| ●| □| □| □| □| □| □ □| □| □| □| ●| □| □| □ ●| □| ●| □| □| □| ●| □ □| ●| □| ●| □| □| ●| □ 図1| 図2 爆弾が爆発すると、その爆弾の上下左右 3 マスに爆風の影響が及び、それらのマスに置かれている爆弾も連鎖的に爆発します。たとえば、図 3 に示す爆弾が爆発すると図 4 の■のマスに爆風の影響が及びます。 | □| □| □| □| □| □| □| □ ---|---|---|---|---|---|---|--- □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| ●| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ | □| □| □| □| □| □| □| □ ---|---|---|---|---|---|---|--- □| □| □| □| □| □| □| □ □| □| □| ■| □| □| □| □ □| □| □| ■| □| □| □| □ □| □| □| ■| □| □| □| □ ■| ■| ■| ●| ■| ■| ■| □ □| □| □| ■| □| □| □| □ □| □| □| ■| □| □| □| □ 図3| 図4 爆弾が置かれている状態と最初に爆発する爆弾の位置を読み込んで、最終的な平面の状態を出力するプログラムを作成してください。
|
def e(x,y):
A[y][x]='0'
for dx,dy in[[-3,0],[-2,0],[-1,0],[1,0],[2,0],[3,0],[0,-3],[0,-2],[0,-1],[0,1],[0,2],[0,3]]:
if 0<=x+dx<8 and 0<=y+dy<8 and A[y+dy][x+dx]=='1':e(x+dx,y+dy)
for i in range(int(input())):
print(f'Data {i+1}')
input()
A=[list(input())for _ in[0]*8]
e(int(input())-1,int(input())-1)
for r in A:print(*r,sep='')
|
s579799178
|
Accepted
| 20
| 5,632
| 273
|
def b(x,y):
A[y][x]='0'
for d in range(-3,4):
0<=x+d<8and'1'==A[y][x+d]and b(x+d,y)
0<=y+d<8and'1'==A[y+d][x]and b(x,y+d)
e=input
for i in range(int(e())):
print(f'Data {i+1}:')
e()
A=[list(e())for _ in[0]*8]
b(int(e())-1,int(e())-1)
for r in A:print(*r,sep='')
|
s006046937
|
p03739
|
u882370611
| 2,000
| 262,144
|
Wrong Answer
| 229
| 14,644
| 735
|
You are given an integer sequence of length N. The i-th term in the sequence is a_i. In one operation, you can select a term and either increment or decrement it by one. At least how many operations are necessary to satisfy the following conditions? * For every i (1≤i≤n), the sum of the terms from the 1-st through i-th term is not zero. * For every i (1≤i≤n-1), the sign of the sum of the terms from the 1-st through i-th term, is different from the sign of the sum of the terms from the 1-st through (i+1)-th term.
|
n = int(input())
a = list(map(int, input().split()))
cnt_1 = 0
base = 0
for i in range(n):
if ((base + a[i] >= 0) and (i % 2 == 1)) or ((base + a[i] <= 0) and (i % 2 == 0)):
cnt_1 += (1 + abs(base + a[i]))
base += (1 + abs(base + a[i])) * (-1) ** i
else:
base += a[i]
cnt_2 = 0
base = 0
for i in range(n):
if ((base + a[i] <= 0) and (i % 2 == 1)) or ((base + a[i] >= 0) and (i % 2 == 0)):
cnt_2 += (1 + abs(base + a[i]))
base -= (1 + abs(base + a[i])) * (-1) ** i
else:
base += a[i]
print(min(cnt_1, cnt_2))
|
s307987339
|
Accepted
| 233
| 14,468
| 757
|
n = int(input())
a = list(map(int, input().split()))
cnt_1 = 0
base = 0
for i in range(n):
if ((base + a[i] >= 0) and (i % 2 == 1)) or ((base + a[i] <= 0) and (i % 2 == 0)):
cnt_1 += (1 + abs(base + a[i]))
base += a[i] + (1 + abs(base + a[i])) * (-1) ** i
else:
base += a[i]
cnt_2 = 0
base = 0
for i in range(n):
if ((base + a[i] <= 0) and (i % 2 == 1)) or ((base + a[i] >= 0) and (i % 2 == 0)):
cnt_2 += (1 + abs(base + a[i]))
base += a[i] + (1 + abs(base + a[i])) * (-1) ** (i + 1)
else:
base += a[i]
print(min(cnt_1, cnt_2))
|
s536656324
|
p03433
|
u587213169
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 90
|
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
N = int(input())
A = int(input())
if (N-A)%500==0:
print("Yes")
else:
print("No")
|
s942924476
|
Accepted
| 18
| 2,940
| 86
|
N = int(input())
A = int(input())
if N%500<=A:
print("Yes")
else:
print("No")
|
s202563180
|
p03385
|
u505420467
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 46
|
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
a=list(input());print("YNeos"[len(set(a))!=3])
|
s513729085
|
Accepted
| 17
| 2,940
| 50
|
a=list(input());print("YNeos"[len(set(a))!=3::2])
|
s379857860
|
p04013
|
u480300350
| 2,000
| 262,144
|
Wrong Answer
| 62
| 5,364
| 3,387
|
Tak has N cards. On the i-th (1 \leq i \leq N) card is written an integer x_i. He is selecting one or more cards from these N cards, so that the average of the integers written on the selected cards is exactly A. In how many ways can he make his selection?
|
#!/usr/bin/env python3
import sys
# import math
# from string import ascii_lowercase, ascii_uppercase, ascii_letters, digits, hexdigits
# import re # re.compile(pattern) => ptn obj; p.search(s), p.match(s), p.finditer(s) => match obj; p.sub(after, s)
# from operator import itemgetter # itemgetter(1), itemgetter('key')
# from collections import defaultdict # subclass of dict. defaultdict(facroty)
# from collections import Counter # subclass of dict. Counter(iter): c.elements(), c.most_common(n), c.subtract(iter)
# from heapq import heapify, heappush, heappop # built-in list. heapify(L) changes list in-place to min-heap in O(n), heappush(heapL, x) and heappop(heapL) in O(lgn).
# from heapq import nlargest, nsmallest # nlargest(n, iter[, key]) returns k-largest-list in O(n+klgn).
# from itertools import count, cycle, repeat # count(start[,step]), cycle(iter), repeat(elm[,n])
# from itertools import groupby # [(k, list(g)) for k, g in groupby('000112')] returns [('0',['0','0','0']), ('1',['1','1']), ('2',['2'])]
# from itertools import starmap # starmap(pow, [[2,5], [3,2]]) returns [32, 9]
# from itertools import product, permutations # product(iter, repeat=n), permutations(iter[,r])
# from itertools import combinations, combinations_with_replacement
# from itertools import accumulate # accumulate(iter[, f])
# from functools import reduce # reduce(f, iter[, init])
# from functools import lru_cache # @lrucache ...arguments of functions should be able to be keys of dict (e.g. list is not allowed)
# from bisect import bisect_left, bisect_right # bisect_left(a, x, lo=0, hi=len(a)) returns i such that all(val<x for val in a[lo:i]) and all(val>-=x for val in a[i:hi]).
# from copy import deepcopy # to copy multi-dimentional matrix without reference
# from fractions import gcd # for Python 3.4 (previous contest @AtCoder)
def main():
mod = 1000000007 # 10^9+7
inf = float('inf') # sys.float_info.max = 1.79...e+308
# inf = 2 ** 64 - 1 # (for fast JIT compile in PyPy) 1.84...e+19
sys.setrecursionlimit(10**6) # 1000 -> 1000000
def input(): return sys.stdin.readline().rstrip()
def ii(): return int(input())
def mi(): return map(int, input().split())
def mi_0(): return map(lambda x: int(x)-1, input().split())
def lmi(): return list(map(int, input().split()))
def lmi_0(): return list(map(lambda x: int(x)-1, input().split()))
def li(): return list(input())
n, a = mi()
L = lmi()
diff = [elm - a for elm in L]
dp = [[0] * 5001 for _ in range(n + 1)]
dp[0][2500] = 1
for i in range(n):
for j in range(5001):
if dp[i][j]:
dp[i+1][j+diff[i]] += dp[i][j]
dp[i+1][j] = dp[i][j]
print(dp[n][2500] - 1)
if __name__ == "__main__":
main()
|
s637023397
|
Accepted
| 71
| 5,620
| 3,394
|
#!/usr/bin/env python3
import sys
# import math
# from string import ascii_lowercase, ascii_uppercase, ascii_letters, digits, hexdigits
# import re # re.compile(pattern) => ptn obj; p.search(s), p.match(s), p.finditer(s) => match obj; p.sub(after, s)
# from operator import itemgetter # itemgetter(1), itemgetter('key')
# from collections import defaultdict # subclass of dict. defaultdict(facroty)
# from collections import Counter # subclass of dict. Counter(iter): c.elements(), c.most_common(n), c.subtract(iter)
# from heapq import heapify, heappush, heappop # built-in list. heapify(L) changes list in-place to min-heap in O(n), heappush(heapL, x) and heappop(heapL) in O(lgn).
# from heapq import nlargest, nsmallest # nlargest(n, iter[, key]) returns k-largest-list in O(n+klgn).
# from itertools import count, cycle, repeat # count(start[,step]), cycle(iter), repeat(elm[,n])
# from itertools import groupby # [(k, list(g)) for k, g in groupby('000112')] returns [('0',['0','0','0']), ('1',['1','1']), ('2',['2'])]
# from itertools import starmap # starmap(pow, [[2,5], [3,2]]) returns [32, 9]
# from itertools import product, permutations # product(iter, repeat=n), permutations(iter[,r])
# from itertools import combinations, combinations_with_replacement
# from itertools import accumulate # accumulate(iter[, f])
# from functools import reduce # reduce(f, iter[, init])
# from functools import lru_cache # @lrucache ...arguments of functions should be able to be keys of dict (e.g. list is not allowed)
# from bisect import bisect_left, bisect_right # bisect_left(a, x, lo=0, hi=len(a)) returns i such that all(val<x for val in a[lo:i]) and all(val>-=x for val in a[i:hi]).
# from copy import deepcopy # to copy multi-dimentional matrix without reference
# from fractions import gcd # for Python 3.4 (previous contest @AtCoder)
def main():
mod = 1000000007 # 10^9+7
inf = float('inf') # sys.float_info.max = 1.79...e+308
# inf = 2 ** 64 - 1 # (for fast JIT compile in PyPy) 1.84...e+19
sys.setrecursionlimit(10**6) # 1000 -> 1000000
def input(): return sys.stdin.readline().rstrip()
def ii(): return int(input())
def mi(): return map(int, input().split())
def mi_0(): return map(lambda x: int(x)-1, input().split())
def lmi(): return list(map(int, input().split()))
def lmi_0(): return list(map(lambda x: int(x)-1, input().split()))
def li(): return list(input())
n, a = mi()
L = lmi()
diff = [elm - a for elm in L]
dp = [[0] * 5001 for _ in range(n + 1)]
dp[0][2500] = 1
for i in range(n):
for j in range(5001):
if dp[i][j]:
dp[i+1][j + diff[i]] += dp[i][j]
dp[i+1][j] += dp[i][j]
print(dp[n][2500] - 1)
if __name__ == "__main__":
main()
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.