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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s452894294
|
p03386
|
u385309449
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 172
|
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 = [i for i in range(a,a+k+1) if i <= b]
bb = [i for i in range(b-k+1,b+1) if i >= a]
c = list(sorted(set(aa+bb)))
for i in c:
print(i)
|
s930121978
|
Accepted
| 17
| 3,060
| 170
|
a,b,k = map(int,input().split())
aa = [i for i in range(a,a+k) if i <= b]
bb = [i for i in range(b-k+1,b+1) if i >= a]
c = list(sorted(set(aa+bb)))
for i in c:
print(i)
|
s831426262
|
p02407
|
u661284763
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,340
| 73
|
Write a program which reads a sequence and prints it in the reverse order.
|
input()
a = input().split()
a.reverse()
for b in a:
print(b, end='_')
|
s984230832
|
Accepted
| 50
| 7,452
| 58
|
input()
a = input().split()
a.reverse()
print(" ".join(a))
|
s914920600
|
p03779
|
u064963667
| 2,000
| 262,144
|
Wrong Answer
| 114
| 3,960
| 101
|
There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0. During the period between time i-1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right. That is, if his coordinate at time i-1 is x, he can be at coordinate x-i, x or x+i at time i. The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible. Find the earliest possible time to reach coordinate X.
|
x = int(input())
num = 1
while num**2+num < 2*x:
print(num,num**2-num)
num += 1
print(num)
|
s332212356
|
Accepted
| 39
| 2,940
| 88
|
import math
x = int(input())
num = 1
while num**2+num < 2*x:
num += 1
print(num)
|
s515527188
|
p03089
|
u720829795
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 458
|
Snuke has an empty sequence a. He will perform N operations on this sequence. In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1). You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
|
N = int(input())
tmp = input().split()
b = []
for i in range(N):
b.append(tmp[i])
for index, item in enumerate(b):
if int(item) > index+1:
print('-1')
exit()
idx_top = 0
idx_bottom = len(b)-1
print(b[idx_top])
idx_top += 1
for i in range(len(b)-1):
if b[idx_top] >= b[idx_bottom]:
print(b[idx_top])
if idx_top < idx_bottom:
idx_top += 1
else:
print(b[idx_bottom])
if idx_bottom > idx_top:
idx_bottom -= 1
|
s867756645
|
Accepted
| 17
| 3,064
| 292
|
N = int(input())
tmp = input().split()
b = []
ans = []
for i in range(N):
b.append(tmp[i])
for index, item in enumerate(b):
if int(item) > index+1:
print('-1')
exit()
dist = index + 1 - int(item)
ans.insert(index-dist, int(item))
for i in range(len(ans)):
print(ans[i])
|
s519204929
|
p02742
|
u771007149
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 260
|
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:
|
h,w = map(int,input().split())
if h%2 == 0 or w%2 == 0:
if h%2 == 0:
result = (h/2) * w
print(result)
else:
result = (w/2) * h
print(result)
else:
result = h//2 * (w//2) + (h-h//2) * (w -w//2)
print(result)
|
s842520054
|
Accepted
| 18
| 3,064
| 299
|
h,w = map(int,input().split())
if h == 1 or w == 1:
print(1)
elif h%2 == 0 or w%2 == 0:
if h%2 == 0:
result = (h//2) * w
print(result)
else:
result = (w//2) * h
print(result)
else:
result = (h//2) * (w//2) + (h - h//2) * (w - w//2)
print(result)
|
s659181129
|
p03816
|
u311442150
| 2,000
| 262,144
|
Wrong Answer
| 38
| 16,484
| 88
|
Snuke has decided to play a game using cards. He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written. He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept. Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.
|
n=int(input())
a=list(input().split())
ans=len(set(a))
if ans%2==0:
ans-1
print(ans)
|
s111957400
|
Accepted
| 37
| 16,312
| 93
|
n=int(input())
a=list(input().split())
ans=len(set(a))
if ans%2 == 0:
ans -= 1
print(ans)
|
s680402166
|
p03698
|
u374082254
| 2,000
| 262,144
|
Wrong Answer
| 16
| 2,940
| 128
|
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
S = input()
a = []
result = "Yes"
for s in S:
if s in a:
result = "No"
break
a.append(s)
print(result)
|
s879811663
|
Accepted
| 18
| 2,940
| 128
|
S = input()
a = []
result = "yes"
for s in S:
if s in a:
result = "no"
break
a.append(s)
print(result)
|
s210559509
|
p03957
|
u579299997
| 1,000
| 262,144
|
Wrong Answer
| 22
| 3,064
| 223
|
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.
|
s = input()
print(s)
state = 0
for c in s:
if c == 'C':
if state == 0:
state = 1
if c == 'F':
if state == 1:
state = 2
if state == 2:
print('Yes')
else:
print('No')
|
s674766059
|
Accepted
| 22
| 3,064
| 214
|
s = input()
state = 0
for c in s:
if c == 'C':
if state == 0:
state = 1
if c == 'F':
if state == 1:
state = 2
if state == 2:
print('Yes')
else:
print('No')
|
s814897077
|
p03067
|
u733774002
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 107
|
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 <= b <= c or c <= b <= a):
print("Yes")
else:
print("No")
|
s619152046
|
Accepted
| 17
| 2,940
| 107
|
a, b, c = map(int, input().split())
if(a <= c <= b or b <= c <= a):
print("Yes")
else:
print("No")
|
s964602332
|
p03369
|
u945065638
| 2,000
| 262,144
|
Wrong Answer
| 30
| 8,904
| 55
|
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().split()
s = s.count('o')
print(700 + 100*s)
|
s523004078
|
Accepted
| 25
| 8,864
| 47
|
s = input()
s = s.count('o')
print(700 + 100*s)
|
s681587947
|
p02747
|
u399481362
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 12
|
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.
|
print(input)
|
s012358443
|
Accepted
| 17
| 2,940
| 209
|
s = input()
isEmpty = False
if 'hi' not in s:
print("No")
isEmpty = True
s = s.split('hi')
s = set(s)
if not isEmpty:
if '' in s and len(s) == 1:
print("Yes")
else:
print("No")
|
s518818221
|
p03860
|
u355661029
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 47
|
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.
|
s = input().split(' ')
print('A' + s[0] + 'C')
|
s677653395
|
Accepted
| 21
| 2,940
| 50
|
s = input().split(' ')
print('A' + s[1][0] + 'C')
|
s638716085
|
p03251
|
u063052907
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 246
|
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.
|
# coding: utf-8
import sys
input = sys.stdin.readline
_, _, X, Y = map(int, input().split())
lst_x = list(map(int, input().split()))
lst_y = list(map(int, input().split()))
if max(lst_x) < min(lst_y):
print("No war")
else:
print("War")
|
s397304783
|
Accepted
| 18
| 3,060
| 206
|
# coding: utf-8
N, M, X, Y = map(int, input().split())
x = max(list(map(int, input().split())) + [X])
y = min(list(map(int, input().split())) + [Y])
ans = "War"
if x < y:
ans = "No War"
print(ans)
|
s861016067
|
p02422
|
u328199937
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,604
| 629
|
Write a program which performs a sequence of commands to a given string $str$. The command is one of: * print a b: print from the a-th character to the b-th character of $str$ * reverse a b: reverse from the a-th character to the b-th character of $str$ * replace a b p: replace from the a-th character to the b-th character of $str$ with p Note that the indices of $str$ start with 0.
|
def Reverse(str, A, B):
a = int(A)
b = int(B)
str = str[0:a] + str[b::-1] + str[b + 1:]
print(str)
return str
def Replace(str, a, b, c):
str2 = ''
Str = list(str)
for i in range(int(b) - int(a) + 1):
Str[i + int(a)] = c[i]
for i in range(len(Str)):
str2 += Str[i]
return str2
str = input()
n = int(input())
for i in range(n):
a = input()
A = a.split()
if A[0] == 'print':
print(str[int(A[1]):int(A[2]) + 1])
elif A[0] == 'reverse':
str = Reverse(str, A[1], A[2])
elif A[0] == 'replace':
str = Replace(str, A[1], A[2], A[3])
|
s187858381
|
Accepted
| 20
| 5,624
| 696
|
def Reverse(str, A, B):
a = int(A)
b = int(B)
if a == 0:
str = str[0:a] + str[b::-1] + str[b + 1:]
else:
str = str[0:a] + str[b:a - 1:-1] + str[b + 1:]
return str
def Replace(str, a, b, c):
str2 = ''
Str = list(str)
for i in range(int(b) - int(a) + 1):
Str[i + int(a)] = c[i]
for i in range(len(Str)):
str2 += Str[i]
return str2
str = input()
n = int(input())
for i in range(n):
a = input()
A = a.split()
if A[0] == 'print':
print(str[int(A[1]):int(A[2]) + 1])
elif A[0] == 'reverse':
str = Reverse(str, A[1], A[2])
elif A[0] == 'replace':
str = Replace(str, A[1], A[2], A[3])
|
s576082538
|
p03854
|
u581603131
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 3,188
| 228
|
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
#9
S = input()
while len(S)!=0:
if S[0:5] == 'dream' or S[0:5] == 'erase':
S = S[5:]
elif S[0:6] == 'eraser':
S = S[6:]
elif S[0:7] == 'dreamer':
S = S[7:]
print('Yes' if len(S)==0 else 'No')
|
s245201523
|
Accepted
| 69
| 3,188
| 273
|
#9
S = input()[::-1]
for i in range(0, len(S)//5+1):
if S[0:5] == 'maerd' or S[0:5] == 'esare':
S = S[5:]
elif S[0:6] == 'resare':
S = S[6:]
elif S[0:7] == 'remaerd':
S = S[7:]
else:
break
print('YES' if len(S)==0 else 'NO')
|
s683395017
|
p03555
|
u073139376
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 43
|
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.
|
print(['No','Yes'][input()==input()[::-1]])
|
s023120040
|
Accepted
| 17
| 2,940
| 43
|
print(['NO','YES'][input()==input()[::-1]])
|
s537742420
|
p03549
|
u078349616
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 77
|
Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`. When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases. Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds. Now, he goes through the following process: * Submit the code. * Wait until the code finishes execution on all the cases. * If the code fails to correctly solve some of the M cases, submit it again. * Repeat until the code correctly solve all the cases in one submission. Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).
|
N, M = map(int, input().split())
ans = pow(2, M) * (1900 * M + 100 * (N - M))
|
s156446288
|
Accepted
| 17
| 2,940
| 79
|
N, M = map(int, input().split())
print(pow(2, M) * (1900 * M + 100 * (N - M)))
|
s511563962
|
p03730
|
u371467115
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 87
|
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())
if abs(b-c)%a==0:
print("YES")
else:
print("NO")
|
s375427207
|
Accepted
| 17
| 2,940
| 139
|
a,b,c=map(int,input().split())
for i in range(1,b+1):
if (a*i)%b==c:
print("YES")
break
else:
print("NO")
|
s739054504
|
p03251
|
u958612488
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,060
| 183
|
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.
|
n,m,x,y=list(map(int,input().split(" ")))
X=list(map(int,input().split(" ")))
Y=list(map(int,input().split(" ")))
if x<y and max(X)<min(Y):
print("War")
else:
print("No War")
|
s537985414
|
Accepted
| 18
| 3,060
| 209
|
n,m,x,y=list(map(int,input().split(" ")))
X=list(map(int,input().split(" ")))
Y=list(map(int,input().split(" ")))
if x<y and max(X)<min(Y) and max(X)<y and x<min(Y):
print("No War")
else:
print("War")
|
s145263383
|
p03494
|
u661983922
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 54
|
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()
lst = map(int,input().split())
print(lst)
|
s572916027
|
Accepted
| 19
| 2,940
| 201
|
N = input()
lst = list(map(int,input().split()))
counter = 0
while all([x%2 == 0 for x in lst]):
counter += 1
for k in range(int(N)):
a = lst[k] / 2
lst[k] = a
print(counter)
|
s152537072
|
p03386
|
u706695185
| 2,000
| 262,144
|
Wrong Answer
| 21
| 3,316
| 782
|
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.
|
from collections import deque
a, b, c = map(int, input().split())
def bfs(a, b, c):
cnt = 0
queue = deque([[a, b, c, cnt]])
while queue:
a, b, c, cnt = queue.popleft()
if a == b and b == c:
return cnt
if a >= 52 or b >= 52 or c >= 52:
return cnt
if b > a and b > c:
queue.append([a+1, b, c+1, cnt+1])
elif a > b and a > b:
queue.append([a, b+1, c+1, cnt+1])
elif c > a and c > b:
queue.append([a+1, b+1, c, cnt+1])
elif a < b or a < c:
queue.append([a+2, b, c, cnt+1])
elif b < a or b < c:
queue.append([a, b+2, c, cnt+1])
elif c < a or c < b:
queue.append([a, b, c+2, cnt+1])
print(bfs(a, b, c))
|
s911026121
|
Accepted
| 17
| 3,060
| 235
|
A, B, K = map(int, input().split())
if B - A < K*2:
for i in range(A, B+1):
print(i)
else:
for a in range(A, A+K):
if a > B:
break
print(a)
for b in range(B-K+1, B+1):
print(b)
|
s167464412
|
p03759
|
u583507988
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 96
|
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful.
|
a, b, c = map(int, input().split())
if abs(b-a) == abs(c-b):
print("Yes")
else:
print("No")
|
s394997340
|
Accepted
| 24
| 9,128
| 86
|
a, b, c = map(int, input().split())
if b-a == c-b:
print('YES')
else:
print('NO')
|
s153266278
|
p03337
|
u989326345
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,064
| 121
|
You are given two integers A and B. Find the largest value among A+B, A-B and A \times B.
|
A,B=map(int,input().split())
S=A+B
T=A-B
U=A*B
if S>T & S>U:
print(S)
elif T>S & T>U:
print(T)
else:
print(U)
|
s264700457
|
Accepted
| 18
| 3,060
| 133
|
A,B=map(int,input().split())
S=A+B
T=A-B
U=A*B
if (S>=T) & (S>=U):
print(S)
elif (T>=S) & (T>=U):
print(T)
else:
print(U)
|
s134361279
|
p04029
|
u993161647
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 88
|
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
N = int(input('Input N > '))
list_N = [ x for x in range(1, N+1) ]
print( sum(list_N) )
|
s026314676
|
Accepted
| 17
| 3,064
| 75
|
N = int(input())
list_N = [ x for x in range(1, N+1) ]
print( sum(list_N) )
|
s894212047
|
p02262
|
u316584871
| 6,000
| 131,072
|
Wrong Answer
| 20
| 5,608
| 908
|
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 $\lceil n^{1.5}\rceil$
|
def insertionSort(A, n, g):
cnti = 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 -= g
cnti += 1
A[j+g] = v
return cnti
def shellSort(A,n):
cnt = 0
G = []
for k in range(int((n+2)/3)):
h = 3*k + 1
if (h < n and len(G) <= 100):
G.append(h)
elif(len(G) > 100):
break
m = len(G)
for i in range(m):
cnt += insertionSort(A, n, G[m-1-i])
print(m)
for i in range(m):
if (i == m-1):
print('{}'.format(G[i]), end = '')
else:
print('{}'.format(G[i]), end = ' ')
return cnt
n = int(input())
nlist = []
for i in range(n):
x = int(input())
nlist.append(x)
c = shellSort(nlist,n)
print()
print(c)
for i in nlist:
print(i)
|
s270458733
|
Accepted
| 18,410
| 45,532
| 1,034
|
def insertionSort(A, n, g):
cnti = 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 -= g
cnti += 1
A[j+g] = v
return cnti
def shellSort(A,n):
cnt = 0
G = [1]
h = 1
for k in range(int(n**0.33333333)+1):
h = 3*h + 1
if (h <= n and len(G) < 100):
G.append(h)
elif(len(G) == 100):
break
G.reverse()
m = len(G)
for i in range(m):
cnt += insertionSort(A, n, G[i])
print(m)
for i in range(m):
if (i == m-1):
print('{}'.format(G[i]), end = '')
else:
print('{}'.format(G[i]), end = ' ')
return cnt
n = int(input())
nlist = []
for i in range(n):
x = int(input())
nlist.append(x)
c = shellSort(nlist,n)
print()
print(c)
for i in nlist:
print(i)
|
s045656727
|
p03139
|
u325264482
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 104
|
We conducted a survey on newspaper subscriptions. More specifically, we asked each of the N respondents the following two questions: * Question 1: Are you subscribing to Newspaper X? * Question 2: Are you subscribing to Newspaper Y? As the result, A respondents answered "yes" to Question 1, and B respondents answered "yes" to Question 2. What are the maximum possible number and the minimum possible number of respondents subscribing to both newspapers X and Y? Write a program to answer this question.
|
N, A, B = list(map(int, input().split()))
ans1 = min(A, B)
ans2 = max(N - A - B, 0)
print(ans1, ans2)
|
s274555189
|
Accepted
| 17
| 2,940
| 80
|
N, A, B = list(map(int, input().split()))
print(min(A, B), max(- N + A + B, 0))
|
s062140962
|
p03435
|
u652081898
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 466
|
We have a 3 \times 3 grid. A number c_{i, j} is written in the square (i, j), where (i, j) denotes the square at the i-th row from the top and the j-th column from the left. According to Takahashi, there are six integers a_1, a_2, a_3, b_1, b_2, b_3 whose values are fixed, and the number written in the square (i, j) is equal to a_i + b_j. Determine if he is correct.
|
x = []
for i in range(3):
x.append(list(map(int, input().split())))
cond1 = bool(x[0][1]-x[0][0] == x[0][2]-x[0][1])
cond2 = bool(x[1][1]-x[1][0] == x[1][2]-x[1][1])
cond3 = bool(x[2][1]-x[2][0] == x[2][2]-x[2][1])
cond4 = bool(x[1][0]-x[0][0] == x[2][0]-x[1][0])
cond5 = bool(x[1][1]-x[0][1] == x[2][1]-x[1][1])
cond6 = bool(x[1][2]-x[0][2] == x[2][2]-x[1][2])
if cond1 and cond2 and cond3 and cond4 and cond5 and cond6:
print("Yes")
else:
print("No")
|
s570269783
|
Accepted
| 18
| 3,064
| 525
|
x = []
for i in range(3):
x.append(list(map(int, input().split())))
A = x[0][1]-x[0][0]
B = x[1][1]-x[1][0]
C = x[2][1]-x[2][0]
D = x[0][2]-x[0][1]
E = x[1][2]-x[1][1]
F = x[2][2]-x[2][1]
G = x[1][0]-x[0][0]
H = x[1][1]-x[0][1]
I = x[1][2]-x[0][2]
J = x[2][0]-x[1][0]
K = x[2][1]-x[1][1]
L = x[2][2]-x[1][2]
cond1 = bool(A == B and B == C)
cond2 = bool(D == E and E == F)
cond3 = bool(G == H and H == I)
cond4 = bool(J == K and K == L)
if cond1 and cond2 and cond3 and cond4:
print("Yes")
else:
print("No")
|
s958156222
|
p03448
|
u561992253
| 2,000
| 262,144
|
Wrong Answer
| 49
| 3,060
| 204
|
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.
|
a = int(input())
b = int(input())
c = int(input())
v = int(input())
ans = 0
for i in range(a):
for j in range(b):
for k in range(c):
if 500*a + 100*b + 50*c == v:
ans += 1
print(ans)
|
s204444626
|
Accepted
| 52
| 3,060
| 210
|
a = int(input())
b = int(input())
c = int(input())
v = int(input())
ans = 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 == v:
ans += 1
print(ans)
|
s120290229
|
p03351
|
u933341648
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 115
|
Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.
|
a, b, c, d = map(int, input().split())
print('yes' if abs(a-c) <= d or (abs(a-b) <= d and abs(b-c) <= d) else 'No')
|
s551956347
|
Accepted
| 17
| 2,940
| 115
|
a, b, c, d = map(int, input().split())
print('Yes' if abs(a-c) <= d or (abs(a-b) <= d and abs(b-c) <= d) else 'No')
|
s537550796
|
p02850
|
u151625340
| 2,000
| 1,048,576
|
Wrong Answer
| 2,106
| 37,044
| 481
|
Given is a tree G with N vertices. The vertices are numbered 1 through N, and the i-th edge connects Vertex a_i and Vertex b_i. Consider painting the edges in G with some number of colors. We want to paint them so that, for each vertex, the colors of the edges incident to that vertex are all different. Among the colorings satisfying the condition above, construct one that uses the minimum number of colors.
|
N = int(input())
G = [[]for i in range(N)]
for i in range(N-1):
a,b = map(int,input().split())
G[a-1].append(b-1)
G[b-1].append(a-1)
import copy
ans = [0]*N
def dfs(i,visited_i,color):
ans[i] = color
visited = copy.deepcopy(visited_i)
visited[i] += 1
c = 1
for j in G[i]:
if visited[j]:
continue
if c==color:
c += 1
dfs(j,visited,c)
c += 1
dfs(0,[0]*N,1)
for i in range(N):
print(ans[i])
|
s029220727
|
Accepted
| 443
| 80,628
| 696
|
import sys, copy, math, heapq, bisect
from itertools import accumulate
from collections import deque, defaultdict, Counter
input = sys.stdin.readline
sys.setrecursionlimit(500000)
N = int(input())
G = [dict() for i in range(N)]
for i in range(N-1):
a,b = map(int,input().split())
G[a-1][b-1] = i
G[b-1][a-1] = i
ans = [0]*(N-1)
visited = [0]*N
def dfs(i,color):
visited[i] += 1
c = 1
for j in G[i]:
if visited[j]:
continue
if c==color:
c += 1
ans[G[i][j]] = c
dfs(j,c)
c += 1
dfs(0,0)
print(max(ans))
for i in range(N-1):
print(ans[i])
|
s564052572
|
p03729
|
u973108807
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 94
|
You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`.
|
a,b,c = input().split()
if a[-1] == b[0] and b[-1] == c[0]:
print('Yes')
else:
print('No')
|
s158098243
|
Accepted
| 17
| 2,940
| 94
|
a,b,c = input().split()
if a[-1] == b[0] and b[-1] == c[0]:
print('YES')
else:
print('NO')
|
s299357422
|
p02396
|
u138628845
| 1,000
| 131,072
|
Wrong Answer
| 90
| 5,908
| 215
|
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.
|
v = 0
i = 0
x = []
while v == 0:
a = input()
if 0 == int(a):
v = 1
else:
x.append(int(a))
i = int(i) + 1
for j in range(int(i)):
print('case %d: %d' % (int(j)+1,x[int(j)] ))
|
s247774483
|
Accepted
| 90
| 5,904
| 215
|
v = 0
i = 0
a = []
while v == 0:
x = input()
if 0 == int(x):
v = 1
else:
a.append(int(x))
i = int(i) + 1
for j in range(int(i)):
print('Case %d: %d' % (int(j)+1,a[int(j)] ))
|
s990407849
|
p03413
|
u596276291
| 2,000
| 262,144
|
Wrong Answer
| 28
| 4,204
| 2,220
|
You have an integer sequence of length N: a_1, a_2, ..., a_N. You repeatedly perform the following operation until the length of the sequence becomes 1: * First, choose an element of the sequence. * If that element is at either end of the sequence, delete the element. * If that element is not at either end of the sequence, replace the element with the sum of the two elements that are adjacent to it. Then, delete those two elements. You would like to maximize the final element that remains in the sequence. Find the maximum possible value of the final element, and the way to achieve it.
|
from collections import defaultdict, Counter
from itertools import product, groupby, count, permutations, combinations
from math import pi, sqrt
from collections import deque
from bisect import bisect, bisect_left, bisect_right
from string import ascii_lowercase
from functools import lru_cache
import sys
sys.setrecursionlimit(10000)
INF = float("inf")
YES, Yes, yes, NO, No, no = "YES", "Yes", "yes", "NO", "No", "no"
dy4, dx4 = [0, 1, 0, -1], [1, 0, -1, 0]
dy8, dx8 = [0, -1, 0, 1, 1, -1, -1, 1], [1, 0, -1, 0, 1, 1, -1, -1]
def inside(y, x, H, W):
return 0 <= y < H and 0 <= x < W
def ceil(a, b):
return (a + b - 1) // b
def check(A, operations):
for o in operations:
if o == 1:
A = A[1:]
elif o == len(A):
A = A[:len(A) - 1]
else:
o -= 1
A = A[:o - 1] + [A[o - 1] + A[o + 1]] + A[o + 2:]
print(A)
assert(len(A) == 1)
return A[0]
def main():
N = int(input())
A = list(map(int, input().split()))
ans1 = 0
del1 = []
ok1 = False
for i in range(0, N, 2):
if A[i] >= 0:
ans1 += A[i]
ok1 = True
else:
del1.append(i + 1)
ans2 = 0
del2 = [1]
ok2 = False
for i in range(1, N, 2):
if A[i] >= 0:
ans2 += A[i]
ok2 = True
else:
del2.append(i + 1)
del1 = del1[::-1]
del2 = del2[::-1]
ans, dele = None, None
if ok1 and ok2:
if ans1 > ans2:
ans, dele = ans1, del1
else:
ans, dele = ans2, del2
elif ok1:
ans, dele = ans1, del1
elif ok2:
ans, dele = ans2, del2
if not ok1 and not ok2:
ans = max(A)
idx = A.index(ans)
print(ans)
a = []
for i in range(N, 0, -1):
if i != idx + 1:
a.append(i)
print(len(a))
print(*a, sep="\n")
else:
print(ans)
e = int(1 in dele) + int(N in dele)
for i in range((N - (len(dele) - e) * 2 - e) // 2):
dele.append(2)
print(len(dele))
print(*dele, sep="\n")
if __name__ == '__main__':
main()
|
s510372810
|
Accepted
| 28
| 4,204
| 2,362
|
from collections import defaultdict, Counter
from itertools import product, groupby, count, permutations, combinations
from math import pi, sqrt
from collections import deque
from bisect import bisect, bisect_left, bisect_right
from string import ascii_lowercase
from functools import lru_cache
import sys
sys.setrecursionlimit(10000)
INF = float("inf")
YES, Yes, yes, NO, No, no = "YES", "Yes", "yes", "NO", "No", "no"
dy4, dx4 = [0, 1, 0, -1], [1, 0, -1, 0]
dy8, dx8 = [0, -1, 0, 1, 1, -1, -1, 1], [1, 0, -1, 0, 1, 1, -1, -1]
def inside(y, x, H, W):
return 0 <= y < H and 0 <= x < W
def ceil(a, b):
return (a + b - 1) // b
def check(A, operations):
c = A[:]
for o in operations:
if o == 1:
A = A[1:]
elif o == len(A):
A = A[:len(A) - 1]
else:
o -= 1
A = A[:o - 1] + [A[o - 1] + A[o + 1]] + A[o + 2:]
if len(A) != 1:
print(c)
print(operations)
assert(len(A) == 1)
return A[0]
def solve(N, A):
if max(A) < 0:
ans = max(A)
idx = A.index(ans)
operations = []
for i in range(N, idx + 1, -1):
operations.append(i)
for i in range(1, idx + 1):
operations.append(1)
return ans, operations
ans = 0
middle = []
start, end = None, None
for i in range(0, N, 2):
if A[i] >= 0:
ans += A[i]
if start is None:
start = i
end = i
else:
if start is not None:
middle.append(i)
operations = []
num = 0
for i in range(end + 1, N):
operations.append(N - num)
num += 1
for i in middle[::-1]:
if start < i < end:
operations.append(i + 1)
num += 2
for i in range(start):
operations.append(1)
num += 1
for i in range((N - num) // 2):
operations.append(2)
return ans, operations
def main():
N = int(input())
A = list(map(int, input().split()))
ans1, ope1 = solve(N, A[:])
ans2, ope2 = solve(N - 1, A[1:])
ope2 = [1] + ope2
if ans1 > ans2:
print(ans1)
print(len(ope1))
print(*ope1, sep="\n")
else:
print(ans2)
print(len(ope2))
print(*ope2, sep="\n")
if __name__ == '__main__':
main()
|
s567938705
|
p02394
|
u556014061
| 1,000
| 131,072
|
Wrong Answer
| 40
| 7,532
| 120
|
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
|
W, H, r, x, y = map(int, input().split())
if r <= x <= W - r and r <= y <= H - r:
print("Yes")
else:
print("No")
|
s034879945
|
Accepted
| 20
| 7,676
| 112
|
w,h,x,y,r = map(int, input().split())
if r <= x <= w-r and r <= y <= h-r:
print("Yes")
else:
print("No")
|
s959689406
|
p02690
|
u129898499
| 2,000
| 1,048,576
|
Wrong Answer
| 22
| 9,444
| 250
|
Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X.
|
X = int(input())
d = int(X**(1/5))+2
list = [0]*d
for i in range(d):
list[i] = i**5
for i in range(d):
for j in range(i,d):
if list[j]-list[i] == X:
print(i,j)
break
elif list[i]+list[j] == X:
print(i,-j)
break
|
s258053525
|
Accepted
| 157
| 9,200
| 235
|
X = int(input())
d= 10**3
list = [0]*d
for i in range(d):
list[i] = i**5
for i in range(d):
for j in range(i,d):
if list[j]-list[i] == X:
print(j,i)
break
elif list[i]+list[j] == X:
print(j,-i)
break
|
s380009759
|
p02831
|
u772816188
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,060
| 320
|
Takahashi is organizing a party. At the party, each guest will receive one or more snack pieces. Takahashi predicts that the number of guests at this party will be A or B. Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted. We assume that a piece cannot be divided and distributed to multiple guests.
|
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10000)
A, B = map(int, input().split())
tmp_a, tmp_b = A, B
print(tmp_a, tmp_b)
while tmp_a % tmp_b != 0:
tmp_a, tmp_b = tmp_b, tmp_a % tmp_b
# temp = B
# B = A % B
# A = temp
print(int(A * B / tmp_b))
|
s929837312
|
Accepted
| 17
| 3,060
| 254
|
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10000)
A, B = map(int, input().split())
tmp_a, tmp_b = A, B
while tmp_a % tmp_b != 0:
tmp_a, tmp_b = tmp_b, tmp_a % tmp_b
print(int(A * B / tmp_b))
|
s035565203
|
p03599
|
u127499732
| 3,000
| 262,144
|
Wrong Answer
| 3,159
| 3,064
| 783
|
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.
|
def main():
a, b, c, d, e, f = map(int, input().split())
a, b = min(a, c), max(a, b)
c, d = min(c, d), max(c, d)
ans = (0, 0)
p = 0
for w in range(1, f + 1):
for s in range(1, f - w + 1):
x = fnc(100 * a, 100 * b, w)
y = fnc(c, d, s)
z = s * 100 <= e * w
if x and y and z:
q = 100 * s / (s + w)
if p <= q:
p = q
ans = (s+w, s)
print(ans)
def fnc(x, y, z):
if z % x == 0:
return True
if z % y == 0:
return True
n = z // x
for i in range(1, n + 1):
rem = z - x * i
if rem % y == 0:
return True
else:
return False
if __name__ == '__main__':
main()
|
s558077021
|
Accepted
| 2,883
| 24,400
| 694
|
def main():
from itertools import combinations
a, b, c, d, e, f = map(int, input().split())
a *= 100
b *= 100
water = [a * i + b * j for i in range(f // a + 1) for j in range(f // b + 1)]
sugar = [c * i + d * j for i in range(f // c + 1) for j in range(f // d + 1)]
water.sort()
sugar.sort(reverse=True)
tmp = 0
x, y = 0, 0
for w in water:
if w == 0:
continue
for s in sugar:
if s + w > f:
continue
if s * 100 <= e * w and tmp <= s / (s + w):
tmp = s / (s + w)
x = s + w
y = s
print(x,y)
if __name__ == '__main__':
main()
|
s886724868
|
p03044
|
u644778646
| 2,000
| 1,048,576
|
Wrong Answer
| 697
| 85,756
| 398
|
We have a tree with N vertices numbered 1 to N. The i-th edge in the tree connects Vertex u_i and Vertex v_i, and its length is w_i. Your objective is to paint each vertex in the tree white or black (it is fine to paint all vertices the same color) so that the following condition is satisfied: * For any two vertices painted in the same color, the distance between them is an even number. Find a coloring of the vertices that satisfies the condition and print it. It can be proved that at least one such coloring exists under the constraints of this problem.
|
import sys
sys.setrecursionlimit(1000000)
N = int(input())
e = [ [] for i in range(N+1)]
for i in range(N-1):
u, v, w = map(int, input().split())
e[u].append((v,w))
e[v].append((u,w))
col = [ -1 for i in range(N+1)]
def dfs(x,c):
col[x] = c
for v,w in e[x]:
if col[v] == -1:
dfs(v , c+w)
dfs(1,0)
print(col)
for i in range(1,N+1):
print(col[i] % 2)
|
s448289872
|
Accepted
| 689
| 82,300
| 388
|
import sys
sys.setrecursionlimit(1000000)
N = int(input())
e = [ [] for i in range(N+1)]
for i in range(N-1):
u, v, w = map(int, input().split())
e[u].append((v,w))
e[v].append((u,w))
col = [ -1 for i in range(N+1)]
def dfs(x,c):
col[x] = c
for v,w in e[x]:
if col[v] == -1:
dfs(v , c+w)
dfs(1,0)
for i in range(1,N+1):
print(col[i] % 2)
|
s248618168
|
p03377
|
u317711717
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,060
| 89
|
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
import math
a,b,c=map(int,input().split())
print("Yes" if a<=c and c<=a+b else "No" )
|
s722498062
|
Accepted
| 19
| 3,060
| 83
|
import math
a,b,c=map(int,input().split())
print("YES" if a<=c<=a+b else "NO" )
|
s476860557
|
p03251
|
u554954744
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,060
| 198
|
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.
|
N,M,X,Y=map(int,input().split())
x=list(map(int,input().split()))
y=list(map(int,input().split()))
x.sort(reverse=True)
y.sort(reverse=True)
if x[0]<y[-1]:
print('War')
else:
print('No War')
|
s038484135
|
Accepted
| 18
| 3,060
| 183
|
N,M,X,Y=map(int,input().split())
x=list(map(int,input().split()))
y=list(map(int,input().split()))
x.append(X)
y.append(Y)
if max(x)<min(y):
print('No War')
else:
print('War')
|
s589143636
|
p03605
|
u050708958
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 53
|
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?
|
s = input()
print('yes') if '9' in s else print('no')
|
s165366053
|
Accepted
| 17
| 2,940
| 53
|
s = input()
print('Yes') if '9' in s else print('No')
|
s391407545
|
p03862
|
u705418271
| 2,000
| 262,144
|
Wrong Answer
| 107
| 19,960
| 202
|
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()))
ans=0
if A[0]>x:
ans+=A[0]-x
A[0]=x
for i in range(1,N):
if A[i]+A[i-1]>x:
ans+-A[i]+A[i-1]-x
A[i]-=A[i]+A[i-1]-x
print(ans)
|
s957835425
|
Accepted
| 111
| 20,124
| 242
|
N, x = map(int, input().split())
A = list(map(int, input().split()))
ans = 0
if A[0] > x:
ans += A[0] - x
A[0] = x
for i in range(1, N):
if A[i]+A[i-1] > x:
ans += A[i]+A[i-1]-x
A[i] -= A[i]+A[i-1]-x
print(ans)
|
s540018041
|
p03546
|
u442877951
| 2,000
| 262,144
|
Wrong Answer
| 37
| 3,444
| 585
|
Joisino the magical girl has decided to turn every single digit that exists on this world into 1. Rewriting a digit i with j (0≤i,j≤9) costs c_{i,j} MP (Magic Points). She is now standing before a wall. The wall is divided into HW squares in H rows and W columns, and at least one square contains a digit between 0 and 9 (inclusive). You are given A_{i,j} that describes the square at the i-th row from the top and j-th column from the left, as follows: * If A_{i,j}≠-1, the square contains a digit A_{i,j}. * If A_{i,j}=-1, the square does not contain a digit. Find the minimum total amount of MP required to turn every digit on this wall into 1 in the end.
|
def warshall_floyd(c,d):
for k in range(10):
for i in range(10):
for j in range(10):
c[j][k] = min(c[j][k],c[j][i] + c[i][k])
if k == 1:
d[j] = c[j][k]
return d
H,W = map(int,input().split())
c = [list(map(int,input().split())) for _ in range(10)]
A = [list(map(int,input().split())) for _ in range(H)]
d_list = [0]*10
ans = 0
warshall_floyd(c,d_list)
for i in range(H):
for j in range(W):
if A[i][j] != -1:
ans += d_list[A[i][j]]
print(ans)
|
s620283084
|
Accepted
| 37
| 3,444
| 585
|
def warshall_floyd(c,d):
for i in range(10):
for j in range(10):
for k in range(10):
c[j][k] = min(c[j][k],c[j][i] + c[i][k])
if k == 1:
d[j] = c[j][k]
return d
H,W = map(int,input().split())
c = [list(map(int,input().split())) for _ in range(10)]
A = [list(map(int,input().split())) for _ in range(H)]
d_list = [0]*10
ans = 0
warshall_floyd(c,d_list)
for i in range(H):
for j in range(W):
if A[i][j] != -1:
ans += d_list[A[i][j]]
print(ans)
|
s008178012
|
p03478
|
u075628913
| 2,000
| 262,144
|
Wrong Answer
| 38
| 3,060
| 252
|
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
inputs = input().split()
n, a, b = list(map(int, inputs))
print(n,a,b)
count=0
for num in range(n+1):
num_str = str(num)
all = 0
for i in range(len(num_str)):
all += int(num_str[i])
if a <= all <= b:
count+=num
print(count)
|
s537611808
|
Accepted
| 37
| 3,060
| 239
|
inputs = input().split()
n, a, b = list(map(int, inputs))
count=0
for num in range(n+1):
num_str = str(num)
all = 0
for i in range(len(num_str)):
all += int(num_str[i])
if a <= all <= b:
count+=num
print(count)
|
s925304378
|
p03338
|
u735008991
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 95
|
You are given a string S of length N consisting of lowercase English letters. We will cut this string at one position into two strings X and Y. Here, we would like to maximize the number of different letters contained in both X and Y. Find the largest possible number of different letters contained in both X and Y when we cut the string at the optimal position.
|
N = int(input())
S = input()
print(max([len(set(S[:i])) + len(set(S[i:])) for i in range(N)]))
|
s152584338
|
Accepted
| 17
| 2,940
| 102
|
N = int(input())
S = input()
print(max([len(set(S[:i]).intersection(set(S[i:]))) for i in range(N)]))
|
s079062257
|
p03658
|
u127499732
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 81
|
Snuke has N sticks. The length of the i-th stick is l_i. Snuke is making a snake toy by joining K of the sticks together. The length of the toy is represented by the sum of the individual sticks that compose it. Find the maximum possible length of the toy.
|
n,k=map(int,input().split())
s=sum( sorted(list(map(int,input().split())))[-k:] )
|
s203179429
|
Accepted
| 17
| 2,940
| 85
|
n,k=map(int,input().split())
print(sum(sorted(list(map(int,input().split())))[-k:] ))
|
s329167098
|
p02613
|
u282103490
| 2,000
| 1,048,576
|
Wrong Answer
| 138
| 9,124
| 259
|
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.
|
import math
def main():
mp = {
"AC": 0,
"TLE": 0,
"WA": 0,
"RE": 0
}
n = int(input())
for _ in range(n):
s = input()
mp[s]+=1
for key in mp.keys():
print(key, "x", mp[key])
main()
|
s407328896
|
Accepted
| 138
| 9,168
| 259
|
import math
def main():
mp = {
"AC": 0,
"WA": 0,
"TLE": 0,
"RE": 0
}
n = int(input())
for _ in range(n):
s = input()
mp[s]+=1
for key in mp.keys():
print(key, "x", mp[key])
main()
|
s399444408
|
p02612
|
u767797498
| 2,000
| 1,048,576
|
Wrong Answer
| 26
| 9,124
| 24
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
print(int(input())%1000)
|
s506088964
|
Accepted
| 28
| 9,124
| 48
|
a=int(input())%1000
print(0 if a==0 else 1000-a)
|
s467690208
|
p02406
|
u144068724
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,672
| 128
|
In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops. Note that use of goto statement is highly discouraged, because it is difficult to trace the control flow of a program which includes goto. Write a program which does precisely the same thing as the following program (this example is wrtten in C++). Let's try to write the program without goto statements. void call(int n){ int i = 1; CHECK_NUM: int x = i; if ( x % 3 == 0 ){ cout << " " << i; goto END_CHECK_NUM; } INCLUDE3: if ( x % 10 == 3 ){ cout << " " << i; goto END_CHECK_NUM; } x /= 10; if ( x ) goto INCLUDE3; END_CHECK_NUM: if ( ++i <= n ) goto CHECK_NUM; cout << endl; }
|
n = int(input())
result = ""
for i in range(n+1):
if i % 3 == 0 or i % 10 == 3:
result += " " + str(i)
print(result)
|
s448030252
|
Accepted
| 20
| 7,684
| 266
|
n = int(input())
result = ""
for i in range(1,n+1):
if i % 3 == 0 or i % 10 == 3:
result += " " + str(i)
continue
x = i
while(x):
if x % 10 == 3:
result += " " + str(i)
break
x //= 10
print(result)
|
s341676945
|
p03852
|
u533084327
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 662
|
Given a lowercase English letter c, determine whether it is a vowel. Here, there are five vowels in the English alphabet: `a`, `e`, `i`, `o` and `u`.
|
S = input()
str_list = ['dream', 'erase', 'dreamer', 'eraser']
start = 0
success_flag = False
continue_flag = False
def reverse_str(var: str):
return ''.join(list(reversed(var)))
reverse_S = reverse_str(S)
while True:
for str_name in str_list:
if reverse_S[start:start + len(str_name)] == reverse_str(str_name):
if len(S) == start + len(str_name):
success_flag = True
break
start += len(str_name)
continue_flag = True
if success_flag:
print('YES')
break
if not continue_flag:
print('NO')
break
else:
continue_flag = False
|
s174987025
|
Accepted
| 270
| 20,448
| 130
|
import numpy as np
c = input()
boin_list = ['a','e','i','o','u']
if c in boin_list:
print ('vowel')
else: print ('consonant')
|
s978489096
|
p03997
|
u553070631
| 2,000
| 262,144
|
Wrong Answer
| 21
| 3,316
| 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)
|
s078312586
|
Accepted
| 17
| 2,940
| 69
|
a=int(input())
b=int(input())
h=int(input())
print(int(((a+b)*h/2)))
|
s272134256
|
p04029
|
u823885866
| 2,000
| 262,144
|
Wrong Answer
| 30
| 9,100
| 32
|
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
a=int(input())
print(a*(a-1)//2)
|
s568322725
|
Accepted
| 26
| 9,156
| 33
|
print(sum(range(int(input())+1)))
|
s143634738
|
p03679
|
u175590965
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 129
|
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though. Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious. Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it. Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
|
x,a,b = map(int,input().split())
if x < b :
print("delicious")
elif x + a > b:
print("safe")
else:
print("dangerous")
|
s742710333
|
Accepted
| 17
| 2,940
| 131
|
x,a,b = map(int,input().split())
if b <= a :
print("delicious")
elif x + a >= b:
print("safe")
else:
print("dangerous")
|
s645370228
|
p03449
|
u446531810
| 2,000
| 262,144
|
Wrong Answer
| 21
| 3,064
| 350
|
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j). You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down. The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them. At most how many candies can you collect when you choose the best way to travel?
|
N=int(input())
row_1=list(map(lambda x:int(x),input().split()))
row_2=list(map(lambda x:int(x),input().split()))
max_value=0
for i in range(1,N):
sum_row_1=sum(row_1[0:i])
sum_row_2=sum(row_2[i-1:N])
print(row_1[0:i],row_2[i-1:N],sum_row_1,sum_row_2)
if max_value<sum_row_1+sum_row_2:max_value=sum_row_1+sum_row_2
print(max_value)
|
s043262864
|
Accepted
| 18
| 3,060
| 293
|
N=int(input())
row_1=list(map(lambda x:int(x),input().split()))
row_2=list(map(lambda x:int(x),input().split()))
max_value=0
for i in range(N):
sum_row_1=sum(row_1[0:i+1])
sum_row_2=sum(row_2[i:N])
if max_value<sum_row_1+sum_row_2:max_value=sum_row_1+sum_row_2
print(max_value)
|
s800703666
|
p03455
|
u846652026
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 95
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
i = list(map(int, input().split()))
if (i[0] * i[1]) % 2:
print("Even")
else:
print("Odd")
|
s313705867
|
Accepted
| 17
| 2,940
| 100
|
i = list(map(int, input().split()))
if (i[0] * i[1]) % 2 == 0:
print("Even")
else:
print("Odd")
|
s569152621
|
p02843
|
u244416763
| 2,000
| 1,048,576
|
Wrong Answer
| 29
| 3,060
| 111
|
AtCoder Mart sells 1000000 of each of the six items below: * Riceballs, priced at 100 yen (the currency of Japan) each * Sandwiches, priced at 101 yen each * Cookies, priced at 102 yen each * Cakes, priced at 103 yen each * Candies, priced at 104 yen each * Computers, priced at 105 yen each Takahashi wants to buy some of them that cost exactly X yen in total. Determine whether this is possible. (Ignore consumption tax.)
|
x = int(input())
for i in range(100000):
if(100*i <= x and 105*i >= x):
print(1)
else:
print(0)
|
s423034839
|
Accepted
| 29
| 3,060
| 126
|
x = int(input())
for i in range(100000):
if(100*i <= x and 105*i >= x):
print(1)
exit()
else:
print(0)
|
s164334785
|
p03854
|
u005899289
| 2,000
| 262,144
|
Wrong Answer
| 40
| 9,184
| 341
|
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
a = input()
b = a[::-1]
i = 0
while i < len(b):
if b[i:i+5] == "esare":
i += 5
elif b[i:i+5] == "maerd":
i += 5
elif b[i:i+6] == "resare":
i += 6
elif b[i:i+7] == "remaerd":
i += 7
else:
ans = "No"
break
if i == len(b):
ans = "Yes"
else:
ans = "No"
print(ans)
|
s014829818
|
Accepted
| 37
| 9,120
| 341
|
a = input()
b = a[::-1]
i = 0
while i < len(b):
if b[i:i+5] == "esare":
i += 5
elif b[i:i+5] == "maerd":
i += 5
elif b[i:i+6] == "resare":
i += 6
elif b[i:i+7] == "remaerd":
i += 7
else:
ans = "NO"
break
if i == len(b):
ans = "YES"
else:
ans = "NO"
print(ans)
|
s029392864
|
p03861
|
u940342887
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 128
|
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(float, input().split())
import math
c = a//x
d = b//x
print(d)
if a%x == 0:
print(d-c+1)
else:
print(d-c)
|
s724001874
|
Accepted
| 17
| 2,940
| 117
|
a, b, x = map(int, input().split())
import math
c = a//x
d = b//x
if a%x == 0:
print(d-c+1)
else:
print(d-c)
|
s905337480
|
p03590
|
u333945892
| 2,000
| 262,144
|
Wrong Answer
| 1,184
| 4,584
| 551
|
_Seisu-ya_ , a store specializing in non-negative integers, sells N non- negative integers. The i-th integer is A_i and has a _utility_ of B_i. There may be multiple equal integers with different utilities. Takahashi will buy some integers in this store. He can buy a combination of integers whose _bitwise OR_ is less than or equal to K. He wants the sum of utilities of purchased integers to be as large as possible. Find the maximum possible sum of utilities of purchased integers.
|
from collections import defaultdict,deque
import sys,heapq,bisect,math,itertools,string,queue,datetime
sys.setrecursionlimit(10**8)
INF = float('inf')
mod = 10**9+7
def inpl(): return list(map(int, input().split()))
def inpls(): return list(input().split())
N,K = inpl()
L = K.bit_length()
print(bin(K))
koho = [K]
tmp = 0
for b in reversed(range(L)):
if (K>>b) & 1:
print(b)
tmp += (1<<b)
koho.append(tmp-1)
nums = [0]*len(koho)
for _ in range(N):
A,B = inpl()
for i,k in enumerate(koho):
if k|A == k:
nums[i] += B
print(max(nums))
|
s630778410
|
Accepted
| 1,017
| 4,336
| 526
|
from collections import defaultdict,deque
import sys,heapq,bisect,math,itertools,string,queue,datetime
sys.setrecursionlimit(10**8)
INF = float('inf')
mod = 10**9+7
def inpl(): return list(map(int, input().split()))
def inpls(): return list(input().split())
N,K = inpl()
L = K.bit_length()
koho = [K]
tmp = 0
for b in reversed(range(L)):
if (K>>b) & 1:
tmp += (1<<b)
koho.append(tmp-1)
nums = [0]*len(koho)
for _ in range(N):
A,B = inpl()
for i,k in enumerate(koho):
if k|A == k:
nums[i] += B
print(max(nums))
|
s206654622
|
p03644
|
u292810930
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 47
|
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.
|
N = bin(int(input()))[::-1]
print(N.find('1'))
|
s045299128
|
Accepted
| 17
| 2,940
| 42
|
N = bin(int(input()))
print(2**(len(N)-3))
|
s463928314
|
p02235
|
u063056051
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,608
| 336
|
For given two sequences $X$ and $Y$, a sequence $Z$ is a common subsequence of $X$ and $Y$ if $Z$ is a subsequence of both $X$ and $Y$. For example, if $X = \\{a,b,c,b,d,a,b\\}$ and $Y = \\{b,d,c,a,b,a\\}$, the sequence $\\{b,c,a\\}$ is a common subsequence of both $X$ and $Y$. On the other hand, the sequence $\\{b,c,a\\}$ is not a longest common subsequence (LCS) of $X$ and $Y$, since it has length 3 and the sequence $\\{b,c,b,a\\}$, which is also common to both $X$ and $Y$, has length 4. The sequence $\\{b,c,b,a\\}$ is an LCS of $X$ and $Y$, since there is no common subsequence of length 5 or greater. Write a program which finds the length of LCS of given two sequences $X$ and $Y$. The sequence consists of alphabetical characters.
|
import sys
n=int(input())
for _ in range(n):
x=sys.stdin.readline().rstrip('\r\n')
y=sys.stdin.readline().rstrip('\r\n')
list=[0]
for z in y:
for i in range(len(list)-1,-1,-1):
tmp=x.find(z,list[i])
if tmp+1:
if i+1<len(list):
list[i+1]=min(tmp+1,list[i+1])
else:
list.append(tmp+1)
print(len(list)-1)
|
s453401954
|
Accepted
| 2,880
| 5,616
| 294
|
def f(x,y):
list=[0]
for z in y:
for i in range(len(list)-1,-1,-1):
tmp=x.find(z,list[i])+1
if tmp:
if i+1 < len(list):
list[i+1]=min(list[i+1],tmp)
else:
list.append(tmp)
return len(list)-1
n=int(input())
for i in range(n):
x=input()
y=input()
print(f(x,y))
|
s008162341
|
p04030
|
u218843509
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 185
|
Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?
|
s_list = list(input())
ans_list = []
for s in s_list:
if s == "1" or s == "0":
ans_list.append(s)
elif ans_list != []:
ans_list.pop()
"".join(ans_list)
|
s313767659
|
Accepted
| 17
| 2,940
| 192
|
s_list = list(input())
ans_list = []
for s in s_list:
if s == "1" or s == "0":
ans_list.append(s)
elif ans_list != []:
ans_list.pop()
print("".join(ans_list))
|
s194916983
|
p03760
|
u234631479
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 121
|
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions. You are given two strings O and E. O contains the characters at the odd- numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
|
A = input()
B = input()
print(A,B)
S = list(A)
for i in range(len(B)):
S.insert(i*2+1,B[i])
print(''.join(map(str, S)))
|
s936131332
|
Accepted
| 17
| 3,064
| 110
|
A = input()
B = input()
S = list(A)
for i in range(len(B)):
S.insert(i*2+1,B[i])
print(''.join(map(str, S)))
|
s388058637
|
p03090
|
u785773061
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 613
|
You are given an integer N. Build an undirected graph with N vertices with indices 1 to N that satisfies the following two conditions: * The graph is simple and connected. * There exists an integer S such that, for every vertex, the sum of the indices of the vertices adjacent to that vertex is S. It can be proved that at least one such graph exists under the constraints of this problem.
|
#b
# n = int(input())
n = 3
pairs = []
if n % 2 == 0:
for x in range(1, int(n/2) + 1):
pairs.append([x, n-x + 1])
else:
for x in range(1, int((n-1)/2) +1):
pairs.append([x, int((n-1))-x+1])
pairs.append([n])
# connect nodes
edges = []
for x in range(1, n+1):
for p in pairs:
# print(f'x: {x}, p: {p}')
if x in p:
break
for node in p:
edges.append(sorted([x, node]))
# print(f'edges >>>>> {edges}')
for e in edges:
print(' '.join([str(i) for i in e]))
|
s190756532
|
Accepted
| 28
| 3,860
| 623
|
#b
n = int(input())
pairs = []
if n % 2 == 0:
for x in range(1, int(n/2) + 1):
pairs.append([x, n-x + 1])
else:
for x in range(1, int((n-1)/2) +1):
pairs.append([x, int((n-1))-x+1])
pairs.append([n])
# connect nodes
edges = []
for x in range(1, n+1):
for p in pairs:
# print(f'x: {x}, p: {p}')
if x in p:
break
for node in p:
edges.append(sorted([x, node]))
# print(f'edges >>>>> {edges}')
print(len(edges))
for e in edges:
print(' '.join([str(i) for i in e]))
|
s796735561
|
p03385
|
u585604152
| 2,000
| 262,144
|
Wrong Answer
| 20
| 3,316
| 158
|
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
S = list(input())
import collections
count = collections.Counter(S)
if count["a"]==1 and count["b"]==1 and count["c"]==1:
print('YES')
else:
print('NO')
|
s155357093
|
Accepted
| 21
| 3,316
| 158
|
S = list(input())
import collections
count = collections.Counter(S)
if count["a"]==1 and count["b"]==1 and count["c"]==1:
print('Yes')
else:
print('No')
|
s225711159
|
p02612
|
u454524105
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 9,100
| 24
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
print(int(input())%1000)
|
s766735351
|
Accepted
| 33
| 9,084
| 58
|
m = int(input()) % 1000
print(1000 - m) if m else print(0)
|
s141857996
|
p02608
|
u808280993
| 2,000
| 1,048,576
|
Wrong Answer
| 842
| 9,264
| 459
|
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 >= 6
import math
N = int(input())
ans = [0] * N
lim = int(math.sqrt(N))
for x in range(1, lim):
for y in range(1, lim):
for z in range(1, lim):
#calc = ((x+y)**2 + (y+z)**2 + (x+z)**2) /2
calc = x**2 + y**2 + z**2 + x*y + y*z + z*x
if calc <= N:
ans[calc -1] += 1
print(ans)
|
s220362885
|
Accepted
| 814
| 9,144
| 488
|
# n >= 6
import math
N = int(input())
ans = [0] * N
lim = int(math.sqrt(N))
for x in range(1, lim):
for y in range(1, lim):
for z in range(1, lim):
#calc = ((x+y)**2 + (y+z)**2 + (x+z)**2) /2
calc = x**2 + y**2 + z**2 + x*y + y*z + z*x
if calc <= N:
ans[calc -1] += 1
#print(ans)
for a in ans:
print(a)
|
s209265859
|
p04043
|
u368882459
| 2,000
| 262,144
|
Wrong Answer
| 16
| 2,940
| 120
|
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.
|
ary = list(map(int, input().split()))
if ary.count(5) == 2 and ary.count(7) == 1:
print('Yes')
else:
print('No')
|
s815379525
|
Accepted
| 17
| 2,940
| 120
|
ary = list(map(int, input().split()))
if ary.count(5) == 2 and ary.count(7) == 1:
print('YES')
else:
print('NO')
|
s194076410
|
p03920
|
u426108351
| 2,000
| 262,144
|
Wrong Answer
| 22
| 3,316
| 184
|
The problem set at _CODE FESTIVAL 20XX Finals_ consists of N problems. The score allocated to the i-th (1≦i≦N) problem is i points. Takahashi, a contestant, is trying to score exactly N points. For that, he is deciding which problems to solve. As problems with higher scores are harder, he wants to minimize the highest score of a problem among the ones solved by him. Determine the set of problems that should be solved.
|
N = int(input())
n = N
now = 0
count = 0
while n > 0:
n -= count
now += count
count += 1
ignore = now - N
for i in range(1, count+1):
if i == ignore:
continue
print(i)
|
s672884860
|
Accepted
| 22
| 3,316
| 183
|
N = int(input())
n = N
now = 0
count = 0
while n > 0:
n -= count
now += count
count += 1
ignore = now - N
for i in range(1, count):
if i == ignore:
continue
print(i)
|
s318790301
|
p03861
|
u731368968
| 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(a//x-b//x)
|
s170158591
|
Accepted
| 17
| 2,940
| 56
|
a, b, x = map(int, input().split())
print(b//x-(a-1)//x)
|
s361289177
|
p02646
|
u189513668
| 2,000
| 1,048,576
|
Wrong Answer
| 23
| 9,180
| 162
|
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
A, V = map(int, input().split())
B, W = map(int, input().split())
T = int(input())
b = abs(A - B) + W*T
a = V*T
if a < b:
print("No")
else:
print("Yes")
|
s605222525
|
Accepted
| 24
| 9,180
| 162
|
A, V = map(int, input().split())
B, W = map(int, input().split())
T = int(input())
b = abs(A - B) + W*T
a = V*T
if a < b:
print("NO")
else:
print("YES")
|
s878416691
|
p03408
|
u869919400
| 2,000
| 262,144
|
Wrong Answer
| 21
| 3,316
| 291
|
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.
|
n = int(input())
blues = [input() for _ in range(n)]
m = int(input())
reds = [input() for _ in range(m)]
words = set(blues)
_max = 0
for w in words:
score = 0
score += blues.count(w)
score -= reds.count(w)
print(w, score)
if _max < score:
_max = score
print(_max)
|
s692170400
|
Accepted
| 17
| 3,064
| 271
|
n = int(input())
blues = [input() for _ in range(n)]
m = int(input())
reds = [input() for _ in range(m)]
words = set(blues)
_max = 0
for w in words:
score = 0
score += blues.count(w)
score -= reds.count(w)
if _max < score:
_max = score
print(_max)
|
s480705666
|
p00015
|
u886122084
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,596
| 309
|
A country has a budget of more than 81 trillion yen. We want to process such data, but conventional integer type which uses signed 32 bit can represent up to 2,147,483,647. Your task is to write a program which reads two integers (more than or equal to zero), and prints a sum of these integers. If given integers or the sum have more than 80 digits, print "overflow".
|
# python template for atcoder1
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline
n = int(input())
ans = []
for _ in range(n):
a = int(input())
b = int(input())
s = a+b
if s >= 10**79:
ans.append("overflow")
else:
ans.append(str(s))
print("\n".join(ans))
|
s736807739
|
Accepted
| 20
| 5,600
| 310
|
# python template for atcoder1
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline
n = int(input())
ans = []
for _ in range(n):
a = int(input())
b = int(input())
s = a+b
if s > 10**80-1:
ans.append("overflow")
else:
ans.append(str(s))
print("\n".join(ans))
|
s699601909
|
p02694
|
u135265051
| 2,000
| 1,048,576
|
Wrong Answer
| 28
| 9,148
| 186
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
#a = list(map(int,input().split()))
#b = list(map(int,input().split()))
a = int(input())
x = int(100)
count= 0
while a >= x:
x += int(x/100)
count+=1
print(count)
print(type(x))
|
s131857908
|
Accepted
| 29
| 9,140
| 166
|
#a = list(map(int,input().split()))
#b = list(map(int,input().split()))
a = int(input())
x = int(100)
count= 0
while a > x:
count+=1
x += x//100
print(count)
|
s094381936
|
p03998
|
u595893956
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,104
| 3,064
| 517
|
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.
|
s=input()
t=input()
u=input()
a=b=c=0
now = 0
while 1:
if now == 0:
if s[a] == 'a':
now = 0
elif s[a] == 'b':
now = 1
else:
now = 2
if a==len(s):
print('A')
break
elif now == 1:
if s[b] == 'a':
now = 0
elif s[b] == 'b':
now = 1
else:
now = 2
if b==len(t):
print('B')
break
else:
if s[c] == 'a':
now = 0
elif s[c] == 'b':
now = 1
else:
now = 2
if c==len(u):
print('C')
break
|
s436153931
|
Accepted
| 18
| 3,064
| 540
|
s=input()
t=input()
u=input()
a=b=c=0
now = 0
while 1:
if now == 0:
if a==len(s):
print('A')
break
if s[a] == 'a':
now = 0
elif s[a] == 'b':
now = 1
else:
now = 2
a+=1
elif now == 1:
if b==len(t):
print('B')
break
if t[b]=='a':
now = 0
elif t[b]== 'b':
now = 1
else:
now = 2
b+=1
else:
if c==len(u):
print('C')
break
if u[c]== 'a':
now = 0
elif u[c]== 'b':
now = 1
else:
now = 2
c+=1
|
s683756298
|
p03471
|
u498575211
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 3,064
| 343
|
The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.
|
N, Y = map(int, input().split())
nn = Y // 10000
mm = Y // 5000
ll = Y // 1000
p = []
ans = "-1, -1, -1"
for i in range(nn):
for j in range(mm):
if i + j >= N:
break
for k in range(ll):
if i + j + k >= N:
break
if i * 10000 + j * 5000 + k * 1000 == Y:
ans = str(i)+", "+str(j)+", "+str(k)
print(ans)
|
s410594351
|
Accepted
| 1,009
| 3,188
| 406
|
N, Y = map(int, input().split())
nn = Y // 10000 if (Y // 10000) < N else N
mm = Y // 5000 if (Y // 5000) < N else N
ll = Y // 1000 if (Y // 1000) < N else N
ans = "-1 -1 -1"
for i in range(nn+1):
for j in range(mm+1):
if i + j > N:
break
if (i * 10000 + j * 5000 + (N-i-j) * 1000) == Y:
ans = str(i)+" "+str(j)+" "+str(N-i-j)
break
print(ans)
|
s999386228
|
p03945
|
u296518383
| 2,000
| 262,144
|
Wrong Answer
| 45
| 3,188
| 84
|
Two foxes Jiro and Saburo are playing a game called _1D Reversi_. This game is played on a board, using black and white stones. On the board, stones are placed in a row, and each player places a new stone to either end of the row. Similarly to the original game of Reversi, when a white stone is placed, all black stones between the new white stone and another white stone, turn into white stones, and vice versa. In the middle of a game, something came up and Saburo has to leave the game. The state of the board at this point is described by a string S. There are |S| (the length of S) stones on the board, and each character in S represents the color of the i-th (1 ≦ i ≦ |S|) stone from the left. If the i-th character in S is `B`, it means that the color of the corresponding stone on the board is black. Similarly, if the i-th character in S is `W`, it means that the color of the corresponding stone is white. Jiro wants all stones on the board to be of the same color. For this purpose, he will place new stones on the board according to the rules. Find the minimum number of new stones that he needs to place.
|
S=input()
res=0
for i in range(len(S)-1):
if S[i]!=S[i+1]:
res+=res
print(res)
|
s581793818
|
Accepted
| 70
| 13,300
| 218
|
def run_length(s) -> list:
res = [[s[0], 1]]
for i in range(1, len(s)):
if res[-1][0] == s[i]:
res[-1][1] += 1
else:
res.append([s[i], 1])
return res
R = run_length(input())
print(len(R) - 1)
|
s610239326
|
p02542
|
u102461423
| 2,000
| 1,048,576
|
Wrong Answer
| 377
| 57,708
| 942
|
There is a board with N rows and M columns. The information of this board is represented by N strings S_1,S_2,\ldots,S_N. Specifically, the state of the square at the i-th row from the top and the j-th column from the left is represented as follows: * S_{i,j}=`.` : the square is empty. * S_{i,j}=`#` : an obstacle is placed on the square. * S_{i,j}=`o` : a piece is placed on the square. Yosupo repeats the following operation: * Choose a piece and move it to its right adjecent square or its down adjacent square. Moving a piece to squares with another piece or an obstacle is prohibited. Moving a piece out of the board is also prohibited. Yosupo wants to perform the operation as many times as possible. Find the maximum possible number of operations.
|
import sys
import networkx as nx
import numpy as np
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
H, W = map(int, readline().split())
S = np.frombuffer(read(), 'S1').reshape(H, -1)[:, :W].astype('U1')
INF = 1 << 32
total_o = (S == 'o').sum()
G = nx.DiGraph()
sink = 'sink'
G.add_node(sink, demand=total_o)
for i in range(H):
for j in range(W):
if S[i, j] == '#':
continue
if S[i, j] == 'o':
G.add_node((i, j), demand=-1)
G.add_edge((i, j), sink, capacity=1, weight=0)
elif S[i, j] == '.':
G.add_node((i, j), demand=0)
G.add_edge((i, j), sink, capacity=1, weight=0)
if i + 1 < H and S[i + 1, j] != '#':
G.add_edge((i, j), (i + 1, j), capacity=INF, weight=-1)
if j + 1 < W and S[i, j + 1] != '#':
G.add_edge((i, j), (i, j + 1), capacity=INF, weight=-1)
|
s552155751
|
Accepted
| 930
| 60,600
| 997
|
import sys
import networkx as nx
import numpy as np
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
H, W = map(int, readline().split())
S = np.frombuffer(read(), 'S1').reshape(H, -1)[:, :W].astype('U1')
INF = 1 << 32
total_o = (S == 'o').sum()
G = nx.DiGraph()
sink = 'sink'
G.add_node(sink, demand=total_o)
for i in range(H):
for j in range(W):
if S[i, j] == '#':
continue
if S[i, j] == 'o':
G.add_node((i, j), demand=-1)
G.add_edge((i, j), sink, capacity=1, weight=0)
elif S[i, j] == '.':
G.add_node((i, j), demand=0)
G.add_edge((i, j), sink, capacity=1, weight=0)
if i + 1 < H and S[i + 1, j] != '#':
G.add_edge((i, j), (i + 1, j), capacity=INF, weight=-1)
if j + 1 < W and S[i, j + 1] != '#':
G.add_edge((i, j), (i, j + 1), capacity=INF, weight=-1)
mcfc = nx.min_cost_flow_cost(G)
ans = -mcfc
print(ans)
|
s787236523
|
p03719
|
u556477263
| 2,000
| 262,144
|
Wrong Answer
| 24
| 9,068
| 90
|
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<=b and b<=c:
print('Yes')
else:
print('No')
|
s001730283
|
Accepted
| 30
| 9,120
| 91
|
a,b,c = map(int,input().split())
if a<= c and b>=c:
print('Yes')
else:
print('No')
|
s089146968
|
p03605
|
u117193815
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 67
|
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?
|
n=input()
if n[0]==9 or n[1]==9:
print("Yes")
else:
print("No")
|
s302510986
|
Accepted
| 16
| 2,940
| 71
|
n=input()
if n[0]=="9" or n[1]=="9":
print("Yes")
else:
print("No")
|
s108994945
|
p02260
|
u672822075
| 1,000
| 131,072
|
Wrong Answer
| 30
| 6,720
| 224
|
Write a program of the Selection Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: SelectionSort(A) 1 for i = 0 to A.length-1 2 mini = i 3 for j = i to A.length-1 4 if A[j] < A[mini] 5 mini = j 6 swap A[i] and A[mini] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 6 of the pseudocode in the case where i ≠ mini.
|
cnt = 0
n = int(input())
a = list(map(int,input().split()))
for i in range(0,n):
mini = i
for j in range(i,n):
if a[j]<a[mini]:
mini = j
cnt += 1
a[j],a[mini] = a[mini],a[j]
print(" ".join(map(str,a)))
print(cnt)
|
s370331437
|
Accepted
| 30
| 6,724
| 236
|
cnt = 0
n = int(input())
a = list(map(int,input().split()))
for i in range(0,n):
mini = i
for j in range(i,n):
if a[j]<a[mini]:
mini = j
a[i],a[mini] = a[mini],a[i]
if i!=mini:
cnt += 1
print(" ".join(map(str,a)))
print(cnt)
|
s283815106
|
p03836
|
u993435350
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 452
|
Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him.
|
sx,sy,tx,ty = map(int,input().split())
dx = abs(tx - sx)
dy = abs(ty - sy)
move_x = "R" * dx
move_y = "U" * dy
f_move = move_x + move_y
if "R" in f_move:
s_move = f_move.replace("R","L")
if "U" in s_move:
s_move = s_move.replace("U","D")
t_move = "L" + "U" + f_move + "D"
if "R" in t_move:
ff_move = t_move.replace("R","L")
if "U" in s_move:
ff_move = ff_move.replace("U","D")
total = f_move + s_move + t_move + ff_move
print(total)
|
s158681236
|
Accepted
| 17
| 3,064
| 390
|
sx,sy,tx,ty = map(int,input().split())
dx = abs(tx - sx)
dy = abs(ty - sy)
move_x = "R" * dx
move_y = "U" * dy
f_move = move_y + move_x
if "R" in f_move:
s_move = f_move.replace("R","L")
if "U" in s_move:
s_move = s_move.replace("U","D")
t_move = "L" + "U" + f_move + "R" + "D"
ff_move = "R" + "D" + s_move + "L" + "U"
total = f_move + s_move + t_move + ff_move
print(total)
|
s369028461
|
p03448
|
u170183831
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 603
|
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.
|
def solve(a, b, c, x):
count_all_50yen = x // 50 # 50
print(count_all_50yen)
count_50yen_pattern = min(c, count_all_50yen)
count_100yen_pattern = min(b, count_all_50yen // 2)
count_500yen_pattern = min(a, count_all_50yen // 10)
print(count_50yen_pattern, count_100yen_pattern, count_500yen_pattern)
return count_50yen_pattern + count_100yen_pattern + count_500yen_pattern
_a = int(input())
_b = int(input())
_c = int(input())
_x = int(input())
print(solve(_a, _b, _c, _x))
|
s862893798
|
Accepted
| 49
| 3,060
| 303
|
def solve(a, b, c, x):
return sum(
x == 500 * i + 100 * j + 50 * k
for i in range(a + 1)
for j in range(b + 1)
for k in range(c + 1)
)
_a = int(input())
_b = int(input())
_c = int(input())
_x = int(input())
print(solve(_a, _b, _c, _x))
|
s381043651
|
p01315
|
u286751622
| 8,000
| 131,072
|
Wrong Answer
| 70
| 6,056
| 1,154
|
You are passionate about the popular web game "Moonlight Ranch". The purpose of this game is to grow crops in the fields, sell them to earn income, and use the income to grow the ranch. You wanted to grow your field quickly. Therefore, we decided to start by arranging the crops that can be grown in the game based on their income efficiency per hour. You have to buy seeds to grow crops. Here the species name of crop i is given by Li and the price by Pi. When a seed is planted in a field, it sprouts after time Ai. Young leaves emerge after time Bi from the emergence of buds. Leaves grow thick after time Ci from the emergence of young leaves. After the time Di has passed since the leaves grow thick, the flowers bloom. After the time Ei after the flower blooms, the fruit is formed. One seed produces Fi seeds, and each of these seeds can be sold at a price of Si. Some crops are multi-season crops and produce Mi crops in total. Mi = 1 for single cropping. Multi-season crops return to leaves after fruiting until Mi th fruiting. The income for a seed is the amount of money sold for all the fruit of that seed minus the price of the seed. In addition, the income efficiency of the seed is the value obtained by dividing the income by the time from planting the seed until all fruits are produced. Your task is to write a program that takes the input crop information and sorts it in descending order of income efficiency.
|
resultss = []
while True:
N = int(input())
if N == 0:
break
B = [list(map(str, input().split())) for _ in range(N)]
results = []
for i in range(N):
A = []
name = B[i][0]
B[i][0] = B[i][1]
for j in range(10):
A.append(int(B[i][j]))
if A[-1] == 1:
time = A[2] + A[3] + A[4] + A[5] + A[6]
money = A[7] * A[8]
res = money - A[1]
res /= time
else:
time = A[2] + A[3] + A[4] + A[5] + A[6] + (A[5] + A[6]) * (A[-1] - 1)
money = A[7] * A[8] * A[-1]
res = money - A[1]
res /= time
results.append([res, name])
tmp = results[0][0]
flag =False
for i in range(len(results)):
if results[i][0] != tmp:
flag = True
if flag:
results.sort()
else:
results = sorted(results, key=lambda x: x[1], reverse=True)
resultss.append(results)
for i in range(len(resultss)):
for j in range(len(resultss[i])):
print(resultss[i][-j-1][1])
print('#')
|
s930958619
|
Accepted
| 60
| 6,060
| 1,039
|
resultss = []
while True:
N = int(input())
if N == 0:
break
B = [list(map(str, input().split())) for _ in range(N)]
results = []
for i in range(N):
A = []
name = B[i][0]
B[i][0] = B[i][1]
for j in range(10):
A.append(int(B[i][j]))
if A[-1] == 1:
time = A[2] + A[3] + A[4] + A[5] + A[6]
money = A[7] * A[8]
res = money - A[1]
res /= time
else:
time = A[2] + A[3] + A[4] + A[5] + A[6] + (A[5] + A[6]) * (A[-1] - 1)
money = A[7] * A[8] * A[-1]
res = money - A[1]
res /= time
results.append([res, name])
results = sorted(results, key=lambda x: x[1], reverse = True)
results = sorted(results, key=lambda x: x[0])
resultss.append(results)
for i in range(len(resultss)):
for j in range(len(resultss[i])):
print(resultss[i][-j-1][1])
print('#')
|
s295390120
|
p03494
|
u829094246
| 2,000
| 262,144
|
Wrong Answer
| 21
| 3,060
| 225
|
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())
An=list(map(int, input().rstrip().split()))
cnt=0
while True:
if len(list(filter(lambda a:(a+1)%2, An))) < len(An):
break
An=list(map(lambda a:int(a/2), An))
print(An)
cnt+=1
print(cnt)
|
s097675307
|
Accepted
| 19
| 2,940
| 211
|
N=int(input())
An=list(map(int, input().rstrip().split()))
cnt=0
while True:
if len(list(filter(lambda a:(a+1)%2, An))) < len(An):
break
An=list(map(lambda a:int(a/2), An))
cnt+=1
print(cnt)
|
s628713346
|
p02398
|
u933096856
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,696
| 97
|
Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.
|
a,b,c=map(int, input().split())
d=0
for i in range(a+1,b+1):
if c%i==0:
d+=1
print(d)
|
s883015764
|
Accepted
| 50
| 7,704
| 95
|
a,b,c=map(int, input().split())
d=0
for i in range(a,b+1):
if c%i==0:
d+=1
print(d)
|
s217387343
|
p03385
|
u382431597
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 97
|
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
s=input()
s = s.replace("a","").replace("b","").replace("c","")
print("Yes" if s == 0 else "No")
|
s694362201
|
Accepted
| 18
| 2,940
| 85
|
s=input()
print("Yes" if s.count("a") == s.count("b") == s.count("c") == 1 else "No")
|
s472885368
|
p03943
|
u396391104
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 99
|
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
a,b,c = map(int,input().split())
d = (a+b+c)//2
print("Yes") if d==a | d==b | d==c else print("No")
|
s955700012
|
Accepted
| 18
| 2,940
| 99
|
a,b,c = map(int,input().split())
if a==b+c or b==c+a or c==a+b:
print("Yes")
else:
print("No")
|
s986113668
|
p03501
|
u797016134
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 88
|
You are parking at a parking lot. You can choose from the following two fee plans: * Plan 1: The fee will be A×T yen (the currency of Japan) when you park for T hours. * Plan 2: The fee will be B yen, regardless of the duration. Find the minimum fee when you park for N hours.
|
n,a,b = map(int, input().split())
if a*n >= b:
print(a*n)
if a*n <= b:
print(b)
|
s397507361
|
Accepted
| 17
| 3,064
| 99
|
n,a,b = map(int, input().split())
if a*n >= b:
print(b)
exit()
if a*n <= b:
print(a*n)
|
s833175402
|
p02608
|
u690781906
| 2,000
| 1,048,576
|
Wrong Answer
| 2,211
| 201,556
| 238
|
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).
|
from collections import defaultdict
ans = defaultdict(int)
N = int(input())
for i in range(1, N):
for j in range(1, N):
for k in range(1, N):
ans[i*i+j*j+k*k+i*j+j*k+k*i] += 1
for i in range(N):
print(ans[i])
|
s297719981
|
Accepted
| 506
| 11,640
| 246
|
from collections import defaultdict
ans = defaultdict(int)
N = int(input())
for i in range(1, 100):
for j in range(1, 100):
for k in range(1, 100):
ans[i*i+j*j+k*k+i*j+j*k+k*i] += 1
for i in range(N):
print(ans[i+1])
|
s309499212
|
p03006
|
u125545880
| 2,000
| 1,048,576
|
Wrong Answer
| 219
| 9,208
| 793
|
There are N balls in a two-dimensional plane. The i-th ball is at coordinates (x_i, y_i). We will collect all of these balls, by choosing two integers p and q such that p \neq 0 or q \neq 0 and then repeating the following operation: * Choose a ball remaining in the plane and collect it. Let (a, b) be the coordinates of this ball. If we collected a ball at coordinates (a - p, b - q) in the previous operation, the cost of this operation is 0. Otherwise, including when this is the first time to do this operation, the cost of this operation is 1. Find the minimum total cost required to collect all the balls when we optimally choose p and q.
|
import itertools
def main():
N = int(input())
z = []
ans = 51
for i in range(N):
z.append(tuple(map(int, input().split())))
z.sort(key=lambda x: x[0])
for i, j in itertools.combinations(range(N), 2):
p = z[i][0] - z[j][0]
q = z[i][1] - z[j][1]
ans = min(ans, solve(N, z, p, q))
print(ans)
def solve(N, z, p, q):
if p < 0:
p, q = -p, -q
passlist = set()
cnt = 0
for i in range(N):
if z[i] in passlist:
continue
x, y = z[i][0], z[i][1]
for j in range(N):
s, t = x-p, y-q
if (s, t) not in z:
cnt += 1
break
passlist.add((s, t))
x, y = s, t
return cnt
if __name__ == "__main__":
main()
|
s960413677
|
Accepted
| 114
| 9,264
| 854
|
import itertools
def main():
N = int(input())
z = []
ans = 51
for i in range(N):
z.append(tuple(map(int, input().split())))
z.sort(key=lambda x: x[1])
z.sort(key=lambda x: x[0], reverse=True)
if len(z) == 1:
print(1)
return
for i, j in itertools.combinations(range(N), 2):
p = z[i][0] - z[j][0]
q = z[i][1] - z[j][1]
ans = min(ans, solve(N, z, p, q))
print(ans)
def solve(N, z, p, q):
passlist = set()
cnt = 0
for i in range(N):
if z[i] in passlist:
continue
x, y = z[i][0], z[i][1]
for j in range(N):
s, t = x-p, y-q
if (s, t) not in z:
cnt += 1
break
passlist.add((s, t))
x, y = s, t
return cnt
if __name__ == "__main__":
main()
|
s917616858
|
p03861
|
u045939752
| 2,000
| 262,144
|
Wrong Answer
| 23
| 3,064
| 64
|
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))
|
s538584551
|
Accepted
| 25
| 3,064
| 70
|
a, b, x = map(int, input().split())
a -= 1
print((b // x) - (a // x))
|
s771242470
|
p03080
|
u296518383
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 92
|
There are N people numbered 1 to N. Each person wears a red hat or a blue hat. You are given a string s representing the colors of the people. Person i wears a red hat if s_i is `R`, and a blue hat if s_i is `B`. Determine if there are more people wearing a red hat than people wearing a blue hat.
|
N=int(input())
S=input()
if S.count("r")-S.count("b")>0:
print("Yes")
else:
print("No")
|
s701995581
|
Accepted
| 17
| 2,940
| 72
|
N,S=input(),input()
print("Yes" if S.count("R")>S.count("B") else "No")
|
s582630314
|
p02271
|
u564398841
| 5,000
| 131,072
|
Wrong Answer
| 20
| 7,700
| 764
|
Write a program which reads a sequence _A_ of _n_ elements and an integer _M_ , and outputs "yes" if you can make _M_ by adding elements in _A_ , otherwise "no". You can use an element only once. You are given the sequence _A_ and _q_ questions where each question contains _M i_.
|
class Solver():
def __init__(self, A: [int]):
self.A = A
self.max_depth = len(A)
def solve(self, tgt):
self.tgt = tgt
return self._rec_solve(0, 0)
def _rec_solve(self, _sum, index):
if index >= self.max_depth or _sum > self.tgt:
return False
elif _sum == self.tgt:
return True
else:
return self._rec_solve(_sum + self.A[index], index + 1) or self._rec_solve(_sum, index + 1)
if __name__ == '__main__':
n = int(input())
A = [int(i) for i in input().rstrip().split()]
print(A)
q = int(input())
m_ary = [int(i) for i in input().rstrip().split()]
print(m_ary)
s = Solver(A)
for i in m_ary:
print(s.solve(i))
|
s132586952
|
Accepted
| 1,420
| 8,396
| 1,172
|
class Solver():
def __init__(self, A: [int]):
self.A = A
self.max_depth = len(A)
def solve(self, tgt):
self.tgt = tgt
self.history = (self.max_depth+1)*[4000 * [None]]
self.history = [[None for i in range(4000)] for x in range(self.max_depth + 1)]
if self._rec_solve(0, 0):
return 'yes'
else:
return 'no'
def _rec_solve(self, _sum, index):
if self.history[index][_sum] is not None:
return self.history[index][_sum]
if _sum == self.tgt:
self.history[index][_sum] = True
return True
if index >= self.max_depth or _sum > self.tgt:
self.history[index][_sum] = False
return False
self.history[index][_sum] = self._rec_solve(_sum + self.A[index], index + 1) or self._rec_solve(_sum, index + 1)
return self.history[index][_sum]
if __name__ == '__main__':
n = int(input())
A = [int(i) for i in input().rstrip().split()]
q = int(input())
m_ary = [int(i) for i in input().rstrip().split()]
s = Solver(A)
for val in m_ary:
print(s.solve(val))
|
s540814510
|
p04011
|
u517630860
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 93
|
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.
|
# -*- coding: utf-8 -*-
w = input()
print(['Yes', 'No'][any(w.count(c) % 2 for c in set(w))])
|
s369948746
|
Accepted
| 17
| 2,940
| 178
|
# -*- coding: utf-8 -*-
N = int(input())
K = int(input())
X = int(input())
Y = int(input())
total = X * (N if N <= K else K)
total += Y * (0 if N <= K else N - K)
print(total)
|
s019050165
|
p03435
|
u075012704
| 2,000
| 262,144
|
Wrong Answer
| 249
| 3,064
| 415
|
We have a 3 \times 3 grid. A number c_{i, j} is written in the square (i, j), where (i, j) denotes the square at the i-th row from the top and the j-th column from the left. According to Takahashi, there are six integers a_1, a_2, a_3, b_1, b_2, b_3 whose values are fixed, and the number written in the square (i, j) is equal to a_i + b_j. Determine if he is correct.
|
C = [list(map(int, input().split())) for i in range(3)]
for a1 in range(101):
for a2 in range(101):
for a3 in range(101):
if (C[0][0]-a1) == (C[1][0]-a2) == (C[2][0]-a3):
if (C[0][1]-a1) == (C[1][1]-a2) == (C[2][1]-a3):
if (C[0][2] - a1) == (C[1][2] - a2) == (C[2][2] - a3):
print("YES")
exit()
print("NO")
|
s691238505
|
Accepted
| 261
| 3,064
| 415
|
C = [list(map(int, input().split())) for i in range(3)]
for a1 in range(101):
for a2 in range(101):
for a3 in range(101):
if (C[0][0]-a1) == (C[1][0]-a2) == (C[2][0]-a3):
if (C[0][1]-a1) == (C[1][1]-a2) == (C[2][1]-a3):
if (C[0][2] - a1) == (C[1][2] - a2) == (C[2][2] - a3):
print("Yes")
exit()
print("No")
|
s562111107
|
p03493
|
u686485517
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 80
|
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=input()
c = 0
for i in range(3):
if s[i] == "1":
c += 1
print("c")
|
s679932210
|
Accepted
| 17
| 2,940
| 78
|
s=input()
c = 0
for i in range(3):
if s[i] == "1":
c += 1
print(c)
|
s438464525
|
p02406
|
u435158342
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,584
| 79
|
In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops. Note that use of goto statement is highly discouraged, because it is difficult to trace the control flow of a program which includes goto. Write a program which does precisely the same thing as the following program (this example is wrtten in C++). Let's try to write the program without goto statements. void call(int n){ int i = 1; CHECK_NUM: int x = i; if ( x % 3 == 0 ){ cout << " " << i; goto END_CHECK_NUM; } INCLUDE3: if ( x % 10 == 3 ){ cout << " " << i; goto END_CHECK_NUM; } x /= 10; if ( x ) goto INCLUDE3; END_CHECK_NUM: if ( ++i <= n ) goto CHECK_NUM; cout << endl; }
|
a=int(input())
b=3
while True:
if b>a:
break
print(b)
b+=3
|
s900459252
|
Accepted
| 20
| 5,872
| 104
|
a=int(input())
for i in range(1, a+1):
if i%3==0 or '3' in str(i):print(' '+ str(i),end='')
print()
|
s024224919
|
p02393
|
u074747865
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,580
| 78
|
Write a program which reads three integers, and prints them in ascending order.
|
li=list(map(int,input().split()))
li.sort()
for i in li:
print("i", sep=" ")
|
s854552212
|
Accepted
| 20
| 5,576
| 47
|
li=sorted(map(int,input().split()))
print(*li)
|
s781052527
|
p02389
|
u838759969
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,592
| 172
|
Write a program which calculates the area and perimeter of a given rectangle.
|
__author__= 'CIPHER'
str = input()
numList = str.split(' ')
numList = [ int(x) for x in numList]
#print(numList)
result = 1
for x in numList:
result *= x
print(result)
|
s328764652
|
Accepted
| 30
| 7,668
| 209
|
__author__= 'CIPHER'
str = input()
numList = str.split(' ')
numList = [ int(x) for x in numList]
#print(numList)
result = 1
length = 0
for x in numList:
result *= x
length += x
print(result,2*length)
|
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