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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s331353476
|
p02603
|
u253527353
| 2,000
| 1,048,576
|
Wrong Answer
| 133
| 27,228
| 1,366
|
To become a millionaire, M-kun has decided to make money by trading in the next N days. Currently, he has 1000 yen and no stocks - only one kind of stock is issued in the country where he lives. He is famous across the country for his ability to foresee the future. He already knows that the price of one stock in the next N days will be as follows: * A_1 yen on the 1-st day, A_2 yen on the 2-nd day, ..., A_N yen on the N-th day. In the i-th day, M-kun can make the following trade **any number of times** (possibly zero), **within the amount of money and stocks that he has at the time**. * Buy stock: Pay A_i yen and receive one stock. * Sell stock: Sell one stock for A_i yen. What is the maximum possible amount of money that M-kun can have in the end by trading optimally?
|
import numpy as np
import itertools
#from scipy import signal
n = int(input())
ajs = input().split()
ajs = [int(aj) for aj in ajs]
ajs = np.array(ajs)
#actions = itertools.product([-1, 0, 1], repeat=n)
#for action in actions:
ajs_delta = ajs[1:] - ajs[:-1]
print(ajs_delta)
for i in range(1, ajs_delta.shape[0]):
if ajs_delta[i] == 0:
ajs_delta[i] = ajs_delta[i-1]
print(ajs_delta)
kyokuchi = {'lower':[], 'upper':[]}
first_value = 0
for aj_delta in ajs_delta:
if aj_delta != 0:
first_value = aj_delta
break
if ajs_delta[0] > 0:
kyokuchi['lower'].append(0)
elif ajs_delta[0] < 0:
kyokuchi['upper'].append(0)
elif ajs_delta[0] == 0 and first_value > 0:
kyokuchi['lower'].append(0)
for i in range(0, ajs_delta.shape[0]-1):
if ajs_delta[i] * ajs_delta[i+1] < 0:
if ajs_delta[i] > 0:
kyokuchi['upper'].append(i+1)
else:
kyokuchi['lower'].append(i+1)
if ajs_delta[-1] > 0:
kyokuchi['upper'].append(n-1)
else:
kyokuchi['lower'].append(n-1)
kyokuchi_s = sorted(kyokuchi['upper'] + kyokuchi['lower'])
current_money = 1000
pos = 0
for idx in kyokuchi_s:
if pos >= 0:
stock = int(current_money / ajs[idx])
current_money = current_money % ajs[idx]
pos = -1
else:
current_money += stock * ajs[idx]
stock = 0
pos = 1
print(current_money)
|
s002998081
|
Accepted
| 119
| 26,988
| 1,348
|
import numpy as np
import itertools
#from scipy import signal
n = int(input())
ajs = input().split()
ajs = [int(aj) for aj in ajs]
ajs = np.array(ajs)
#actions = itertools.product([-1, 0, 1], repeat=n)
#for action in actions:
ajs_delta = ajs[1:] - ajs[:-1]
for i in range(1, ajs_delta.shape[0]):
if ajs_delta[i] == 0:
ajs_delta[i] = ajs_delta[i-1]
kyokuchi = {'lower':[], 'upper':[]}
first_value = 0
for aj_delta in ajs_delta:
if aj_delta != 0:
first_value = aj_delta
break
if ajs_delta[0] > 0:
kyokuchi['lower'].append(0)
elif ajs_delta[0] < 0:
#kyokuchi['upper'].append(0)
pass
elif ajs_delta[0] == 0 and first_value > 0:
kyokuchi['lower'].append(0)
for i in range(0, ajs_delta.shape[0]-1):
if ajs_delta[i] * ajs_delta[i+1] < 0:
if ajs_delta[i] > 0:
kyokuchi['upper'].append(i+1)
else:
kyokuchi['lower'].append(i+1)
if ajs_delta[-1] > 0:
kyokuchi['upper'].append(n-1)
else:
#kyokuchi['lower'].append(n-1)
pass
kyokuchi_s = sorted(kyokuchi['upper'] + kyokuchi['lower'])
current_money = 1000
pos = 0
for idx in kyokuchi_s:
if pos >= 0:
stock = int(current_money / ajs[idx])
current_money = current_money % ajs[idx]
pos = -1
else:
current_money += stock * ajs[idx]
stock = 0
pos = 1
print(current_money)
|
s586726982
|
p03386
|
u282228874
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 207
|
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())
if a+k >= b:
ans = [i for i in range(a,b+1)]
else:
ans = []
for i in range(k):
ans.append(a+i)
ans.append(b-i)
ans = sorted(set(ans))
print(ans)
|
s217277186
|
Accepted
| 17
| 3,060
| 223
|
a,b,k=map(int,input().split())
if a+k >= b:
ans = [i for i in range(a,b+1)]
else:
ans = []
for i in range(k):
ans.append(a+i)
ans.append(b-i)
ans = sorted(set(ans))
for x in ans:
print(x)
|
s572193771
|
p03543
|
u173329233
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 545
|
We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**?
|
N = input()
num_list = []
num_list.append(N[:1])
num_list.append(N[1:2])
num_list.append(N[2:3])
num_list.append(N[3:4])
n_1 = num_list[1:4]
n_2 = num_list[2:4]
same_list_1 = [0]
for m in n_1:
if num_list[0] == m:
same_list_1.append(m)
same_list_2 = [0]
for n in n_2:
if num_list[1] == n:
same_list_2.append(n)
len_1 = len(same_list_1)
len_2 = len(same_list_2)
if len_1 == 3:
print('yes')
elif len_2 == 3:
print('yes')
elif len_1 == 4:
print('yes')
elif len_2 == 4:
print('yes')
else:
print('no')
|
s828996819
|
Accepted
| 18
| 2,940
| 100
|
n = input()
if n[0] == n[1] == n[2] or n[1] == n[2] == n[3]:
print('Yes')
else:
print('No')
|
s717928540
|
p03643
|
u304058693
| 2,000
| 262,144
|
Wrong Answer
| 29
| 9,112
| 287
|
This contest, _AtCoder Beginner Contest_ , is abbreviated as _ABC_. When we refer to a specific round of ABC, a three-digit number is appended after ABC. For example, ABC680 is the 680th round of ABC. What is the abbreviation for the N-th round of ABC? Write a program to output the answer.
|
n = int(input())
ans = []
for i in range(1, n + 1):
x = i
for j in range(100):
if x % 2 == 0:
x = int(x // 2)
#ans.append(j)
else:
#ans.append(j)
break
ans.append(j)
#print(ans)
print(ans.index(max(ans)) + 1)
|
s823035961
|
Accepted
| 27
| 8,972
| 39
|
n = int(input())
print("ABC" + str(n))
|
s543058822
|
p02389
|
u892219101
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,604
| 56
|
Write a program which calculates the area and perimeter of a given rectangle.
|
a,b=map(int,input("input:").split())
print(a*b,a+a+b+b)
|
s079089513
|
Accepted
| 40
| 7,520
| 47
|
a,b=map(int,input().split())
print(a*b,a+a+b+b)
|
s671831714
|
p03693
|
u094191970
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 63
|
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
r,g,b=input().split()
print('Yes' if int(r+g+b)%4==0 else 'No')
|
s395874789
|
Accepted
| 27
| 9,016
| 71
|
r,g,b=input().split()
num=int(r+g+b)
print('YES' if num%4==0 else 'NO')
|
s337833855
|
p03545
|
u535171899
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 664
|
Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.
|
a,b,c,d = map(int,list(input()))
sum_abcd=a+b+c+d
gap = (sum_abcd-7)/2
ans = [a,'+',b,'+',c,'+',d]
if gap==0:
print(''.join(map(str,ans)))
elif gap==(b+c+d):
for i in range(1,len(ans),2):
ans[i]='-'
print(''.join(map(str,ans)))
elif gap in [b,c,d]:
for i in range(2,len(ans),2):
if ans[i]==gap:
ans[i-1]='-'
break
print(''.join(map(str,ans)))
else:
for i in range(2,len(ans)-1,2):
for j in range(i+2,len(ans),2):
if ans[i]+ans[j]==gap:
ans[i-1]='-'
ans[j-1]='-'
break
print(''.join(map(str,ans)))
|
s907368910
|
Accepted
| 18
| 3,064
| 343
|
s = list(map(int,list(input())))
a = s.pop(0)
bit_n = 3
for i in range(2**bit_n):
sign = ['+']*3
tmp = a
for j in range(bit_n):
if ((i>>j)&1):
tmp+=s[j]
else:
tmp-=s[j]
sign[j]='-'
if tmp==7:
print("{0}{1[0]}{2[0]}{1[1]}{2[1]}{1[2]}{2[2]}=7".format(a,sign,s));exit()
|
s125183771
|
p03448
|
u091051505
| 2,000
| 262,144
|
Wrong Answer
| 71
| 8,276
| 253
|
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())
x = int(input())
num = []
for a_ in range(a+1):
for b_ in range(b+1):
for c_ in range(c+1):
cal = a_ * 500 + b_ * 100 + c_ * 50
num.append(cal)
print(num.count(6000))
|
s935499426
|
Accepted
| 67
| 8,276
| 250
|
a = int(input())
b = int(input())
c = int(input())
x = int(input())
num = []
for a_ in range(a+1):
for b_ in range(b+1):
for c_ in range(c+1):
cal = a_ * 500 + b_ * 100 + c_ * 50
num.append(cal)
print(num.count(x))
|
s717533031
|
p03140
|
u418527037
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 271
|
You are given three strings A, B and C. Each of these is a string of length N consisting of lowercase English letters. Our objective is to make all these three strings equal. For that, you can repeatedly perform the following operation: * Operation: Choose one of the strings A, B and C, and specify an integer i between 1 and N (inclusive). Change the i-th character from the beginning of the chosen string to some other lowercase English letter. What is the minimum number of operations required to achieve the objective?
|
N = int(input())
A = input()
B = input()
C = input()
a = list(A)
b = list(B)
c = list(C)
ans = 0
for i in range(N):
if a[i] == b[i] or a[i] == c[i] or b[i] == c[i]:
ans += 1
elif a[i] == b[i] == c[i]:
pass
else:
ans += 2
print(ans)
|
s843068152
|
Accepted
| 17
| 3,064
| 304
|
N = int(input())
A = input()
B = input()
C = input()
a = list(A)
b = list(B)
c = list(C)
ans = 0
for i,j in enumerate(a):
if j == b[i] or j == c[i] or b[i] == c[i]:
if j == b[i] == c[i]:
continue
ans += 1
elif a[i] != b[i] != c[i]:
ans += 2
print(ans)
|
s684037119
|
p03386
|
u362560965
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 124
|
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 = (int(i) for i in input().split())
for i in range(A, A+K):
print(i)
for j in range(B, B-K, -1):
print(j)
|
s707500016
|
Accepted
| 17
| 3,060
| 262
|
A, B, K = (int(i) for i in input().split())
ans = []
for i in range(A, A+K):
ans.append(i)
if i == B:
break
for j in range(B, B-K, -1):
ans.append(j)
if j == A:
break
ans = list(set(ans))
ans.sort()
for x in ans:
print(x)
|
s102650739
|
p02678
|
u486074261
| 2,000
| 1,048,576
|
Wrong Answer
| 2,206
| 63,028
| 1,050
|
There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.
|
from sys import stdin
from collections import deque
def main():
N, M = [int(x) for x in stdin.readline().rstrip().split()]
ipt = [[int(y) for y in x.rstrip().split()] for x in stdin.readlines()]
MOD = 100100100100100100
pl = [0] * (N + 1)
for a, b in ipt:
if pl[a] == 0:
pl[a] = [b]
else:
pl[a].append(b)
if pl[b] == 0:
pl[b] = [a]
else:
pl[b].append(a)
print(pl)
que = deque([1])
deep = [MOD] * (N + 1)
deep[1] = 0
deep[0] = 0
arrow = [0] * (N + 1)
cnt = 1
while not not que:
q = que.popleft()
tmp = False
for i in pl[q]:
if deep[i] != MOD:
continue
deep[i] = deep[q] + 1
arrow[i] = q
tmp = True
cnt += 1
if tmp:
que.extend(pl[q])
if N != cnt:
print("No")
else:
print("Yes")
for i in arrow[2:]:
print(i)
if __name__ == "__main__":
main()
|
s800259673
|
Accepted
| 593
| 58,204
| 972
|
from sys import stdin
from collections import deque
def main():
N, M = [int(x) for x in stdin.readline().rstrip().split()]
ipt = [[int(y) for y in x.rstrip().split()] for x in stdin.readlines()]
MOD = 100100100100100100
pl = [0] * (N + 1)
for a, b in ipt:
if pl[a] == 0:
pl[a] = [b]
else:
pl[a].append(b)
if pl[b] == 0:
pl[b] = [a]
else:
pl[b].append(a)
que = deque([1])
deep = [MOD] * (N + 1)
deep[1] = 0
deep[0] = 0
arrow = [0] * (N + 1)
cnt = 1
while not not que:
q = que.popleft()
for i in pl[q]:
if deep[i] != MOD:
continue
deep[i] = deep[q] + 1
arrow[i] = q
cnt += 1
que.append(i)
if N != cnt:
print("No")
else:
print("Yes")
for i in arrow[2:]:
print(i)
if __name__ == "__main__":
main()
|
s626996132
|
p03416
|
u425366405
| 2,000
| 262,144
|
Wrong Answer
| 20
| 3,064
| 486
|
Find the number of _palindromic numbers_ among the integers between A and B (inclusive). Here, a palindromic number is a positive integer whose string representation in base 10 (without leading zeros) reads the same forward and backward.
|
# -*- coding: utf-8 -*-
# n1, n2 = map(int, input().split())
n1, n2 = map(int, "31415 92653".split())
l1 = [int(x) for x in list(str(n1))]
l2 = [int(x) for x in list(str(n2))]
n1Low = n1%100
n2Low = n2%100
n1HighRev = int(str(l1[1])+str(l1[0]))
n2HighRev = int(str(l2[1])+str(l2[0]))
ans = 0
if n1Low - n1HighRev <= 0:
ans += 1
elif n2Low - n2HighRev >= 0:
ans += 1
ans += 9 - l1[2] + l2[2]
ans += (l2[0] - l1[0] - 1) * 100
ans += (9 - l1[1] + l2[1]) * 10
print(ans)
|
s892624101
|
Accepted
| 17
| 3,064
| 604
|
# -*- coding: utf-8 -*-
n1, n2 = map(int, input().split())
l1 = [int(x) for x in list(str(n1))]
l2 = [int(x) for x in list(str(n2))]
n1Low = n1%100
n2Low = n2%100
n1HighRev = int(str(l1[1])+str(l1[0]))
n2HighRev = int(str(l2[1])+str(l2[0]))
ans = 0
if n1Low <= n1HighRev:
ans += 1
if n2Low >= n2HighRev:
ans += 1
if l1[0] != l2[0]:
ans += 9 - l1[2] + l2[2]
ans += (l2[0] - l1[0] - 1) * 100
ans += (9 - l1[1] + l2[1]) * 10
else:
if l1[1] != l2[1]:
ans += 9 - l1[2] + l2[2]
ans += (l2[1] - l1[1] - 1) * 10
else:
ans += l2[2] - l1[2]-1
print(ans)
|
s141605966
|
p03994
|
u065089468
| 2,000
| 262,144
|
Wrong Answer
| 168
| 7,060
| 292
|
Mr. Takahashi has a string s consisting of lowercase English letters. He repeats the following operation on s exactly K times. * Choose an arbitrary letter on s and change that letter to the next alphabet. Note that the next letter of `z` is `a`. For example, if you perform an operation for the second letter on `aaz`, `aaz` becomes `abz`. If you then perform an operation for the third letter on `abz`, `abz` becomes `aba`. Mr. Takahashi wants to have the lexicographically smallest string after performing exactly K operations on s. Find the such string.
|
import string
import sys
s = input()
k = int(input())
a = [ord(_) for _ in s]
for i in range(len(s)-1):
df = ord('z') - a[i] + 1
if k >= df:
a[i] = 97
k -= df
if k == 0:
break
if k != 0:
k = k % 26
a[len(a)-1] = 97+k
[print(chr(_), end="") for _ in a]
|
s418896506
|
Accepted
| 189
| 7,024
| 385
|
import string
import sys
s = input()
k = int(input())
a = [ord(_) for _ in s]
for i in range(len(s)-1):
if a[i] != ord('a'):
df = ord('z') - a[i] + 1
if k >= df:
a[i] = ord('a')
k -= df
if k == 0: break
if k != 0:
k = k % 26
a[len(a)-1] += k - 26 if ord('z') < k + a[len(a)-1] else k
[print(chr(_), end="") for _ in a]
print("")
|
s608580382
|
p03623
|
u033524082
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,316
| 65
|
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x,a,b=map(int,input().split())
print("A" if (a-x)<(b-x) else "B")
|
s708591251
|
Accepted
| 20
| 3,316
| 75
|
x,a,b=map(int,input().split())
print("A" if abs((a-x))<abs((b-x)) else "B")
|
s922554169
|
p02796
|
u782930273
| 2,000
| 1,048,576
|
Wrong Answer
| 660
| 37,116
| 526
|
In a factory, there are N robots placed on a number line. Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative. We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect. Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints. Find the maximum number of robots that we can keep.
|
N = int(input())
X = [0] * N
for i in range(N):
x, l = map(int, input().split())
X[i] = x, l, x - l, x + l
X.sort(key=lambda x: x[1], reverse=True)
X = list(enumerate(X))
removed = [False] * N
X2 = sorted(X, key=lambda x: x[1][2])
for i, t in X:
x, l, rl, rr = t
if i <= N - 2 and rr <= X2[i + 1][1][2] and not removed[X2[i + 1][0]]:
removed[i] = True
if i >= 1 and rl >= X2[i - 1][1][3] and not removed[X2[i - 1][0]]:
removed[i] = True
print(N - sum(removed))
|
s725240662
|
Accepted
| 454
| 26,152
| 285
|
N = int(input())
X = [0] * N
for i in range(N):
x, l = map(int, input().split())
X[i] = x, l, x - l, x + l
X.sort(key=lambda x: x[3])
right = X[0][3]
count = 1
for i in range(1, N):
if right <= X[i][2]:
count += 1
right = X[i][3]
print(count)
|
s634270333
|
p03416
|
u057079894
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 237
|
Find the number of _palindromic numbers_ among the integers between A and B (inclusive). Here, a palindromic number is a positive integer whose string representation in base 10 (without leading zeros) reads the same forward and backward.
|
def Is_cyc(x):
s = str(x)
for i in range(len(s)):
if s[i] != s[-i-1]:
return False
return True
def main():
a,b = map(int,input().split())
ans = 0
for i in range(a,b+1):
if Is_cyc(i):
ans += 1
print(ans)
|
s509851041
|
Accepted
| 86
| 3,060
| 284
|
def Is_cyc(x):
s = str(x)
for i in range(len(s)):
if s[i] != s[-i-1]:
return False
return True
def main():
a,b = map(int,input().split())
ans = 0
for i in range(a,b+1):
if Is_cyc(i):
ans += 1
print(ans)
return
if __name__ == "__main__":
main()
|
s252908583
|
p03587
|
u268792407
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 66
|
Snuke prepared 6 problems for a upcoming programming contest. For each of those problems, Rng judged whether it can be used in the contest or not. You are given a string S of length 6. If the i-th character of s is `1`, it means that the i-th problem prepared by Snuke is accepted to be used; `0` means that the problem is not accepted. How many problems prepared by Snuke are accepted to be used in the contest?
|
s=input()
a=0
for i in range(6):
if s[i]=="1":
a+-1
print(a)
|
s524964438
|
Accepted
| 17
| 2,940
| 67
|
s=input()
a=0
for i in range(6):
if s[i]=="1":
a+=1
print(a)
|
s514385277
|
p03474
|
u179111046
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 199
|
The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.
|
a, b = map(int, input().split())
s = input()
print(s[a])
if(s[a] == "-"):
try:
int(s[:a])
int(s[a + 1:])
print("Yes")
except:
print("No")
else:
print("No")
|
s210302321
|
Accepted
| 17
| 2,940
| 187
|
a, b = map(int, input().split())
s = input()
if(s[a] == "-"):
try:
int(s[:a])
int(s[a + 1:])
print("Yes")
except:
print("No")
else:
print("No")
|
s639193802
|
p03351
|
u670180528
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 91
|
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());A=abs;print("YNeos"[min(A(a-c),max(A(a-b),A(b-c)))<=d::2])
|
s140552263
|
Accepted
| 18
| 2,940
| 90
|
a,b,c,d=map(int,input().split());A=abs;print("YNeos"[min(A(a-c),max(A(a-b),A(b-c)))>d::2])
|
s816057302
|
p02257
|
u148477094
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,656
| 303
|
A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. For example, the first four prime numbers are: 2, 3, 5 and 7. Write a program which reads a list of _N_ integers and prints the number of prime numbers in the list.
|
import math
n=int(input("n:"))
R=[0]*n
sum=0
for i in range(n):
R[i]=int(input(""))
for i in range(n):
s=math.sqrt(R[i])
s=int(s)+1
for j in range(2,s+1):
if R[i]==j:
sum=sum+1
if R[i]%j==0:
break
if j==s:
sum=sum+1
print(sum)
|
s026964121
|
Accepted
| 1,740
| 6,044
| 329
|
import math
n=int(input())
R=[0]*n
sum=0
for i in range(n):
R[i]=int(input())
for i in range(n):
if R[i]<0:
continue
s=math.sqrt(R[i])
s=int(s)+1
for j in range(2,s+1):
if R[i]==j:
sum=sum+1
if R[i]%j==0:
break
if j==s:
sum=sum+1
print(sum)
|
s778489468
|
p03129
|
u782098901
| 2,000
| 1,048,576
|
Wrong Answer
| 20
| 3,316
| 77
|
Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.
|
N, K = map(int, input().split())
print("Yes" if (N + 1) // 2 >= K else "No")
|
s072782470
|
Accepted
| 17
| 2,940
| 77
|
N, K = map(int, input().split())
print("YES" if (N + 1) // 2 >= K else "NO")
|
s483319069
|
p02613
|
u221280246
| 2,000
| 1,048,576
|
Wrong Answer
| 149
| 16,204
| 219
|
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.
|
N=int(input())
S=[]
for i in range(N):
S.append(input())
print('AC x {}'.format(S.count('AC')))
print('WA x {}'.format(S.count('WA')))
print('TLE x {}'.format(S.count('TLE')))
print('AC x {}'.format(S.count('RE')))
|
s815650731
|
Accepted
| 146
| 16,232
| 219
|
N=int(input())
S=[]
for i in range(N):
S.append(input())
print('AC x {}'.format(S.count('AC')))
print('WA x {}'.format(S.count('WA')))
print('TLE x {}'.format(S.count('TLE')))
print('RE x {}'.format(S.count('RE')))
|
s372323463
|
p02612
|
u770558697
| 2,000
| 1,048,576
|
Wrong Answer
| 31
| 9,160
| 89
|
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.
|
n = int(input())
if n%1000:
print(0)
else:
oturi = (int(n/1000)+1)*1000-n
print(oturi)
|
s203946208
|
Accepted
| 29
| 9,168
| 95
|
n = int(input())
if n%1000 == 0:
print(0)
else:
oturi = (int(n/1000)+1)*1000-n
print(oturi)
|
s500605938
|
p03485
|
u063614215
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 88
|
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
a,b = map(int, input().split())
ans = a+b
if ans%2==0:
print(ans)
else:
print(ans+1)
|
s646486208
|
Accepted
| 17
| 2,940
| 94
|
a,b = map(int, input().split())
ans = a+b
if ans%2==0:
print(ans//2)
else:
print(ans//2+1)
|
s635523227
|
p02392
|
u098047375
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,580
| 67
|
Write a program which reads three integers a, b and c, and prints "Yes" if a < b < c, otherwise "No".
|
A = map(int, input().split())
set_A = set(A)
print(sorted(set_A))
|
s922514406
|
Accepted
| 20
| 5,584
| 96
|
a, b, c = map(int, input().split())
if a < b and b < c:
print('Yes')
else:
print('No')
|
s561476803
|
p02646
|
u667687319
| 2,000
| 1,048,576
|
Wrong Answer
| 24
| 9,168
| 146
|
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())
if a + v * t > b + w * t:
print('Yes')
else:
print('No')
|
s351834243
|
Accepted
| 22
| 9,176
| 263
|
a,v = map(int, input().split())
b,w = map(int, input().split())
t = int(input())
if a <= b:
if a + v * t >= b + w * t:
print('YES')
else:
print('NO')
else:
if a - v * t <= b - w * t:
print('YES')
else:
print('NO')
|
s118858342
|
p03351
|
u126823513
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 263
|
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.
|
int_a, int_b, int_c, int_d = map(int, input().split())
result = False
if abs(int_a - int_c) <= int_d:
result = True
elif abs(int_a - int_b) <= int_d and abs(int_b - int_c) <= int_d:
result = True
if result:
print('YES')
else:
print('No')
|
s191525066
|
Accepted
| 17
| 2,940
| 258
|
int_a, int_b, int_c, int_d = map(int, input().split())
result = False
if abs(int_a - int_c) <= int_d:
result = True
elif abs(int_a - int_b) <= int_d and abs(int_b - int_c) <= int_d:
result = True
if result:
print('Yes')
else:
print('No')
|
s915121057
|
p02843
|
u239653493
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 126
|
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())
if x>=2100:
print("YES")
else:
if x%100<=5*(x//100):
print("YES")
else:
print("NO")
|
s019959088
|
Accepted
| 17
| 2,940
| 127
|
x=int(input())
if x>=2000:
print(1)
else:
if (x//100)*100<=x<=(x//100)*105:
print(1)
else:
print(0)
|
s442974590
|
p03992
|
u112002050
| 2,000
| 262,144
|
Wrong Answer
| 22
| 3,192
| 41
|
This contest is `CODE FESTIVAL`. However, Mr. Takahashi always writes it `CODEFESTIVAL`, omitting the single space between `CODE` and `FESTIVAL`. So he has decided to make a program that puts the single space he omitted. You are given a string s with 12 letters. Output the string putting a single space between the first 4 letters and last 8 letters in the string s.
|
s = input()
print(s[0:3] + " " + s[4:-1])
|
s134206441
|
Accepted
| 23
| 3,064
| 39
|
s = input()
print(s[0:4] + " " + s[4:])
|
s328866123
|
p03721
|
u945181840
| 2,000
| 262,144
|
Wrong Answer
| 318
| 25,972
| 315
|
There is an empty array. The following N operations will be performed to insert integers into the array. In the i-th operation (1≤i≤N), b_i copies of an integer a_i are inserted into the array. Find the K-th smallest integer in the array after the N operations. For example, the 4-th smallest integer in the array \\{1,2,2,3,3,3\\} is 3.
|
import sys
input = sys.stdin.readline
N, K = map(int, input().split())
a = [0] * N
b = [0] * N
for i in range(N):
a[i], b[i] = map(int, input().split())
a, b = zip(*sorted(zip(a, b)))
count = 0
for i in range(N):
K -= b[i]
if K <= 0:
print(a[count])
exit()
else:
continue
|
s692321328
|
Accepted
| 297
| 25,952
| 274
|
import sys
input = sys.stdin.readline
N, K = map(int, input().split())
a = [0] * N
b = [0] * N
for i in range(N):
a[i], b[i] = map(int, input().split())
a, b = zip(*sorted(zip(a, b)))
for i in range(N):
K -= b[i]
if K <= 0:
print(a[i])
exit()
|
s394274776
|
p03486
|
u846150137
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 250
|
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
|
a=input()
b=input()
aa=[]
for i in a:
aa.append(i)
aa.sort()
aaa="".join(aa)
bb=[]
for i in b:
bb.append(i)
bb.sort(reverse=True)
bbb="".join(bb)
ccc=[aaa,bbb]
ccc.sort()
print(a,b,aaa,bbb,ccc)
if bbb==ccc[0]:
print('No')
else:
print('Yes')
|
s216389118
|
Accepted
| 17
| 3,060
| 244
|
a=input().strip()
b=input().strip()
aa=[]
for i in a:
aa.append(i)
aa.sort()
aaa="".join(aa)
bb=[]
for i in b:
bb.append(i)
bb.sort(reverse=True)
bbb="".join(bb)
ccc=[aaa,bbb]
ccc.sort()
if bbb==ccc[0]:
print('No')
else:
print('Yes')
|
s003617238
|
p03623
|
u225388820
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 60
|
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x,a,b=map(int,input().split())
print(min(abs(x-a),abs(x-b)))
|
s606747674
|
Accepted
| 17
| 2,940
| 85
|
x,a,b=map(int,input().split())
if abs(x-a)<abs(x-b):
print('A')
else:
print('B')
|
s771255846
|
p03610
|
u687574784
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,192
| 20
|
You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.
|
print(input()[1::2])
|
s551731643
|
Accepted
| 17
| 3,192
| 19
|
print(input()[::2])
|
s263591313
|
p02272
|
u481571686
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,608
| 512
|
Write a program of a Merge Sort algorithm implemented by the following pseudocode. You should also report the number of comparisons in the Merge function. Merge(A, left, mid, right) n1 = mid - left; n2 = right - mid; create array L[0...n1], R[0...n2] for i = 0 to n1-1 do L[i] = A[left + i] for i = 0 to n2-1 do R[i] = A[mid + i] L[n1] = SENTINEL R[n2] = SENTINEL i = 0; j = 0; for k = left to right-1 if L[i] <= R[j] then A[k] = L[i] i = i + 1 else A[k] = R[j] j = j + 1 Merge-Sort(A, left, right){ if left+1 < right then mid = (left + right)/2; call Merge-Sort(A, left, mid) call Merge-Sort(A, mid, right) call Merge(A, left, mid, right)
|
def merge(A,left,center,right):
global cnt
l=A[left:center]
r=A[center:right]
l.append(float("inf"))
r.append(float("inf"))
j=0
k=0
for i in range(left,right):
cnt+=1
if l[j]<=r[k]:
A[i]=l[j]
j+=1
else:
A[i]=r[j]
k+=1
def mergesort(A,left,right):
if left+1<right:
center=(left+right)//2
mergesort(A,left,center)
mergesort(A,center,right)
merge(A,left,center,right)
n=int(input())
A=list(map(int,input().split()))
cnt=0
mergesort(A,0,n)
print(" ".join(map(str,A)))
print(cnt)
|
s997051593
|
Accepted
| 4,250
| 69,564
| 668
|
cnt=0
def merge(a,left,mid,right):
n1=mid-left
n2=right-mid
l=a[left:mid]
r=a[mid:right]
l.append(float("inf"))
r.append(float("inf"))
i=0
j=0
for k in range(left,right):
global cnt
cnt+=1
if(l[i] <= r[j]):
a[k] = l[i]
i+=1
else:
a[k]=r[j]
j+=1
def merge_sort(a,left,right):
if(left+1<right):
mid=(left+right)//2
merge_sort(a,left,mid)
merge_sort(a,mid,right)
merge(a,left,mid,right)
n=int(input())
s=input()
a=list(map(int,s.split()))
merge_sort(a,0,n)
a=map(str,a)
print(" ".join(list(a)))
print(cnt)
|
s846235191
|
p03407
|
u117193815
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 79
|
An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.
|
a,b,c = map(int ,input().split())
if a+b<=c:
print("Yes")
else:
print("No")
|
s351929738
|
Accepted
| 17
| 2,940
| 80
|
a,b,c = map(int ,input().split())
if a+b>=c:
print("Yes")
else:
print("No")
|
s007965244
|
p02603
|
u766646838
| 2,000
| 1,048,576
|
Wrong Answer
| 26
| 9,208
| 349
|
To become a millionaire, M-kun has decided to make money by trading in the next N days. Currently, he has 1000 yen and no stocks - only one kind of stock is issued in the country where he lives. He is famous across the country for his ability to foresee the future. He already knows that the price of one stock in the next N days will be as follows: * A_1 yen on the 1-st day, A_2 yen on the 2-nd day, ..., A_N yen on the N-th day. In the i-th day, M-kun can make the following trade **any number of times** (possibly zero), **within the amount of money and stocks that he has at the time**. * Buy stock: Pay A_i yen and receive one stock. * Sell stock: Sell one stock for A_i yen. What is the maximum possible amount of money that M-kun can have in the end by trading optimally?
|
n,*a=map(int,open(0).read().split())
ans=1000
num= 0
for i in range(n):
if i == n-1:
if a[i-1]<a[i]:
ans += a[i]*num
elif a[i]<a[i+1] and a[i+1]<ans:
num = ans//a[i]
ans = ans%a[i]
print(i,num,ans)
elif a[i]<a[i+1] and a[i+1]>ans:
continue
elif a[i] == a[i+1]:
continue
else:
ans += num*a[i]
print(ans)
|
s902039344
|
Accepted
| 35
| 9,184
| 404
|
N = int(input())
a = list(map(int,input().split()))
num = 1000
flag = True
flag2 = True
for i in range(1,N):
if a[i-1]<a[i] and flag == True:
mi = a[i-1]
ma = a[i]
flag = False
elif a[i-1]<a[i] and flag == False:
ma = a[i]
elif a[i-1]>a[i] and flag == False:
num = (num//mi)*ma+(num%mi)
flag = True
if i == N-1 and flag == False:
num = (num//mi)*ma+(num%mi)
print(num)
|
s933540168
|
p02928
|
u518064858
| 2,000
| 1,048,576
|
Wrong Answer
| 1,227
| 3,388
| 409
|
We have a sequence of N integers A~=~A_0,~A_1,~...,~A_{N - 1}. Let B be a sequence of K \times N integers obtained by concatenating K copies of A. For example, if A~=~1,~3,~2 and K~=~2, B~=~1,~3,~2,~1,~3,~2. Find the inversion number of B, modulo 10^9 + 7. Here the inversion number of B is defined as the number of ordered pairs of integers (i,~j)~(0 \leq i < j \leq K \times N - 1) such that B_i > B_j.
|
n,k=map(int,input().split())
M=10**9+7
a=list(map(int,input().split()))
cnt=[[0 for i in range(2)] for j in range(n)]
for i in range(n):
for j in range(n):
if j<i and a[j]<a[i]:
cnt[i][0]+=1
elif i<j and a[j]<a[i]:
cnt[i][1]+=1
sum=0
for i in range(n):
s=(1/2)*(k**2)*(cnt[i][0]+cnt[i][1])+(1/2)*(cnt[i][1]-cnt[i][0])*k
s%=M
sum+=s
sum%=M
print(sum)
|
s692011845
|
Accepted
| 1,432
| 3,416
| 391
|
n,k=map(int,input().split())
M=10**9+7
a=list(map(int,input().split()))
cnt=[[0 for i in range(2)] for j in range(n)]
for i in range(n):
for j in range(n):
if j<i and a[j]<a[i]:
cnt[i][0]+=1
elif i<j and a[j]<a[i]:
cnt[i][1]+=1
sum=0
for i in range(n):
s=((k*(cnt[i][0]+cnt[i][1])+(cnt[i][1]-cnt[i][0]))*k)//2
sum+=s
sum%=M
print(sum)
|
s408720416
|
p02558
|
u179376941
| 5,000
| 1,048,576
|
Wrong Answer
| 809
| 71,540
| 267
|
You are given an undirected graph with N vertices and 0 edges. Process Q queries of the following types. * `0 u v`: Add an edge (u, v). * `1 u v`: Print 1 if u and v are in the same connected component, 0 otherwise.
|
n, q = map(int, input().split())
nei = {}
for i in range(n):
nei[i] = set()
for _ in range(q):
t, u, v = map(int, input().split())
if t == 0:
nei[u].add(v)
nei[v].add(u)
else:
if v in nei[u] and u in nei[v]:
print(1)
else:
print(0)
|
s216880476
|
Accepted
| 755
| 11,484
| 795
|
class UnionFind():
def __init__(self, n):
self.n = n
self.parents = [-1] * n
def find(self, x):
if self.parents[x] < 0:
return x
else:
self.parents[x] = self.find(self.parents[x])
return self.parents[x]
def union(self, x, y):
x = self.find(x)
y = self.find(y)
if x == y:
return
if self.parents[x] > self.parents[y]:
x, y = y, x
self.parents[x] += self.parents[y]
self.parents[y] = x
def same(self, x, y):
return self.find(x) == self.find(y)
n, q = map(int, input().split())
un = UnionFind(n)
for _ in range(q):
t, u, v = map(int, input().split())
if t == 0:
un.union(u, v)
else:
print(1 if un.same(u, v) else 0)
|
s377111210
|
p03436
|
u919730120
| 2,000
| 262,144
|
Wrong Answer
| 2,108
| 58,988
| 648
|
We have an H \times W grid whose squares are painted black or white. The square at the i-th row from the top and the j-th column from the left is denoted as (i, j). Snuke would like to play the following game on this grid. At the beginning of the game, there is a character called Kenus at square (1, 1). The player repeatedly moves Kenus up, down, left or right by one square. The game is completed when Kenus reaches square (H, W) passing only white squares. Before Snuke starts the game, he can change the color of some of the white squares to black. However, he cannot change the color of square (1, 1) and (H, W). Also, changes of color must all be carried out before the beginning of the game. When the game is completed, Snuke's score will be the number of times he changed the color of a square before the beginning of the game. Find the maximum possible score that Snuke can achieve, or print -1 if the game cannot be completed, that is, Kenus can never reach square (H, W) regardless of how Snuke changes the color of the squares. The color of the squares are given to you as characters s_{i, j}. If square (i, j) is initially painted by white, s_{i, j} is `.`; if square (i, j) is initially painted by black, s_{i, j} is `#`.
|
from collections import deque
h,w=map(int,input().split())
s=[input() for _ in range(h)]
d=[[1,0],[-1,0],[0,1],[0,-1]]
bcnt=0
for i in range(h):
bcnt+=s[i].count('#')
dist=[[1000]*w for _ in range(h)]
chk=[[0]*w for _ in range(h)]
r=deque()
r.append([0,0])
dist[0][0]=0
while r:
x,y=r.popleft()
chk[y][x]=1
for dx,dy in d:
nx=x+dx
ny=y+dy
if nx>=0 and nx<=w-1 and ny>=0 and ny<=h-1:
if chk[ny][nx]==1:
continue
if s[ny][nx]=='.':
dist[ny][nx]=min(dist[ny][nx],dist[y][x]+1)
r.append([nx,ny])
print(dist)
print(h*w-dist[h-1][w-1]-1-bcnt)
|
s535305220
|
Accepted
| 43
| 9,424
| 787
|
from collections import deque
h,w=map(int,input().split())
dist=[[-1]*(w+2) for _ in range(h+2)]
dist[1][1]=1
maze=[]
maze.append(['#' for _ in range(w+2)])
for i in range(h):
maze.append(['#']+list(input())+['#'])
maze.append(['#' for _ in range(w+2)])
cnt=0
for i in range(h+2):
for j in range(w+2):
if maze[i][j]=='#':
cnt+=1
d=[[-1,0],[1,0],[0,-1],[0,1]]
q=deque()
q.append([1,1])
while q:
x,y=q.popleft()
for dx,dy in d:
if (0<=x+dx<=h+1) and (0<y+dy<w+1):
if maze[x+dx][y+dy]==".":
if dist[x+dx][y+dy]==-1 or dist[x+dx][y+dy]>dist[x][y]+1:
q.append([x+dx,y+dy])
dist[x+dx][y+dy]=dist[x][y]+1
if dist[h][w]==-1:
print(-1)
else:
print((h+2)*(w+2)-cnt-dist[h][w])
|
s260556871
|
p03449
|
u115682115
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,060
| 217
|
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())
a = [list(map(int,input().split())) for i in range(2)]
saidaiti = []
for i in range(N):
saidaiti.append(sum(a[0][:i+1]+a[1][i:]))
print(a[0][:i+1])
print(a[1][i:])
print(max(saidaiti))
|
s943783119
|
Accepted
| 18
| 2,940
| 174
|
N = int(input())
a = [list(map(int,input().split())) for i in range(2)]
saidaiti = []
for i in range(N):
saidaiti.append(sum(a[0][:i+1]+a[1][i:]))
print(max(saidaiti))
|
s894675217
|
p02255
|
u400765446
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,596
| 509
|
Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step.
|
def main():
n = int(input())
Mtx = list(map(int, input().split()))
for i in range(1,n):
for j in range(i,0,-1):
if Mtx[j] < Mtx[j-1]:
Mtx[j-1], Mtx[j] = Mtx[j], Mtx[j-1]
# print(i, j, sep=', ')
showMtx(Mtx)
def showMtx(Mtx):
q = len(Mtx)
for i in range(q):
if i != q - 1:
print(Mtx[i], sep=' ', end=' ')
else:
print(Mtx[i])
if __name__ == '__main__':
main()
|
s184205694
|
Accepted
| 20
| 5,600
| 206
|
n = int(input())
A = input().split()
print(' '.join(A))
for i in range(1,n):
for j in range(i,0,-1):
if int(A[j]) < int(A[j-1]):
A[j], A[j-1] = A[j-1], A[j]
print(' '.join(A))
|
s009372417
|
p03486
|
u068844030
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 214
|
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
|
s = input()
t = input()
s_sort = "".join(sorted(s))
t_sort = "".join(sorted(t,reverse=True))
print("yes" if s_sort < t_sort else "No")
|
s917491103
|
Accepted
| 18
| 2,940
| 222
|
##82_B
s = input()
t = input()
s_sort = "".join(sorted(s))
t_sort = "".join(sorted(t,reverse=True))
print("Yes" if s_sort < t_sort else "No")
|
s871917038
|
p03434
|
u416758623
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 240
|
We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.
|
n = int(input())
a = [int(x) for x in input().split()]
sortA = sorted(a, reverse=True)
Alice = 0
Bob = 0
print(sortA)
for i in range(len(a)):
if i % 2 == 0:
Alice+=sortA[i]
else:
Bob+=sortA[i]
print(Alice - Bob)
|
s205055967
|
Accepted
| 30
| 9,184
| 207
|
n = int(input())
l = sorted(list(map(int, input().split())), reverse=True)
Alice = 0
Bob = 0
for i in range(len(l)):
if i % 2 == 0:
Alice += l[i]
else:
Bob += l[i]
print(Alice - Bob)
|
s703950396
|
p04046
|
u798316285
| 2,000
| 262,144
|
Wrong Answer
| 305
| 18,932
| 442
|
We have a large square grid with H rows and W columns. Iroha is now standing in the top-left cell. She will repeat going right or down to the adjacent cell, until she reaches the bottom-right cell. However, she cannot enter the cells in the intersection of the bottom A rows and the leftmost B columns. (That is, there are A×B forbidden cells.) There is no restriction on entering the other cells. Find the number of ways she can travel to the bottom-right cell. Since this number can be extremely large, print the number modulo 10^9+7.
|
N=2*10**5+3
mod=10**9+7
fac=[1]*(N+1)
for i in range(1,N+1):
fac[i]=fac[i-1]*i%mod
inv_fac=[1]*(N+1)
inv_fac[N]=pow(fac[N],mod-2,mod)
for i in range(N-1,0,-1):
inv_fac[i]=inv_fac[i+1]*(i+1)%mod
def nCr(n,r):
if n<=0 or r<0 or r>n:
return 0
return fac[n]*inv_fac[r]%mod*inv_fac[n-r]%mod
h,w,a,b=map(int,input().split())
ans=nCr(h+w-2,h-1)
for i in range(b):
ans=(ans-nCr(h-a+i-1,i)*nCr(a+w-i-2,a-1))%mod
print(ans)
|
s337405092
|
Accepted
| 309
| 18,804
| 440
|
N=2*10**5+3
mod=10**9+7
fac=[1]*(N+1)
for i in range(1,N+1):
fac[i]=fac[i-1]*i%mod
inv_fac=[1]*(N+1)
inv_fac[N]=pow(fac[N],mod-2,mod)
for i in range(N-1,0,-1):
inv_fac[i]=inv_fac[i+1]*(i+1)%mod
def nCr(n,r):
if n<0 or r<0 or r>n:
return 0
return fac[n]*inv_fac[r]%mod*inv_fac[n-r]%mod
h,w,a,b=map(int,input().split())
ans=nCr(h+w-2,h-1)
for i in range(b):
ans=(ans-nCr(h-a+i-1,i)*nCr(a+w-i-2,a-1))%mod
print(ans)
|
s850977103
|
p03501
|
u853900545
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 53
|
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())
print(min((n+a),b))
|
s102794239
|
Accepted
| 17
| 2,940
| 53
|
n,a,b = map(int,input().split())
print(min((n*a),b))
|
s229204625
|
p03385
|
u239375815
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 54
|
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
a = sorted(input())
print("Yes" if a=='abc' else "No")
|
s674493477
|
Accepted
| 17
| 2,940
| 69
|
a = sorted(input())
a = ''.join(a)
print("Yes" if a=='abc' else "No")
|
s008486463
|
p03129
|
u496815777
| 2,000
| 1,048,576
|
Wrong Answer
| 21
| 3,316
| 86
|
Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.
|
n, k = map(int, input().split())
if n < 2 * k - 1:
print("No")
else:
print("Yes")
|
s207767667
|
Accepted
| 17
| 2,940
| 86
|
n, k = map(int, input().split())
if n < 2 * k - 1:
print("NO")
else:
print("YES")
|
s118303797
|
p03352
|
u021548497
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 141
|
You are given a positive integer X. Find the largest _perfect power_ that is at most X. Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2.
|
from math import sqrt
from sys import exit
n = int(input())
for i in range(n,0):
if sqrt(n).is_integer():
print(i)
exit()
|
s877683305
|
Accepted
| 17
| 3,064
| 352
|
from sys import exit
n = int(input())
powlist = set()
k = 2
chance = True
if n == 1:
print(1)
exit()
while chance:
for i in range(1, n):
key = i**k
if key <= n:
powlist.add(key)
else:
k += 1
judge = i
break
if judge == 2:
chance = False
print(max(powlist))
|
s359409584
|
p02612
|
u969848070
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 8,936
| 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)
|
s280525112
|
Accepted
| 29
| 9,068
| 64
|
a = int(input())%1000
if a ==0:
print(0)
else:
print(1000-a)
|
s872949590
|
p02578
|
u667084803
| 2,000
| 1,048,576
|
Wrong Answer
| 88
| 25,220
| 133
|
N persons are standing in a row. The height of the i-th person from the front is A_i. We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person: Condition: Nobody in front of the person is taller than the person. Here, the height of a person includes the stool. Find the minimum total height of the stools needed to meet this goal.
|
N = int(input())
A = map(int, input().split())
curr = 0
ans = 0
for a in A:
if curr < a:
curr = a
ans += a-curr
print(ans)
|
s128164657
|
Accepted
| 96
| 25,072
| 143
|
N = int(input())
A = map(int, input().split())
curr = 0
ans = 0
for a in A:
if curr > a:
ans += curr - a
else:
curr = a
print(ans)
|
s865473733
|
p02399
|
u476441153
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,692
| 81
|
Write a program which reads two integers a and b, and calculates the following values: * a ÷ b: d (in integer) * remainder of a ÷ b: r (in integer) * a ÷ b: f (in real number)
|
a,b = list(map(int, input().split()))
print (int((a/b)) , round(a%b) ,float(a/b))
|
s339796743
|
Accepted
| 20
| 7,684
| 107
|
a,b = list(map(int, input().split()))
d = a // b
r = a % b
f = a / b
print('{0} {1} {2:.5f}'.format(d,r,f))
|
s810334436
|
p02613
|
u515128734
| 2,000
| 1,048,576
|
Wrong Answer
| 144
| 9,212
| 361
|
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.
|
n = int(input())
ct_ac = 0
ct_wa = 0
ct_tle = 0
ct_re = 0
for i in range(n):
s = input()
if s == 'AC':
ct_ac += 1
if s == 'WA':
ct_wa += 1
if s == 'TLE':
ct_tle += 1
if s == 'RE':
ct_re += 1
print('AC × ' + str(ct_ac))
print('WA × ' + str(ct_wa))
print('TLE × ' + str(ct_tle))
print('RE × ' + str(ct_re))
|
s108386132
|
Accepted
| 152
| 9,236
| 357
|
n = int(input())
ct_ac = 0
ct_wa = 0
ct_tle = 0
ct_re = 0
for i in range(n):
s = input()
if s == 'AC':
ct_ac += 1
if s == 'WA':
ct_wa += 1
if s == 'TLE':
ct_tle += 1
if s == 'RE':
ct_re += 1
print('AC x ' + str(ct_ac))
print('WA x ' + str(ct_wa))
print('TLE x ' + str(ct_tle))
print('RE x ' + str(ct_re))
|
s368551387
|
p02613
|
u845847173
| 2,000
| 1,048,576
|
Wrong Answer
| 167
| 20,292
| 391
|
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.
|
N = int(input())
S = []
for _ in range(N):
S.append(input())
AC = []
WA = []
TLE = []
RE = []
for i in range(N):
if S[i] == "AC":
AC.append(i)
elif S[i] == "WA":
WA.append(i)
elif S[i] == "TLE":
TLE.append(i)
else:
RE.append(i)
print("AC", "×", len(AC))
print("WA", "×", len(WA))
print("TLE", "×", len(TLE))
print("RE", "×", len(RE))
|
s121512141
|
Accepted
| 170
| 20,176
| 395
|
N = int(input())
S = []
for _ in range(N):
S.append(input())
AC = []
WA = []
TLE = []
RE = []
for i in range(N):
if S[i] == "AC":
AC.append(i)
elif S[i] == "WA":
WA.append(i)
elif S[i] == "TLE":
TLE.append(i)
else:
RE.append(i)
print("AC", "x", len(AC))
print("WA", "x", len(WA))
print("TLE", "x", len(TLE))
print("RE", "x", len(RE))
|
s378054178
|
p03544
|
u440129511
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 81
|
It is November 18 now in Japan. By the way, 11 and 18 are adjacent Lucas numbers. You are given an integer N. Find the N-th Lucas number. Here, the i-th Lucas number L_i is defined as follows: * L_0=2 * L_1=1 * L_i=L_{i-1}+L_{i-2} (i≥2)
|
n=int(input())
l=[1,2]
for i in range(n):
l.append(l[-1]+l[-2])
print(l[n-1])
|
s546743818
|
Accepted
| 17
| 3,064
| 79
|
n=int(input())
l=[2,1]
for i in range(n):
l.append(l[-1]+l[-2])
print(l[n])
|
s486867299
|
p03523
|
u644126199
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 453
|
You are given a string S. Takahashi can insert the character `A` at any position in this string any number of times. Can he change S into `AKIHABARA`?
|
data =input()
data = list(data)
deter =""
if len(data) >=9:
print("NO")
if len(data) >=5 and len(data)<=8:
if data[0] !="A":
data.insert(0,"A")
if data[4] !="A":
data.insert(4,"A")
if data[6] !="A":
data.insert(6,"A")
if data[len(data)-1] !="R":
print("NO")
else:
data.append("A")
for s in range(len(data)):
deter +=data[s]
if deter =="AKIHABARA":
print("YES")
print(deter)
if len(data) <=4:
print("NO")
|
s714275943
|
Accepted
| 17
| 3,064
| 859
|
data =input()
data = list(data)
deter =""
cnt2 =0
cnt =[0]*5
cnt3 =0
for t in range(len(data)):
if data[t] =="I":
cnt[0] +=1
if data[t] =="K":
cnt[1] +=1
if data[t] =="H":
cnt[2] +=1
if data[t] =="B":
cnt[3] +=1
if data[t] =="R":
cnt[4] +=1
if cnt[0] ==1 and cnt[1] ==1 and cnt[2] ==1 and cnt[3] ==1 and cnt[4] ==1:
cnt2 +=1
if len(data) >9 or cnt2 ==0 :
print("NO")
cnt3 +=1
if len(data) >=5 and len(data)<=9 and cnt2 ==1:
if data[0] !="A":
data.insert(0,"A")
if data[4] !="A":
data.insert(4,"A")
if data[6] !="A":
data.insert(6,"A")
if len(data) ==8:
data.append("A")
else:
if data[8] !="A":
data.insert(8,"A")
for s in range(len(data)):
deter +=data[s]
if deter =="AKIHABARA":
print("YES")
else:
print("NO")
if len(data) <=4 and cnt3==0:
print("NO")
|
s488139011
|
p02399
|
u078042885
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,568
| 96
|
Write a program which reads two integers a and b, and calculates the following values: * a ÷ b: d (in integer) * remainder of a ÷ b: r (in integer) * a ÷ b: f (in real number)
|
c=input()
a,b=c.split()
print("{} {} {}".format(int(int(a)/int(b)),int(a)%int(b),int(a)/int(b)))
|
s816106235
|
Accepted
| 30
| 7,652
| 74
|
(a,b)=(int(i) for i in input().split())
print('%s %s %.5f'%(a//b,a%b,a/b))
|
s695441839
|
p02613
|
u012820749
| 2,000
| 1,048,576
|
Wrong Answer
| 143
| 16,160
| 267
|
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.
|
def main():
num = int(input())
result = []
for i in range(num):
result.append(input())
print(f'AC X {result.count("AC")}')
print(f'WA X {result.count("WA")}')
print(f'TLE X {result.count("TLE")}')
print(f'RE X {result.count("RE")}')
return
main()
|
s224387812
|
Accepted
| 143
| 16,288
| 267
|
def main():
num = int(input())
result = []
for i in range(num):
result.append(input())
print(f'AC x {result.count("AC")}')
print(f'WA x {result.count("WA")}')
print(f'TLE x {result.count("TLE")}')
print(f'RE x {result.count("RE")}')
return
main()
|
s816153893
|
p03574
|
u953794676
| 2,000
| 262,144
|
Wrong Answer
| 30
| 3,188
| 609
|
You are given an H × W grid. The squares in the grid are described by H strings, S_1,...,S_H. The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W). `.` stands for an empty square, and `#` stands for a square containing a bomb. Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square. (Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.) He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square. Print the strings after the process.
|
dy = [-1, -1, -1, 0, 0, 1, 1, 1]
dx = [-1, 0, 1, -1, 1, -1, 0, 1]
field = []
H, W = map(int, input().split())
for _ in range(H):
field.append(list(input()))
for y in range(H):
for x in range(W):
if field[y][x] == '#':
continue
count = 0
for k in range(8):
nx = x + dx[k]
ny = y + dy[k]
if nx < 0 or W <= nx or ny < 0 or H <= ny:
continue
if field[ny][nx] == '#':
count += 1
field[y][x] = str(count)
for h in range(H):
print(field[h])
|
s064851153
|
Accepted
| 31
| 3,188
| 618
|
dy = [-1, -1, -1, 0, 0, 1, 1, 1]
dx = [-1, 0, 1, -1, 1, -1, 0, 1]
field = []
H, W = map(int, input().split())
for _ in range(H):
field.append(list(input()))
for y in range(H):
for x in range(W):
if field[y][x] == '#':
continue
count = 0
for k in range(8):
nx = x + dx[k]
ny = y + dy[k]
if nx < 0 or W <= nx or ny < 0 or H <= ny:
continue
if field[ny][nx] == '#':
count += 1
field[y][x] = str(count)
for h in range(H):
print("".join(field[h]))
|
s121325431
|
p03681
|
u594329566
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 4,616
| 530
|
Snuke has N dogs and M monkeys. He wants them to line up in a row. As a Japanese saying goes, these dogs and monkeys are on bad terms. _("ken'en no naka", literally "the relationship of dogs and monkeys", means a relationship of mutual hatred.)_ Snuke is trying to reconsile them, by arranging the animals so that there are neither two adjacent dogs nor two adjacent monkeys. How many such arrangements there are? Find the count modulo 10^9+7 (since animals cannot understand numbers larger than that). Here, dogs and monkeys are both distinguishable. Also, two arrangements that result from reversing each other are distinguished.
|
import math
def ini(): return int(input())
def inli(): return list(map(int, input().split()))
def inf(): return float(input())
def inlf(): return list(map(float, input().split()))
def inl(): return list(input())
def pli(): return "".join(list(map(str, ans)))
a = inli()
n=a[0]
m=a[1]
if n-m ==1 or m-n==1:
if n < m:
tmp =math.factorial(n)
print(tmp * tmp * m)
else:
tmp = math.factorial(m)
print(tmp * tmp *n)
elif n==m :
tmp =math.factorial(m)
print(tmp * tmp)
else :
print(0)
|
s215867146
|
Accepted
| 400
| 4,608
| 571
|
import math
def ini(): return int(input())
def inli(): return list(map(int, input().split()))
def inf(): return float(input())
def inlf(): return list(map(float, input().split()))
def inl(): return list(input())
def pli(): return "".join(list(map(str, ans)))
a = inli()
n=a[0]
m=a[1]
if n-m ==1 or m-n==1:
if n < m:
tmp =math.factorial(n)
print((tmp * tmp * m)%1000000007)
else:
tmp = math.factorial(m)
print((tmp * tmp *n)%1000000007)
elif n==m :
tmp =math.factorial(m)
print((tmp * tmp *2)%1000000007)
else :
print(0)
|
s465613018
|
p03353
|
u982762220
| 2,000
| 1,048,576
|
Wrong Answer
| 2,104
| 6,004
| 446
|
You are given a string s. Among the **different** substrings of s, print the K-th lexicographically smallest one. A substring of s is a string obtained by taking out a non-empty contiguous part in s. For example, if s = `ababc`, `a`, `bab` and `ababc` are substrings of s, while `ac`, `z` and an empty string are not. Also, we say that substrings are different when they are different as strings. Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.
|
S = input()
K = int(input())
l = len(S)
topk = []
vocab = set()
for idx in range(K):
ss = S[0:idx + 1]
topk.append(ss)
vocab.add(ss)
for idx in range(1, l):
sblen = 0
for sblen in range(1, l - idx):
ss = S[idx: idx + sblen]
if ss in vocab:
continue
vocab.add(ss)
if ss > topk[-1]:
break
else:
topk[-1] = ss
topk.sort()
print(topk[K-1])
|
s549557907
|
Accepted
| 38
| 5,084
| 227
|
S = input()
K = int(input())
l = len(S)
topk = []
a = []
for i in range(l):
for j in range(i + 1, i + K + 1):
if j > l:
break
ss = S[i: j]
a.append(ss)
print(sorted(list(set(a)))[K-1])
|
s202463886
|
p03475
|
u159994501
| 3,000
| 262,144
|
Wrong Answer
| 65
| 3,064
| 296
|
A railroad running from west to east in Atcoder Kingdom is now complete. There are N stations on the railroad, numbered 1 through N from west to east. Tomorrow, the opening ceremony of the railroad will take place. On this railroad, for each integer i such that 1≤i≤N-1, there will be trains that run from Station i to Station i+1 in C_i seconds. No other trains will be operated. The first train from Station i to Station i+1 will depart Station i S_i seconds after the ceremony begins. Thereafter, there will be a train that departs Station i every F_i seconds. Here, it is guaranteed that F_i divides S_i. That is, for each Time t satisfying S_i≤t and t％F_i=0, there will be a train that departs Station i t seconds after the ceremony begins and arrives at Station i+1 t+C_i seconds after the ceremony begins, where A％B denotes A modulo B, and there will be no other trains. For each i, find the earliest possible time we can reach Station N if we are at Station i when the ceremony begins, ignoring the time needed to change trains.
|
N = int(input())
tc,ts,tf = 0,0,0
C, S, F, =[], [], []
t = 0
for i in range(N-1):
tc,ts,tf = map(int,input().split())
C.append(tc)
S.append(ts)
F.append(tf)
print(C,S,F)
for i in range(N-1):
t = 0
for j in range(i, N-1):
t = max(t,S[j])+C[j]
print(t)
print(0)
|
s656034423
|
Accepted
| 97
| 3,188
| 392
|
import math
N = int(input())
tc,ts,tf = 0,0,0
C, S, F, =[], [], []
t = 0
for i in range(N-1):
tc,ts,tf = map(int,input().split())
C.append(tc)
S.append(ts)
F.append(tf)
#print(C,S,F)
for i in range(N-1):
t = 0
for j in range(i, N-1):
if t <= S[j]:
t = S[j] + C[j]
else:
t = math.ceil(t/F[j]) * F[j] + C[j]
print(t)
print(0)
|
s620114453
|
p03386
|
u372259664
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 192
|
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.
|
temp = input()
a,b,k = temp.split()
a = int(a)
b = int(b)
k = int(k)
box = []
for i in range(k):
box.append(a+i)
box.append(b-i)
out = set(box)
sorted(out)
for i in out:
print(i)
|
s679628237
|
Accepted
| 17
| 3,060
| 254
|
temp = input()
a,b,k = temp.split()
a = int(a)
b = int(b)
k = int(k)
box = []
for i in range(k):
box.append(a+i)
box.append(b-i)
out = set(box)
out = list(out)
out = [i for i in out if i>=a and i<=b]
out = sorted(out)
for i in out:
print(i)
|
s607937332
|
p03385
|
u672898046
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 75
|
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()
l = ["a","b","c"]
if s in l:
print("Yes")
else:
print("No")
|
s396796451
|
Accepted
| 17
| 2,940
| 103
|
s = input()
l = [i for i in s]
r = ["a","b", "c"]
if sorted(l) == r:
print("Yes")
else:
print("No")
|
s444559325
|
p02853
|
u624613992
| 2,000
| 1,048,576
|
Wrong Answer
| 30
| 9,160
| 149
|
We held two competitions: Coding Contest and Robot Maneuver. In each competition, the contestants taking the 3-rd, 2-nd, and 1-st places receive 100000, 200000, and 300000 yen (the currency of Japan), respectively. Furthermore, a contestant taking the first place in both competitions receives an additional 400000 yen. DISCO-Kun took the X-th place in Coding Contest and the Y-th place in Robot Maneuver. Find the total amount of money he earned.
|
x,y = map(int,input().split())
a = [0]*210
a[:3] = [300000, 200000, 100000]
if x == y and x == 1:
print(400000)
else:
print(a[x-1] + a[y-1])
|
s218056648
|
Accepted
| 30
| 9,168
| 150
|
x,y = map(int,input().split())
a = [0]*210
a[:3] = [300000, 200000, 100000]
if x == y and x == 1:
print(1000000)
else:
print(a[x-1] + a[y-1])
|
s504785927
|
p00032
|
u744114948
| 1,000
| 131,072
|
Wrong Answer
| 30
| 6,720
| 203
|
機械に辺・対角線の長さのデータを入力し、プラスティック板の型抜きをしている工場があります。この工場では、サイズは様々ですが、平行四辺形の型のみを切り出しています。あなたは、切り出される平行四辺形のうち、長方形とひし形の製造個数を数えるように上司から命じられました。 「機械に入力するデータ」を読み込んで、長方形とひし形の製造個数を出力するプログラムを作成してください。
|
s=0
n=0
while True:
try:
a,b,c = map(int, input().split())
except:
break
if a == b:
n+=1
elif c**2 == a**2 + b**2:
s+=1
print(str(s) + "\n" + str(n))
|
s557516157
|
Accepted
| 30
| 6,720
| 204
|
s=0
n=0
while True:
try:
a,b,c = map(int, input().split(","))
except:
break
if a == b:
n+=1
elif c**2 == a**2 + b**2:
s+=1
print(str(s) + "\n" + str(n))
|
s304866174
|
p03828
|
u258647915
| 2,000
| 262,144
|
Wrong Answer
| 23
| 3,064
| 280
|
You are given an integer N. Find the number of the positive divisors of N!, modulo 10^9+7.
|
from math import *
def divisor(n):
res = list()
for i in range(1, int(sqrt(n))):
if n % i == 0:
res.append(i)
if i != n // i: res.append(n / i)
return res
MOD = int(1e9 + 7)
n = int(input())
print(len(divisor(factorial(n))) % MOD)
|
s947249067
|
Accepted
| 59
| 3,064
| 322
|
MOD = int(1e9 + 7)
n = int(input())
es = [0] * 1010
def div(x):
if x == 1: return
for i in range(2, x + 1):
while x % i == 0:
es[i] += 1
x //= i
for i in range(1, n + 1):
div(i)
ans = 1
for i in range(1, n + 1):
ans *= es[i] + 1
print(ans % MOD)
|
s465620586
|
p02612
|
u470439462
| 2,000
| 1,048,576
|
Wrong Answer
| 28
| 9,092
| 87
|
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.
|
N = int(input())
if N % 1000 == 0:
print(0)
else:
print(N - (N // 1000) * 1000)
|
s586574925
|
Accepted
| 27
| 8,996
| 91
|
N = int(input())
if N % 1000 == 0:
print(0)
else:
print(((N // 1000)+1) * 1000 - N)
|
s957163170
|
p02850
|
u955251526
| 2,000
| 1,048,576
|
Wrong Answer
| 2,105
| 56,480
| 402
|
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())
edges = [tuple(map(int, input().split())) for _ in range(n-1)]
color = {}
forbidden = [set() for _ in range(n+1)]
for a, b in sorted(edges):
for c in range(n):
if c not in forbidden[a] and c not in forbidden[b]:
forbidden[a].add(c)
forbidden[b].add(c)
color[(a, b)] = c
break
for a, b in edges:
print(color[(a, b)] + 1)
|
s701001866
|
Accepted
| 804
| 64,216
| 717
|
n = int(input())
edge = [tuple(map(int, input().split())) for _ in range(n-1)]
connect = [set() for _ in range(n)]
for a, b in edge:
connect[a-1].add(b-1)
connect[b-1].add(a-1)
ans = {(-1, 0): -1}
que = {(-1, 0)}
while que:
queque = set()
for p, v in que:
c = 1
for w in connect[v]:
if w != p:
if c == ans[(p, v)]:
c += 1
ans[(v, w)] = c
queque.add((v, w))
c += 1
que = queque
pri = [0] * (n-1)
for i, (a, b) in enumerate(edge):
a -= 1
b -= 1
if (a, b) in ans:
pri[i] = ans[(a, b)]
else:
pri[i] = ans[(b, a)]
print(max(pri))
for x in pri:
print(x)
|
s742645693
|
p02865
|
u812867074
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 71
|
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
n = int(input())
if n % 2 == 0:
print(n/2 - 1)
else:
print((n-1)/2)
|
s845616570
|
Accepted
| 17
| 2,940
| 81
|
n = int(input())
if n % 2 == 0:
print(int(n/2 - 1))
else:
print(int((n-1)/2))
|
s424591479
|
p02612
|
u844590940
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 9,088
| 71
|
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.
|
n = input()
keta = len(n)
ans = int(n[keta - 3:keta]) % 1000
print(ans)
|
s389508565
|
Accepted
| 25
| 9,080
| 67
|
n = int(input())
simosan = n % 1000
print((1000 - simosan) % 1000)
|
s591068695
|
p03779
|
u374802266
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,188
| 174
|
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.
|
n=int(input())
l,r=0,10**9
while l+1!=r:
m=(l+r)/2
if m*(m+1)//2==n:
print(m)
exit()
elif m*(m+1)//2>n:
r=m
else:
l=m
print(r)
|
s935977278
|
Accepted
| 25
| 2,940
| 80
|
n=int(input())
a=0
for i in range(1,10**5):
a+=i
if a>=n:
break
print(i)
|
s164246999
|
p02613
|
u934940582
| 2,000
| 1,048,576
|
Wrong Answer
| 147
| 16,328
| 345
|
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.
|
N = int(input())
result = [input() for _ in range(N)]
C_AC,C_TLE,C_WA,C_RE = [0,0,0,0]
for i in result:
if i == "AC":
C_AC += 1
elif i == "WA":
C_WA += 1
elif i == "TLE":
C_TLE += 1
elif i == "RE":
C_RE += 1
print("AC x " + str(C_AC))
print("WA x " + str(C_WA))
print("TLE x " + str(C_TLE))
print("AC x " + str(C_RE))
|
s429013525
|
Accepted
| 147
| 16,260
| 345
|
N = int(input())
result = [input() for _ in range(N)]
C_AC,C_TLE,C_WA,C_RE = [0,0,0,0]
for i in result:
if i == "AC":
C_AC += 1
elif i == "WA":
C_WA += 1
elif i == "TLE":
C_TLE += 1
elif i == "RE":
C_RE += 1
print("AC x " + str(C_AC))
print("WA x " + str(C_WA))
print("TLE x " + str(C_TLE))
print("RE x " + str(C_RE))
|
s939600673
|
p03673
|
u667024514
| 2,000
| 262,144
|
Wrong Answer
| 2,105
| 20,972
| 203
|
You are given an integer sequence of length n, a_1, ..., a_n. Let us consider performing the following n operations on an empty sequence b. The i-th operation is as follows: 1. Append a_i to the end of b. 2. Reverse the order of the elements in b. Find the sequence b obtained after these n operations.
|
n = int(input())
lis = []
li = list(map(str,input().split()))
if n % 2 == 1:
lis.append(li[0])
li.pop(0)
for i in range(n//2):
lis.append(li[i])
lis.insert(0,li[i+1])
print(" ".join(lis))
|
s170351704
|
Accepted
| 66
| 27,384
| 206
|
n = int(input())
lis = list(map(str,input().split()))
li = lis[::2]
l = lis[1::2]
if n % 2 == 0:
l = l[::-1]
ans = l+li
print(" ".join(ans))
else:
li = li[::-1]
ans = li+l
print(" ".join(ans))
|
s084701705
|
p02407
|
u485986915
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,584
| 71
|
Write a program which reads a sequence and prints it in the reverse order.
|
n = int(input())
a = list(map(int,input().split()))
a.sort()
print(a)
|
s826429784
|
Accepted
| 20
| 5,596
| 122
|
n = int(input())
a = list(map(int,input().split()))
a.reverse()
str = ""
for d in a:
str +="%d "%(d)
print(str[:-1])
|
s398514151
|
p03587
|
u698535708
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 75
|
Snuke prepared 6 problems for a upcoming programming contest. For each of those problems, Rng judged whether it can be used in the contest or not. You are given a string S of length 6. If the i-th character of s is `1`, it means that the i-th problem prepared by Snuke is accepted to be used; `0` means that the problem is not accepted. How many problems prepared by Snuke are accepted to be used in the contest?
|
a = input()
ans = 0
for i in a:
if a=='1':
ans += 1
print(ans)
|
s436395626
|
Accepted
| 20
| 2,940
| 76
|
a = input()
ans = 0
for i in a:
if i=='1':
ans += 1
print(ans)
|
s036281698
|
p03351
|
u119578112
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 171
|
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.
|
N = list(map(int, input().split()))
max_N = max(N)
min_N = min(N)
if max_N-min_N <= N[3]:
print('Yes')
elif N[2]-N[0] <= N[3]:
print('Yes')
else :
print('No')
|
s954143565
|
Accepted
| 17
| 3,060
| 171
|
N = list(map(int, input().split()))
if abs(N[0]-N[2])<=N[3]:
print('Yes')
elif abs(N[0]-N[1])<= N[3] and abs(N[1]-N[2])<= N[3]:
print('Yes')
else:
print('No')
|
s525204113
|
p03407
|
u597017430
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 95
|
An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.
|
A = list(map(int, input().split()))
if A[0] + A[1] <= A[2]:
print('Yes')
exit()
print('No')
|
s876872053
|
Accepted
| 17
| 2,940
| 96
|
A = list(map(int, input().split()))
if A[0] + A[1] >= A[2]:
print('Yes')
exit()
print('No')
|
s911469510
|
p02410
|
u343251190
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,696
| 273
|
Write a program which reads a $ n \times m$ matrix $A$ and a $m \times 1$ vector $b$, and prints their product $Ab$. A column vector with m elements is represented by the following equation. \\[ b = \left( \begin{array}{c} b_1 \\\ b_2 \\\ : \\\ b_m \\\ \end{array} \right) \\] A $n \times m$ matrix with $m$ column vectors, each of which consists of $n$ elements, is represented by the following equation. \\[ A = \left( \begin{array}{cccc} a_{11} & a_{12} & ... & a_{1m} \\\ a_{21} & a_{22} & ... & a_{2m} \\\ : & : & : & : \\\ a_{n1} & a_{n2} & ... & a_{nm} \\\ \end{array} \right) \\] $i$-th element of a $m \times 1$ column vector $b$ is represented by $b_i$ ($i = 1, 2, ..., m$), and the element in $i$-th row and $j$-th column of a matrix $A$ is represented by $a_{ij}$ ($i = 1, 2, ..., n,$ $j = 1, 2, ..., m$). The product of a $n \times m$ matrix $A$ and a $m \times 1$ column vector $b$ is a $n \times 1$ column vector $c$, and $c_i$ is obtained by the following formula: \\[ c_i = \sum_{j=1}^m a_{ij}b_j = a_{i1}b_1 + a_{i2}b_2 + ... + a_{im}b_m \\]
|
n, m = map(int, input().split())
A = []
for i in range(n):
A.append(list(map(int, input().split())))
print(A)
B = []
for j in range(m):
B.append(int(input()))
print(B)
for k in range(n):
x = 0
for l in range(m):
x += A[k][l] * B[l]
print(x)
|
s537506644
|
Accepted
| 30
| 7,980
| 253
|
n, m = map(int, input().split())
A = []
for i in range(n):
A.append(list(map(int, input().split())))
B = []
for j in range(m):
B.append(int(input()))
for k in range(n):
x = 0
for l in range(m):
x += A[k][l] * B[l]
print(x)
|
s217718662
|
p03456
|
u965337330
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 88
|
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.
|
A=list(map(int, input().split()))
if A[0]*A[0]==A[1]:
print('Yes')
else:
print('No')
|
s906449789
|
Accepted
| 18
| 3,064
| 243
|
import math
A=list(map(int, input().split()))
b=str(A[0])+str(A[1])
c=int(b)
#print(round(math.sqrt(int(b)),10))
#c=round(math.sqrt(int(b)),10)
d=0
for i in range(10000):
if i*i==c:
print('Yes')
d=1
if d==0:
print('No')
|
s515949289
|
p02412
|
u914146430
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,588
| 322
|
Write a program which identifies the number of combinations of three integers which satisfy the following conditions: * You should select three distinct integers from 1 to n. * A total sum of the three integers is x. For example, there are two combinations for n = 5 and x = 9. * 1 + 3 + 5 = 9 * 2 + 3 + 4 = 9
|
m=0
while True:
n, x = list(map(int,input().split()))
if n==0 and x==0:
break
else:
for i in range(1,n+1):
for j in range(i+1,n+1):
for k in range(j+1, n+1):
print(i,j,k)
if i+j+k==x:
m+=1
print(m)
|
s241761945
|
Accepted
| 530
| 7,600
| 326
|
while True:
n, x = list(map(int,input().split()))
m=0
if n==0 and x==0:
break
else:
for i in range(1,n+1):
for j in range(i+1,n+1):
for k in range(j+1, n+1):
#print(i,j,k)
if i+j+k==x:
m+=1
print(m)
|
s457818389
|
p02612
|
u566685958
| 2,000
| 1,048,576
|
Wrong Answer
| 26
| 8,980
| 31
|
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.
|
N = int(input())
print(N%1000)
|
s779264961
|
Accepted
| 28
| 9,076
| 76
|
N = int(input())
m = N % 1000
if m == 0:
print(0)
else:
print(1000 - m)
|
s975007088
|
p03338
|
u468972478
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 9,124
| 98
|
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()
t = 0
for i in range(n):
if s.count(s[i]) >= 2:
t += 1
print(t)
|
s026655815
|
Accepted
| 30
| 9,172
| 112
|
n = int(input())
s = input()
a = []
for i in range(1, n):
a.append(len(set(s[:i]) & set(s[i:])))
print(max(a))
|
s910065900
|
p04044
|
u090267744
| 2,000
| 262,144
|
Wrong Answer
| 154
| 12,420
| 320
|
Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds: * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i. * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m.
|
import numpy as np
n,l=list(map(int,input().split( )))
s=[]
for i in range(n):
s.append(input())
S=[]
while len(s)>0:
a=0
for i in range(n-a):
if s[i]==min(s):
S.append(s[i])
del s[i]
break
a+=1
print(S)
|
s270107076
|
Accepted
| 17
| 3,060
| 112
|
n,l=map(int,input().split( ))
s=[]
for i in range(n):
s.append(input())
print(''.join(sorted(s)))
|
s307658417
|
p02295
|
u072053884
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,652
| 759
|
For given two segments s1 and s2, print the coordinate of the cross point of them. s1 is formed by end points p0 and p1, and s2 is formed by end points p2 and p3.
|
def cross(c1, c2):
return c1.real * c2.imag - c1.imag * c2.real
def print_cross_point(p1, p2, p3, p4):
# p1 and p2 are end points of a segment.
# p3 and p4 are end points of the other segment.
base = p4 - p3
hypo1 = p1 - p3
hypo2 = p2 - p3
d1 = abs(cross(base, hypo1)) / abs(base)
d2 = abs(cross(base, hypo2)) / abs(base)
cp = p1 + d1 / (d1 + d2) * (p2 - p1)
print("{0:.10f}, {1:.10f}".format(cp.real, cp.imag))
import sys
file_input = sys.stdin
sq = file_input.readline()
for line in file_input:
x_p0, y_p0, x_p1, y_p1, x_p2, y_p2, x_p3, y_p3 = map(int, line.split())
p0 = x_p0 + y_p0 * 1j
p1 = x_p1 + y_p1 * 1j
p2 = x_p2 + y_p2 * 1j
p3 = x_p3 + y_p3 * 1j
print_cross_point(p0, p1, p2, p3)
|
s750174703
|
Accepted
| 30
| 7,784
| 758
|
def cross(c1, c2):
return c1.real * c2.imag - c1.imag * c2.real
def print_cross_point(p1, p2, p3, p4):
# p1 and p2 are end points of a segment.
# p3 and p4 are end points of the other segment.
base = p4 - p3
hypo1 = p1 - p3
hypo2 = p2 - p3
d1 = abs(cross(base, hypo1)) / abs(base)
d2 = abs(cross(base, hypo2)) / abs(base)
cp = p1 + d1 / (d1 + d2) * (p2 - p1)
print("{0:.10f} {1:.10f}".format(cp.real, cp.imag))
import sys
file_input = sys.stdin
sq = file_input.readline()
for line in file_input:
x_p0, y_p0, x_p1, y_p1, x_p2, y_p2, x_p3, y_p3 = map(int, line.split())
p0 = x_p0 + y_p0 * 1j
p1 = x_p1 + y_p1 * 1j
p2 = x_p2 + y_p2 * 1j
p3 = x_p3 + y_p3 * 1j
print_cross_point(p0, p1, p2, p3)
|
s650800154
|
p03729
|
u864900001
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 123
|
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`.
|
#60
a, b, c = map(str, input().split())
if(a[len(a)-1]==b[0] and b[len(b)-1]==c[0]):
print("Yes")
else:
print("No")
|
s884332820
|
Accepted
| 17
| 2,940
| 123
|
#60
a, b, c = map(str, input().split())
if(a[len(a)-1]==b[0] and b[len(b)-1]==c[0]):
print("YES")
else:
print("NO")
|
s792027548
|
p03943
|
u169221932
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 247
|
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 = (int(x) for x in input().split())
candy = [a, b, c]
#print(candy)
candy.sort()
print(candy)
max_candy = candy[2]
middle_candy = candy[1]
min_candy = candy[0]
if max_candy == middle_candy + min_candy:
print('Yes')
else:
print('No')
|
s081918631
|
Accepted
| 17
| 3,060
| 248
|
a,b,c = (int(x) for x in input().split())
candy = [a, b, c]
#print(candy)
candy.sort()
#print(candy)
max_candy = candy[2]
middle_candy = candy[1]
min_candy = candy[0]
if max_candy == middle_candy + min_candy:
print('Yes')
else:
print('No')
|
s275504622
|
p03591
|
u776189585
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 54
|
Ringo is giving a present to Snuke. Ringo has found out that Snuke loves _yakiniku_ (a Japanese term meaning grilled meat. _yaki_ : grilled, _niku_ : meat). He supposes that Snuke likes grilled things starting with `YAKI` in Japanese, and does not like other things. You are given a string S representing the Japanese name of Ringo's present to Snuke. Determine whether S starts with `YAKI`.
|
s = input()
print('YES' if s[:4] == 'YAKI' else 'NO')
|
s917879367
|
Accepted
| 17
| 2,940
| 53
|
s = input()
print('Yes' if s[:4] == 'YAKI' else 'No')
|
s288802737
|
p03385
|
u128914900
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 107
|
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()
#sl=["a","b","c"]
if s[0]!=s[1] and s[0]!=s[2] and s[2]!=s[2]:
print("Yes")
else:
print("No")
|
s822828747
|
Accepted
| 17
| 2,940
| 107
|
s=input()
#sl=["a","b","c"]
if s[0]!=s[1] and s[0]!=s[2] and s[1]!=s[2]:
print("Yes")
else:
print("No")
|
s251405461
|
p02398
|
u587193722
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,704
| 81
|
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 = [int(i) for i in input().split()]
bet = c - a
ans = bet // b
print(ans)
|
s492889485
|
Accepted
| 30
| 7,704
| 154
|
a, b, c = [int(i) for i in input().split()]
nums = 0
for i in range(a, b + 1):
if c%i == 0:
nums = nums + 1
else:
pass
print(nums)
|
s472870374
|
p02742
|
u652583512
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 93
|
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())
ans = H * W // 2
if H & 1 | W & 1:
ans += 1
print(ans)
|
s701515433
|
Accepted
| 17
| 2,940
| 135
|
H, W = map(int, input().split())
ans = H * W // 2
if H == 1 or W == 1:
ans = 1
elif (H & 1) and (W & 1):
ans += 1
print(ans)
|
s198641027
|
p02833
|
u682271925
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 132
|
For an integer n not less than 0, let us define f(n) as follows: * f(n) = 1 (if n < 2) * f(n) = n f(n-2) (if n \geq 2) Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N).
|
n = int(input())
size = len(str(n))
count = 0
i = 1
while 5 ** i <= n:
a = 5 ** i
b = n // a
count += b
i += 1
print(count)
|
s201427019
|
Accepted
| 17
| 2,940
| 87
|
n=int(input())
ans=0
if n%2==0:
n//=2
while(n>0):
n//=5
ans+=n
print(ans)
|
s479421917
|
p03759
|
u130860429
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 110
|
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=input().split()
a1=int(a)
b1=int(b)
c1=int(c)
d=b1-a1
e=c1-b1
if d==e:
print('yes')
else :
print('no')
|
s420818702
|
Accepted
| 17
| 3,060
| 110
|
a,b,c=input().split()
a1=int(a)
b1=int(b)
c1=int(c)
d=b1-a1
e=c1-b1
if d==e:
print('YES')
else :
print('NO')
|
s888176680
|
p03547
|
u216631280
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 50
|
In programming, hexadecimal notation is often used. In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively. In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`. When X and Y are seen as hexadecimal numbers, which is larger?
|
li = list(input().split())
li.sort()
print(li[-1])
|
s489001103
|
Accepted
| 18
| 2,940
| 141
|
li = list(input().split())
if li[0] == li[1]:
print('=')
else:
liTemp = sorted(li)
if liTemp[0] == li[0]:
print('<')
else:
print('>')
|
s872314274
|
p03337
|
u144980750
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 65
|
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())
c=[a-b,b-a,a*b]
c.sort()
print(c[2])
|
s027414798
|
Accepted
| 17
| 2,940
| 65
|
a,b=map(int,input().split())
c=[a+b,a-b,a*b]
c.sort()
print(c[2])
|
s557858242
|
p00354
|
u737311644
| 1,000
| 262,144
|
Wrong Answer
| 20
| 5,584
| 402
|
The 9th day of September 2017 is Saturday. Then, what day of the week is the X-th of September 2017? Given a day in September 2017, write a program to report what day of the week it is.
|
a=int(input())
if a==1 or a==8 or a==15 or a==22 or a==29:
print("mon")
if a==2 or a==9 or a==16 or a==23 or a==30:
print("tue")
if a==3 or a==10 or a==17 or a==24:
print("wed")
if a==4 or a==11 or a==18 or a==25:
print("thu")
if a==5 or a==12 or a==19 or a==26:
print("fri")
if a==6 or a==13 or a==20 or a==27:
print("sat")
if a==7 or a==14 or a==21 or a==28:
print("sun")
|
s346440164
|
Accepted
| 20
| 5,592
| 211
|
a=int(input())
if a%7==1:
print("fri")
if a%7==2:
print("sat")
if a%7==3:
print("sun")
if a%7==4:
print("mon")
if a%7==5:
print("tue")
if a%7==6:
print("wed")
if a%7==0:
print("thu")
|
s463951589
|
p04029
|
u997530672
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 65
|
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())
ans = 0
for i in range(n):
ans += i
print(ans)
|
s976804415
|
Accepted
| 17
| 2,940
| 67
|
n = int(input())
ans = 0
for i in range(n):
ans += i+1
print(ans)
|
s127643324
|
p03377
|
u942868002
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 76
|
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.
|
a,b,x = map(int,input().split())
print("Yes" if a<=x and (a+b)>=x else "No")
|
s154462933
|
Accepted
| 18
| 2,940
| 76
|
a,b,x = map(int,input().split())
print("YES" if a<=x and (a+b)>=x else "NO")
|
s574023086
|
p02600
|
u886571850
| 2,000
| 1,048,576
|
Wrong Answer
| 31
| 9,184
| 360
|
M-kun is a competitor in AtCoder, whose highest rating is X. In this site, a competitor is given a _kyu_ (class) according to his/her highest rating. For ratings from 400 through 1999, the following kyus are given: * From 400 through 599: 8-kyu * From 600 through 799: 7-kyu * From 800 through 999: 6-kyu * From 1000 through 1199: 5-kyu * From 1200 through 1399: 4-kyu * From 1400 through 1599: 3-kyu * From 1600 through 1799: 2-kyu * From 1800 through 1999: 1-kyu What kyu does M-kun have?
|
a = int(input())
if 400 <= a and 599 <= a:
print(8)
elif 600 <= a and 799 <= a:
print(7)
elif 800 <= a and 999 <= a:
print(6)
elif 1000 <= a and 1199 <= a:
print(5)
elif 1200 <= a and 1399 <= a:
print(4)
elif 1400 <= a and 1599 <= a:
print(3)
elif 1600 <= a and 1799 <= a:
print(2)
elif 1800 <= a and 1999 <= a:
print(1)
|
s458824088
|
Accepted
| 28
| 9,188
| 359
|
a = int(input())
if 400 <= a and a <= 599:
print(8)
elif 600 <= a and a <= 799:
print(7)
elif 800 <= a and a <= 999:
print(6)
elif 1000 <= a and a <= 1199:
print(5)
elif 1200 <= a and a <= 1399:
print(4)
elif 1400 <= a and a <= 1599:
print(3)
elif 1600 <= a and a <= 1799:
print(2)
elif 1800 <= a and a <= 1999:
print(1)
|
s902367156
|
p03360
|
u978313283
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 84
|
There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations?
|
A,B,C=map(int,input().split())
K=int(input())
M=max(max(A,B),C)
print(M**K+A+B+C-M)
|
s161107018
|
Accepted
| 17
| 2,940
| 88
|
A,B,C=map(int,input().split())
K=int(input())
M=max(max(A,B),C)
print(M*(2**K)+A+B+C-M)
|
s979160912
|
p02401
|
u435158342
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,560
| 92
|
Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part.
|
while True:
a=input()
if '?' in a:
break
eval('print({0})'.format(a))
|
s337259236
|
Accepted
| 20
| 5,556
| 137
|
ans=[]
while True:
a=input()
if '?' in a:
break
eval('ans.append(int({0}))'.format(a))
for a in ans:
print(a)
|
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