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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s751550287 | p02866 | u367866606 | 2,000 | 1,048,576 | Wrong Answer | 45 | 14,396 | 363 | Given is an integer sequence D_1,...,D_N of N elements. Find the number, modulo 998244353, of trees with N vertices numbered 1 to N that satisfy the following condition: * For every integer i from 1 to N, the distance between Vertex 1 and Vertex i is D_i. | import sys
n=int(input())
d=list(map(int,input().split()))
num_d=[0 for i in range(n)]
if d[0]!=0:
print(0)
sys.exit()
for i in d:
if i==0:
print(0)
sys.exit()
num_d[i]+=1
max_d=0
ans=1
for i in range(1,n):
if num_d[i]!=0:
if max_d!=i-1:
print(0)
sys.exit()
max_d=i
ans*=num_d[i-1]**num_d[i]
ans%=998244353
print(ans) | s823374534 | Accepted | 130 | 14,396 | 395 | import sys
n=int(input())
d=list(map(int,input().split()))
num_d=[0 for i in range(n)]
num_d[0]=1
if d[0]!=0:
print(0)
sys.exit()
for i in range(1,len(d)):
if d[i]==0:
print(0)
sys.exit()
num_d[d[i]]+=1
max_d=0
ans=1
for i in range(1,n):
if num_d[i]!=0:
if max_d!=i-1:
print(0)
sys.exit()
max_d=i
ans*=num_d[i-1]**num_d[i]
ans%=998244353
print(ans)
|
s282607318 | p04043 | u437638594 | 2,000 | 262,144 | Wrong Answer | 16 | 2,940 | 107 | Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order. | a, b, c = sorted(map(len, input().split()))
if [a, b, c] == [5, 5, 7]:
print("YES")
else:
print("NO")
| s132457560 | Accepted | 17 | 2,940 | 107 | a, b, c = sorted(map(int, input().split()))
if [a, b, c] == [5, 5, 7]:
print("YES")
else:
print("NO")
|
s332536668 | p03555 | u333731247 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 125 | 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. | C1=list(input())
C2=list(input())
if C1[0]==C2[2] and C1[1]==C2[1] and C1[2]==C2[0]:
print('Yes')
else :
print('No') | s410321763 | Accepted | 17 | 2,940 | 125 | C1=list(input())
C2=list(input())
if C1[0]==C2[2] and C1[1]==C2[1] and C1[2]==C2[0]:
print('YES')
else :
print('NO') |
s120201794 | p03167 | u140251125 | 2,000 | 1,048,576 | Wrong Answer | 555 | 14,964 | 448 | There is a grid with H horizontal rows and W vertical columns. Let (i, j) denote the square at the i-th row from the top and the j-th column from the left. For each i and j (1 \leq i \leq H, 1 \leq j \leq W), Square (i, j) is described by a character a_{i, j}. If a_{i, j} is `.`, Square (i, j) is an empty square; if a_{i, j} is `#`, Square (i, j) is a wall square. It is guaranteed that Squares (1, 1) and (H, W) are empty squares. Taro will start from Square (1, 1) and reach (H, W) by repeatedly moving right or down to an adjacent empty square. Find the number of Taro's paths from Square (1, 1) to (H, W). As the answer can be extremely large, find the count modulo 10^9 + 7. | # input
h, w = map(int, input().split())
A = [input() for _ in range(h)]
dp = [[0 for _ in range(w)] for _ in range(h)]
for i in range(1, h):
if A[i][0] == '.':
dp[i][0] = dp[i - 1][0]
for j in range(1, w):
if A[0][j] == '.':
dp[0][j] = dp[0][j - 1]
for i in range(1, h):
for j in range(1, w):
if A[i][j] == '.':
dp[i][j] += (dp[i - 1][j] + dp[i][j - 1]) % (10 ** 9 + 7)
print(dp[h - 1][w - 1]) | s292776254 | Accepted | 758 | 158,452 | 462 | # input
h, w = map(int, input().split())
A = [input() for _ in range(h)]
dp = [[0 for _ in range(w)] for _ in range(h)]
dp[0][0] = 1
for i in range(1, h):
if A[i][0] == '.':
dp[i][0] = dp[i - 1][0]
for j in range(1, w):
if A[0][j] == '.':
dp[0][j] = dp[0][j - 1]
for i in range(1, h):
for j in range(1, w):
if A[i][j] == '.':
dp[i][j] += (dp[i - 1][j] + dp[i][j - 1])
print(dp[h - 1][w - 1] % (10 ** 9 + 7)) |
s897522325 | p03672 | u717001163 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 160 | We will call a string that can be obtained by concatenating two equal strings an _even_ string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not. You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input. | s=input()
for i in range(1,len(s)):
news = s[:-1*i]
nv = len(news)
if nv % 2 == 1: continue
if news[0:nv//2] == news[nv//2:]:
print(nv//2)
break | s155744428 | Accepted | 18 | 3,060 | 157 | s=input()
for i in range(1,len(s)):
news = s[:-1*i]
nv = len(news)
if nv % 2 == 1: continue
if news[0:nv//2] == news[nv//2:]:
print(nv)
break |
s109139556 | p03495 | u072717685 | 2,000 | 262,144 | Wrong Answer | 21 | 3,316 | 281 | Takahashi has N balls. Initially, an integer A_i is written on the i-th ball. He would like to rewrite the integer on some balls so that there are at most K different integers written on the N balls. Find the minimum number of balls that Takahashi needs to rewrite the integers on them. | from collections import Counter
def main():
n, k = map(int, input().split())
a = tuple(map(int, input().split()))
col = max(0, len(set(a)) - k)
ac = Counter(a)
acl = ac.most_common()[::-1]
acll = [ace[1] for ace in acl[:col]]
r = sum(acll)
print(r) | s409535309 | Accepted | 174 | 48,664 | 387 | import sys
from collections import Counter
def main():
n, k = map(int, input().split())
a = tuple(map(int, input().split()))
col = max(0, len(set(a)) - k)
if col == 0:
print(0)
sys.exit()
ac = Counter(a)
acl = ac.most_common()[::-1]
acll = [ace[1] for ace in acl[:col]]
r = sum(acll)
print(r)
if __name__ == '__main__':
main()
|
s795282290 | p04029 | u313103408 | 2,000 | 262,144 | Wrong Answer | 27 | 9,020 | 33 | 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())
print(n*(n+1)/2) | s332417114 | Accepted | 25 | 9,052 | 40 | n = int(input())
print(int(n*(n+1)/2))
|
s097075414 | p03139 | u509368316 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 60 | 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=map(int,input().split())
x=min(a,b)
y=a+b-n
print(x,y) | s352717390 | Accepted | 17 | 2,940 | 67 | n,a,b=map(int,input().split())
x=min(a,b)
y=max(0,a+b-n)
print(x,y) |
s220655651 | p03448 | u971673384 | 2,000 | 262,144 | Wrong Answer | 47 | 3,060 | 219 | 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, B, C, X = [int(input()) for i in range(4)]
count = 0
for i in range(A):
for j in range(B):
for k in range(C):
if (i * 500) + (j * 100) + (k * 50) == X:
count += 1
print(count) | s317956185 | Accepted | 50 | 3,060 | 225 | A, B, C, X = [int(input()) for i in range(4)]
count = 0
for i in range(A+1):
for j in range(B+1):
for k in range(C+1):
if (i * 500) + (j * 100) + (k * 50) == X:
count += 1
print(count) |
s767336424 | p02257 | u091332835 | 1,000 | 131,072 | Wrong Answer | 20 | 5,584 | 324 | 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. | n = int(input())
p = 0
for i in range(n):
x = int(input())
if x == 2:
p += 1
elif x%2 == 1:
j=3
f = 0
while j*j < 4*x:
if x%j == 0:
f = 1
break
else:
j += 2
if f == 0:
p += 1
print(p)
| s198991738 | Accepted | 2,740 | 5,672 | 254 | from math import sqrt
n = int(input())
p = 0
for i in range(n):
x = int(input())
ad = 1
j = 2
while j < (int(sqrt(x))+1):
if x%j == 0:
ad = 0
break
else:
j += 1
p += ad
print(p)
|
s148905055 | p01620 | u078042885 | 8,000 | 131,072 | Wrong Answer | 120 | 7,684 | 297 | とある国の偉い王様が、突然友好国の土地を視察することになった。 その国は電車が有名で、王様はいろいろな駅を視察するという。 電車の駅は52個あり、それぞれに大文字か小文字のアルファベット1文字の名前がついている（重なっている名前はない）。 この電車の路線は環状になっていて、aの駅の次はbの駅、bの駅の次はcの駅と順に並んでいてz駅の次はA駅、その次はBの駅と順に進み、Z駅の次はa駅になって元に戻る。 単線であり、逆方向に進む電車は走っていない。 ある日、新聞社の記者が王様が訪れる駅の順番のリストを手に入れた。 「dcdkIlkP…」 最初にd駅を訪れ、次にc駅、次にd駅と順に訪れていくという。これで、偉い国の王様を追跡取材できると思った矢先、思わぬことが発覚した。そのリストは、テロ対策のため暗号化されていたのだ！記者の仲間が、その暗号を解く鍵を入手したという。早速この記者は鍵を譲ってもらい、リストの修正にとりかかった。鍵はいくつかの数字の列で構成されている。 「3 1 4 5 3」 この数字の意味するところは、はじめに訪れる駅は、リストに書いてある駅の3つ前の駅。 2番目に訪れる駅はリストの2番目の駅の前の駅、という風に、実際訪れる駅がリストの駅の何駅前かを示している。 記者は修正に取りかかったが、訪れる駅のリストの数よりも、鍵の数の方が小さい、どうするのかと仲間に聞いたところ、最後の鍵をつかったら、またはじめの鍵から順に使っていけばよいらしい。 そして記者はようやくリストを修正することができた。 「abZfFijL…」 これでもう怖い物は無いだろう、そう思った矢先、さらに思わぬ事態が発覚した。 偉い王様は何日間も滞在し、さらにそれぞれの日程ごとにリストと鍵が存在したのだ。 記者は上司から、すべてのリストを復号するように指示されたが、量が量だけに、彼一人では終わらない。 あなたの仕事は彼を助け、このリストの復号を自動で行うプログラムを作成することである。 | while 1:
n=int(input())
if n==0:break
a=list(map(int,input().split()))
s=list(input())
for i in range(len(s)):
for j in range(a[i%n]):
if s[i]=='A':s[i]='a'
elif s[i]=='a':s[i]='Z'
else:s[i]=s[i]=chr(ord(s[i])-1)
print(*s,sep='') | s239512064 | Accepted | 130 | 7,704 | 297 | while 1:
n=int(input())
if n==0:break
a=list(map(int,input().split()))
s=list(input())
for i in range(len(s)):
for j in range(a[i%n]):
if s[i]=='A':s[i]='z'
elif s[i]=='a':s[i]='Z'
else:s[i]=s[i]=chr(ord(s[i])-1)
print(*s,sep='') |
s358521323 | p03303 | u727787724 | 2,000 | 1,048,576 | Wrong Answer | 21 | 3,064 | 266 | You are given a string S consisting of lowercase English letters. We will write down this string, starting a new line after every w letters. Print the string obtained by concatenating the letters at the beginnings of these lines from top to bottom. | s=list(input())
w=int(input())
x=[]
y=''
ans=''
count=[]
for i in range(len(s)):
y+=s[i]
if len(list(y))==w:
x.append(y)
y=''
if i==len(s)-1:
x.append(y)
for j in range(len(x)-1):
count=list(x[j])
ans+=count[0]
print(ans)
| s130133166 | Accepted | 21 | 3,064 | 274 | s=list(input())
w=int(input())
x=[]
y=''
ans=''
count=[]
for i in range(len(s)):
y+=s[i]
if len(list(y))==w:
x.append(y)
y=''
if i==len(s)-1 and y!='':
x.append(y)
for j in range(len(x)):
count=list(x[j])
ans+=count[0]
print(ans)
|
s651243710 | p03588 | u941753895 | 2,000 | 262,144 | Wrong Answer | 347 | 6,160 | 458 | A group of people played a game. All players had distinct scores, which are positive integers. Takahashi knows N facts on the players' scores. The i-th fact is as follows: the A_i-th highest score among the players is B_i. Find the maximum possible number of players in the game. | import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time
sys.setrecursionlimit(10**7)
inf=10**20
mod=10**9+7
def LI(): return list(map(int,input().split()))
def II(): return int(input())
def LS(): return input().split()
def S(): return input()
def main():
n=II()
a=b=0
for i in range(n):
_a,_b=LI()
if _a>a:
a=_a
b=_b
if n==3 and a==6 and b==2:
exit()
print(a+b)
main()
# print(main())
| s265940455 | Accepted | 331 | 6,296 | 409 | import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time
sys.setrecursionlimit(10**7)
inf=10**20
mod=10**9+7
def LI(): return list(map(int,input().split()))
def II(): return int(input())
def LS(): return input().split()
def S(): return input()
def main():
n=II()
a=b=0
for i in range(n):
_a,_b=LI()
if _a>a:
a=_a
b=_b
return a+b
print(main())
|
s520868960 | p02260 | u539753516 | 1,000 | 131,072 | Wrong Answer | 20 | 5,592 | 209 | 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. | n=int(input())
a=list(map(int,input().split()))
c=0
for i in range(n):
minj=i
for j in range(i,n):
if a[j]<a[minj]:
minj=j
a[i],a[minj]=a[minj],a[i]
c+=1
print(*a)
print(c)
| s843350684 | Accepted | 30 | 5,600 | 220 | n=int(input())
a=list(map(int,input().split()))
c=0
for i in range(n):
minj=i
for j in range(i,n):
if a[j]<a[minj]:
minj=j
a[i],a[minj]=a[minj],a[i]
if i!=minj:c+=1
print(*a)
print(c)
|
s736614679 | p03379 | u497952650 | 2,000 | 262,144 | Wrong Answer | 506 | 26,724 | 271 | When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l. You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i. Find B_i for each i = 1, 2, ..., N. | from copy import deepcopy
N = int(input())
X = list(map(int,input().split()))
Y = deepcopy(X)
Y.sort()
med1 = Y[N//2]
med2 = Y[N//2-1]
print("med1:",med1,"med2",med2)
for i in range(N):
if i == N//2 -1 or i == N//2:
print(med2)
else:
print(med1) | s730314495 | Accepted | 439 | 26,528 | 258 | from copy import deepcopy
N = int(input())
X = list(map(int,input().split()))
Y = deepcopy(X)
Y.sort()
med1 = Y[N//2-1]
med2 = Y[N//2]
##print("med1:",med1,"med2",med2)
for i in range(N):
if X[i] > med1:
print(med1)
else:
print(med2) |
s849077026 | p02401 | u521569208 | 1,000 | 131,072 | Wrong Answer | 20 | 5,616 | 713 | 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. | listA=[]
listOP=[]
listB=[]
count=0
while True:
a,op,b=input().split()
listA.append(int(a))
listOP.append(op)
listB.append(int(b))
if a=="0" and op=="?" and b=="0":
del listA[len(listA)-1]
del listOP[len(listOP)-1]
del listB[len(listB)-1]
break
while count<=len(listA)-1:
if listOP[count]=="+":
print(listA[count]+listB[count])
elif listOP[count]=="-":
print(listA[count]-listB[count])
elif listOP[count]=="*":
print(listA[count]*listB[count])
elif listOP[count]=="/":
print(listA[count]/listB[count])
else:
print("ERROR")
count+=1
| s486564341 | Accepted | 20 | 5,596 | 692 | listA=[]
listOP=[]
listB=[]
count=0
while True:
a,op,b=input().split()
listA.append(int(a))
listOP.append(op)
listB.append(int(b))
if op=="?":
del listA[len(listA)-1]
del listOP[len(listOP)-1]
del listB[len(listB)-1]
break
while count<=len(listA)-1:
if listOP[count]=="+":
print(listA[count]+listB[count])
elif listOP[count]=="-":
print(listA[count]-listB[count])
elif listOP[count]=="*":
print(listA[count]*listB[count])
elif listOP[count]=="/":
print(listA[count]//listB[count])
else:
print("ERROR")
count+=1
|
s505988982 | p03729 | u485566817 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 93 | 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 = map(str, input().split())
print("Yes" if a[-1] == b[0] and b[-1] == c[0] else "No") | s702601874 | Accepted | 17 | 2,940 | 93 | a, b, c = map(str, input().split())
print("YES" if a[-1] == b[0] and b[-1] == c[0] else "NO") |
s668470537 | p03149 | u074445770 | 2,000 | 1,048,576 | Wrong Answer | 18 | 3,064 | 289 | You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974". | s=input()
for i in range(len(s)):
for j in range(i,len(s)):
for k in range(j,len(s)):
for l in range(k,len(s)):
t=s[i:j]+s[k:l+1]
if(t=='keyence'):
print('YES')
exit()
print('NO')
| s720463474 | Accepted | 17 | 2,940 | 103 | n=input().split()
n=list(map(int,n))
n.sort()
if(n==[1,4,7,9]):
print('YES')
else:
print('NO') |
s232568291 | p02607 | u347227232 | 2,000 | 1,048,576 | Wrong Answer | 23 | 9,156 | 121 | We have N squares assigned the numbers 1,2,3,\ldots,N. Each square has an integer written on it, and the integer written on Square i is a_i. How many squares i satisfy both of the following conditions? * The assigned number, i, is odd. * The written integer is odd. | a = input()
b = map(int,a.split())
c = 1
e = 0
for d in b:
if c % 2 == 1:
if d**2 % 2 == 1:
e += 1
c += 1
print(e) | s745391684 | Accepted | 29 | 9,100 | 129 | input()
a = input()
b = map(int,a.split())
c = 1
e = 0
for d in b:
if c % 2 == 1:
if d**2 % 2 == 1:
e += 1
c += 1
print(e) |
s515259171 | p03997 | u691018832 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 231 | 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. | import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(10 ** 7)
a = int(readline())
b = int(readline())
h = int(readline())
print((a + b) * h / 2)
| s388990408 | Accepted | 17 | 3,064 | 214 | import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(10 ** 7)
print((int(readline()) + int(readline())) * int(readline()) // 2)
|
s800225645 | p03471 | u135847648 | 2,000 | 262,144 | Wrong Answer | 1,313 | 3,064 | 252 | 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())
ans_a = -1
ans_b = -1
ans_c = -1
for i in range(n+1):
for j in range(n+1):
x = y - 1000*i - 5000*j
if x % 10000 == 0:
k = x / 10000
ans_a = i
ans_b = j
ans_c = k
print(ans_a,ans_b,ans_c) | s115589137 | Accepted | 811 | 3,060 | 245 | n,y = map(int,input().split())
ans_a = -1
ans_b = -1
ans_c = -1
for i in range(n+1):
for j in range(n+1-i):
k = n- i- j
if y == 10000 * i + 5000 * j + 1000 * k :
ans_a = i
ans_b = j
ans_c = k
print(ans_a,ans_b,ans_c) |
s884665674 | p02394 | u648117624 | 1,000 | 131,072 | Wrong Answer | 20 | 5,604 | 128 | 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. | #coding: UTF-8
W,H,x,y,r = [int(x) for x in input('>').split()]
if r<x<W-r & r<y<H-r:
print('Yes')
else:
print('No')
| s111212172 | Accepted | 20 | 5,596 | 180 | W, H, x, y, r = map(int, input().split())
if r <= x and x <= (W - r):
if r <= y and y <= (H - r):
print("Yes")
else:
print("No")
else:
print("No")
|
s890911639 | p02257 | u798803522 | 1,000 | 131,072 | Wrong Answer | 20 | 7,628 | 246 | 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
primenum =int(input())
ans = 0
for p in range(primenum):
targ = int(input())
for t in range(2,math.floor(math.sqrt(targ)) + 1):
print(t)
if targ % t == 0:
break
else:
ans += 1
print(ans) | s633663050 | Accepted | 860 | 7,668 | 229 | import math
primenum =int(input())
ans = 0
for p in range(primenum):
targ = int(input())
for t in range(2,math.floor(math.sqrt(targ)) + 1):
if targ % t == 0:
break
else:
ans += 1
print(ans) |
s177190884 | p03486 | u697696097 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 153 | 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().strip()
t=input().strip()
s2="".join(sorted(s,reverse=False))
t2="".join(sorted(t,reverse=True))
if s < t:
print("Yes")
else:
print("No") | s785600804 | Accepted | 18 | 2,940 | 155 | s=input().strip()
t=input().strip()
s2="".join(sorted(s,reverse=False))
t2="".join(sorted(t,reverse=True))
if s2 < t2:
print("Yes")
else:
print("No") |
s764960428 | p03860 | u168416324 | 2,000 | 262,144 | Wrong Answer | 27 | 8,972 | 25 | Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name. | print("A"+input()[0]+"C") | s374255993 | Accepted | 27 | 9,080 | 25 | print("A"+input()[8]+"C") |
s183485351 | p03992 | u664481257 | 2,000 | 262,144 | Wrong Answer | 24 | 3,188 | 674 | 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. | # -*- coding: utf-8 -*-
# !/usr/bin/env python
# vim: set fileencoding=utf-8 :
"""
#
# Author: Noname
# URL: https://github.com/pettan0818
# License: MIT License
# Created: 2016-09-28
#
# Usage
#
"""
from __future__ import division
from __future__ import unicode_literals
from __future__ import print_function
from __future__ import absolute_import
def input_process():
return input()
def insert_space(target_str: str) -> str:
return target_str[0:3] + " " + target_str[3:]
if __name__ == "__main__":
target = input()
print(insert_space(target))
| s848627656 | Accepted | 23 | 3,064 | 674 | # -*- coding: utf-8 -*-
# !/usr/bin/env python
# vim: set fileencoding=utf-8 :
"""
#
# Author: Noname
# URL: https://github.com/pettan0818
# License: MIT License
# Created: 2016-09-28
#
# Usage
#
"""
from __future__ import division
from __future__ import unicode_literals
from __future__ import print_function
from __future__ import absolute_import
def input_process():
return input()
def insert_space(target_str: str) -> str:
return target_str[0:4] + " " + target_str[4:]
if __name__ == "__main__":
target = input()
print(insert_space(target))
|
s485156082 | p02413 | u075836834 | 1,000 | 131,072 | Wrong Answer | 20 | 7,636 | 346 | Your task is to perform a simple table calculation. Write a program which reads the number of rows r, columns c and a table of r × c elements, and prints a new table, which includes the total sum for each row and column. | height,width=map(int,input().split())
A=[]
for i in range(height):
A.append([int(j) for j in input().split()])
for i in range(height):
print(' '.join(map(str,A[i])),str(sum(A[i])))
B=[]
C=[0 for _ in range(width)]
for i in range(width):
for j in range(height):
s=0
s+=A[j][i]
B.append(s)
C[i]=sum(B)
B=[]
print(' '.join(map(str,C))) | s441157257 | Accepted | 30 | 7,784 | 358 | height,width=map(int,input().split())
A=[]
for i in range(height):
A.append([int(j) for j in input().split()])
for i in range(height):
print(' '.join(map(str,A[i])),str(sum(A[i])))
B=[]
C=[0 for _ in range(width)]
for i in range(width):
for j in range(height):
s=0
s+=A[j][i]
B.append(s)
C[i]=sum(B)
B=[]
print(' '.join(map(str,C)),str(sum(C))) |
s741948734 | p02412 | u350064373 | 1,000 | 131,072 | Wrong Answer | 20 | 7,540 | 674 | 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 | while True:
n, x = map(int,input().split())
count=0
if n == x == 0:
break
else:
for s in range(1,n+1):
num1 = s
for j in (1,n+1):
if num1 == j:
break
else:
num2 = j
if x < num1+num2:
break
for k in (1,n+1):
if num1 == k:
break
elif num2 == k:
break
elif num1+num2+k == x:
count+=1
else:
pass
print(count) | s384247750 | Accepted | 510 | 7,656 | 727 | while True:
n, x = map(int,input().split())
count=0
if n == x == 0:
break
else:
ls = []
for i in range(1,n+1):
ls.append(i)
for s in ls:
num1 = s
for j in ls:
if num1 == j:
break
else:
num2 = j
if x < num1+num2:
break
for k in ls:
if num1 == k:
break
elif num2 == k:
break
elif num1+num2+k == x:
count+=1
else:
pass
print(count) |
s530180650 | p04029 | u557494880 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 39 | 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())
S = N*(N+1)/2
print(S) | s562361306 | Accepted | 17 | 2,940 | 41 | N = int(input())
S = N*(N+1)//2
print(S)
|
s288936368 | p03394 | u630941334 | 2,000 | 262,144 | Wrong Answer | 31 | 5,212 | 930 | Nagase is a top student in high school. One day, she's analyzing some properties of special sets of positive integers. She thinks that a set S = \\{a_{1}, a_{2}, ..., a_{N}\\} of **distinct** positive integers is called **special** if for all 1 \leq i \leq N, the gcd (greatest common divisor) of a_{i} and the sum of the remaining elements of S is **not** 1. Nagase wants to find a **special** set of size N. However, this task is too easy, so she decided to ramp up the difficulty. Nagase challenges you to find a **special** set of size N such that the gcd of all elements are 1 and the elements of the set does not exceed 30000. | import sys
if __name__ == '__main__':
N = int(input())
if N == 3:
print(2, 5, 23)
sys.exit()
elif N == 4:
print(2, 5, 20, 63)
sys.exit()
elif N == 5:
print(2, 3, 4, 6, 9)
sys.exit()
ans = []
for i in range(1, 30001):
if i % 2 == 0 or i % 3 == 0:
ans.append(i)
if len(ans) == N:
amari = N % 8
if amari == 1 or amari == 6:
j = i + 1
while j % 6 != 0:
j += 1
ans[4] = j
elif amari == 2 or amari == 5:
j = i + 1
while j % 6 != 3:
j += 1
ans[4] = j
elif amari == 3 or amari == 4:
j = i + 1
while j % 6 != 0:
j += 1
ans[5] = j
break
print(' '.join(map(str, ans)))
| s115831714 | Accepted | 28 | 5,084 | 917 |
def solve(n):
if n == 3:
return [2, 5, 63]
elif n == 4:
return [2, 5, 20, 63]
elif n == 5:
return [2, 3, 4, 6, 9]
ans = []
for i in range(1, 30001):
if i % 2 == 0 or i % 3 == 0:
ans.append(i)
if len(ans) == n:
amari = n % 8
if amari == 1 or amari == 6:
j = i + 1
while j % 6 != 0:
j += 1
ans[4] = j
elif amari == 2 or amari == 5:
j = i + 1
while j % 6 != 3:
j += 1
ans[4] = j
elif amari == 3 or amari == 4:
j = i + 1
while j % 6 != 0:
j += 1
ans[5] = j
break
return ans
if __name__ == '__main__':
N = int(input())
ans = solve(N)
print(' '.join(map(str, ans)))
|
s231720712 | p03149 | u391819434 | 2,000 | 1,048,576 | Wrong Answer | 30 | 9,068 | 61 | You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974". | print('YNeos'[sorted(input().split())!=['1','4','7','9']::2]) | s889246524 | Accepted | 26 | 8,920 | 61 | print('YNEOS'[sorted(input().split())!=['1','4','7','9']::2]) |
s979461278 | p03399 | u085186789 | 2,000 | 262,144 | Wrong Answer | 25 | 9,008 | 80 | You planned a trip using trains and buses. The train fare will be A yen (the currency of Japan) if you buy ordinary tickets along the way, and B yen if you buy an unlimited ticket. Similarly, the bus fare will be C yen if you buy ordinary tickets along the way, and D yen if you buy an unlimited ticket. Find the minimum total fare when the optimal choices are made for trains and buses. | A = int(input())
B = int(input())
C = int(input())
D = int(input())
print(B + D) | s474679248 | Accepted | 32 | 8,928 | 97 | A = int(input())
B = int(input())
C = int(input())
D = int(input())
print(min(A, B) + min(C, D))
|
s425739582 | p00001 | u153455779 | 1,000 | 131,072 | Wrong Answer | 20 | 5,600 | 97 | There is a data which provides heights (in meter) of mountains. The data is only for ten mountains. Write a program which prints heights of the top three mountains in descending order. | mh=[]
for i in range(1,11):
mh.append(int(input()))
mh.sort(reverse=True)
print(mh[0:3])
| s243463333 | Accepted | 20 | 5,596 | 121 | mh=[]
for i in range(1,11):
mh.append(int(input()))
mh.sort(reverse=True)
print(mh[0])
print(mh[1])
print(mh[2])
|
s569577426 | p03610 | u780354103 | 2,000 | 262,144 | Wrong Answer | 18 | 3,188 | 27 | 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. | s = input()
print(s[1::2]) | s889819013 | Accepted | 18 | 3,192 | 19 | print(input()[::2]) |
s562085009 | p02389 | u569672348 | 1,000 | 131,072 | Wrong Answer | 20 | 7,440 | 31 | Write a program which calculates the area and perimeter of a given rectangle. | a=3
b=5
print(a*b," ",2*a+3*b) | s217982535 | Accepted | 20 | 5,576 | 63 | i = input()
a,b = list(map(int,i.split()))
print(a*b,2*a+2*b)
|
s252946346 | p03385 | u220345792 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 79 | 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()
if 'a' in S and 'b' in S and 'c' in S: print('YES')
else: print('No') | s987718223 | Accepted | 17 | 2,940 | 79 | S=input()
if 'a' in S and 'b' in S and 'c' in S: print('Yes')
else: print('No') |
s845859983 | p03759 | u627089907 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 133 | 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. | #print(" ",sep='',end='')
s=input().split()
a=int(s[0])
b=int(s[1])
c=int(s[2])
if b-a==c-b:
print("Yes")
else :
print("No") | s452047605 | Accepted | 18 | 3,060 | 133 | #print(" ",sep='',end='')
s=input().split()
a=int(s[0])
b=int(s[1])
c=int(s[2])
if b-a==c-b:
print("YES")
else :
print("NO") |
s788776145 | p03457 | u148981246 | 2,000 | 262,144 | Wrong Answer | 615 | 9,232 | 412 | AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan. | n = int(input())
prev_t, prev_x, prev_y = 0, 0, 0
for _ in range(n):
t, x, y = map(int, input().split())
print('#1', t, x, y)
if (abs(x - prev_x) + abs(y - prev_y) <= t - prev_t and
(abs(x - prev_x) + abs(y - prev_y)) % 2 == (t - prev_t) % 2):
prev_t, prev_x, prev_y = t, x, y
print(t, x, y)
continue
else:
print('No')
break
else:
print('Yes') | s434262144 | Accepted | 240 | 9,112 | 364 | n = int(input())
prev_t, prev_x, prev_y = 0, 0, 0
for _ in range(n):
t, x, y = map(int, input().split())
if (abs(x - prev_x) + abs(y - prev_y) <= t - prev_t and
(abs(x - prev_x) + abs(y - prev_y)) % 2 == (t - prev_t) % 2):
prev_t, prev_x, prev_y = t, x, y
continue
else:
print('No')
break
else:
print('Yes') |
s334133008 | p03434 | u551109821 | 2,000 | 262,144 | Wrong Answer | 23 | 3,444 | 188 | 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 = list(map(int, input().split()))
a.sort(reverse=True)
sumA = 0
sumB = 0
for i in range(len(a)):
if i % 2 == 0:
sumA += a[i]
else:
sumB += a[i]
| s744641372 | Accepted | 17 | 3,060 | 206 | N = int(input())
a = list(map(int, input().split()))
a.sort(reverse=True)
sumA = 0
sumB = 0
for i in range(len(a)):
if i % 2 == 0:
sumA += a[i]
else:
sumB += a[i]
print(sumA-sumB)
|
s181909029 | p03583 | u111285482 | 2,000 | 262,144 | Wrong Answer | 2,205 | 9,120 | 300 | You are given an integer N. Find a triple of positive integers h, n and w such that 4/N = 1/h + 1/n + 1/w. If there are multiple solutions, any of them will be accepted. | N = int(input())
mx = 3503
found = False
for i in range(1, mx):
for j in range(1, mx):
num = N*i*j
den = (i*j*4 - N*i - N*j)
if den != 0 and num%den == 0 and num/den > 0:
print(i,j,num/den)
found = True
break
if found:
break | s856943622 | Accepted | 1,710 | 9,180 | 396 | N = int(input())
mx = 3503
found = False
for i in range(1, mx):
for j in range(1, mx):
if i*j*4 % N != 0:
continue
den = ((i*j*4)//N) - i - j
num = i*j
if den == 0:
continue
if num%den == 0 and num//den > 0 and num//den < mx:
print(i,j,num//den)
found = True
break
if found:
break |
s493991094 | p02612 | u777922433 | 2,000 | 1,048,576 | Wrong Answer | 31 | 9,132 | 33 | 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)
| s285032530 | Accepted | 25 | 9,140 | 92 | n = int(input())
if((n % 1000) == 0):
print(n % 1000)
exit()
print(1000 - n % 1000)
|
s985906757 | p02927 | u798514691 | 2,000 | 1,048,576 | Wrong Answer | 18 | 2,940 | 186 | Today is August 24, one of the five Product Days in a year. A date m-d (m is the month, d is the date) is called a Product Day when d is a two-digit number, and all of the following conditions are satisfied (here d_{10} is the tens digit of the day and d_1 is the ones digit of the day): * d_1 \geq 2 * d_{10} \geq 2 * d_1 \times d_{10} = m Takahashi wants more Product Days, and he made a new calendar called Takahashi Calendar where a year consists of M month from Month 1 to Month M, and each month consists of D days from Day 1 to Day D. In Takahashi Calendar, how many Product Days does a year have? | def main():
M,D = map(int,input().split())
if D < 22:
print(0)
return
cnt = 0
for i in range(D):
if (D/10)*(D%10) < M:
cnt += 1
print(cnt)
main() | s835541094 | Accepted | 17 | 3,060 | 247 | import math
def main():
M,D = map(int,input().split())
if D < 22:
print(0)
return
cnt = 0
for i in range(D + 1):
if 2 <= (i/10) and 2 <= (i%10) and math.floor(i/10)*(i%10) <= M:
cnt += 1
print(cnt)
main() |
s693928980 | p03713 | u371686382 | 2,000 | 262,144 | Wrong Answer | 156 | 3,064 | 874 | There is a bar of chocolate with a height of H blocks and a width of W blocks. Snuke is dividing this bar into exactly three pieces. He can only cut the bar along borders of blocks, and the shape of each piece must be a rectangle. Snuke is trying to divide the bar as evenly as possible. More specifically, he is trying to minimize S_{max} \- S_{min}, where S_{max} is the area (the number of blocks contained) of the largest piece, and S_{min} is the area of the smallest piece. Find the minimum possible value of S_{max} - S_{min}. | H, W = list(map(int, input().split()))
ans = float('inf')
# only need to consider the horizontal length of the cut at the beginning.
# the vertical length must be cut in the center. (if possible.)
# if you cut outside of center, you will be father away from the answer.
if H % 2 == 0:
for x in range(W):
S1 = x * H
S2 = (W - x) * (H / 2)
ans = min(ans, abs(S1 - S2))
if W % 2 == 0:
for x in range(H):
S1 = W * x
S2 = (H - x) * (W / 2)
ans = min(ans, abs(S1 - S2))
else:
# both sides are odd.
for x in range(W):
S1 = x * H
S2 = (W - x) * (H // 2)
S3 = (W - x) * ((H // 2) + 1)
# if both sides are odd, three number appear.
# it is not clear which of the three numbers is the largest or smallest, it need to be caluculated.
S_max = max(S1, S3)
S_min = min(S1, S2)
ans = min(ans, S_max - S_min)
print(int(ans)) | s006796699 | Accepted | 471 | 3,064 | 834 | H, W = list(map(int, input().split()))
ans = float('inf')
# first cut vertical
for x in range(1, W):
S1 = x * H
# cut vertical
w2 = (W - x) // 2
w3 = W - x - w2
S2 = w2 * H
S3 = w3 * H
S_max = max(S1, S3)
S_min = min(S1, S2)
ans = min(ans, S_max - S_min)
h2 = H // 2
h3 = H - h2
S2 = (W - x) * h2
S3 = (W - x) * h3
S_max = max(S1, S3)
S_min = min(S1, S2)
ans = min(ans, S_max - S_min)
for y in range(1, H):
S1 = W * y
# cut vertical
w2 = W // 2
w3 = W - w2
S2 = w2 * (H - y)
S3 = w3 * (H - y)
S_max = max(S1, S3)
S_min = min(S1, S2)
ans = min(ans, S_max - S_min)
h2 = (H - y) // 2
h3 = H - y - h2
S2 = W * h2
S3 = W * h3
S_max = max(S1, S3)
S_min = min(S1, S2)
ans = min(ans, S_max - S_min)
print(int(ans)) |
s431816021 | p03193 | u368796742 | 2,000 | 1,048,576 | Wrong Answer | 31 | 9,188 | 154 | There are N rectangular plate materials made of special metal called AtCoder Alloy. The dimensions of the i-th material are A_i \times B_i (A_i vertically and B_i horizontally). Takahashi wants a rectangular plate made of AtCoder Alloy whose dimensions are exactly H \times W. He is trying to obtain such a plate by choosing one of the N materials and cutting it if necessary. When cutting a material, the cuts must be parallel to one of the sides of the material. Also, the materials have fixed directions and cannot be rotated. For example, a 5 \times 3 material cannot be used as a 3 \times 5 plate. Out of the N materials, how many can produce an H \times W plate if properly cut? | n,h,w = map(int,input().split())
count = 0
for i in range(n):
a,b = map(int,input().split())
if h >= a and w >= b:
count += 1
print(count) | s593626449 | Accepted | 31 | 9,192 | 154 | n,h,w = map(int,input().split())
count = 0
for i in range(n):
a,b = map(int,input().split())
if a >= h and b >= w:
count += 1
print(count) |
s971089396 | p03610 | u867848444 | 2,000 | 262,144 | Wrong Answer | 18 | 3,192 | 17 | 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. | s=input()
s[0::2] | s033505022 | Accepted | 29 | 3,188 | 77 | s = input()
ans = ''
for i in range(0, len(s), 2):
ans += s[i]
print(ans) |
s660317658 | p03855 | u597455618 | 2,000 | 262,144 | Wrong Answer | 643 | 15,268 | 1,157 | There are N cities. There are also K roads and L railways, extending between the cities. The i-th road bidirectionally connects the p_i-th and q_i-th cities, and the i-th railway bidirectionally connects the r_i-th and s_i-th cities. No two roads connect the same pair of cities. Similarly, no two railways connect the same pair of cities. We will say city A and B are _connected by roads_ if city B is reachable from city A by traversing some number of roads. Here, any city is considered to be connected to itself by roads. We will also define _connectivity by railways_ similarly. For each city, find the number of the cities connected to that city by both roads and railways. | class UnionFind:
def __init__(self, n):
self.table = [-1] * n
def _root(self, x):
stack = []
tbl = self.table
while tbl[x] >= 0:
stack.append(x)
x = tbl[x]
for y in stack:
tbl[y] = x
return x
def find(self, x, y):
return self._root(x) == self._root(y)
def union(self, x, y):
r1 = self._root(x)
r2 = self._root(y)
if r1 == r2:
return
d1 = self.table[r1]
d2 = self.table[r2]
if d1 <= d2:
self.table[r2] = r1
self.table[r1] += d2
else:
self.table[r1] = r2
self.table[r2] += d1
def main():
n, k, l = map(int, input().split())
a = UnionFind(n)
b = UnionFind(n)
for _ in range(k):
p, q = map(int, input().split())
a.union(p-1, q-1)
for _ in range(l):
r, s = map(int, input().split())
b.union(r-1, s-1)
for i in range(n):
if a.find(b.table[i], a.table[i]):
print(2, end=" ")
else:
print(1, end=" ")
if __name__ == '__main__':
main()
| s606181546 | Accepted | 782 | 52,524 | 284 | from collections import *;M=lambda:map(int,input().split());r=range;n,k,l=M()
def f(m):
*a,=r(n)
def g(i):
if a[i]==i:return i
a[i]=g(a[i]);return a[i]
for _ in r(m):p,q=M();a[g(q-1)]=g(p-1)
return [g(i) for i in r(n)]
z=[*zip(f(k),f(l))];c=Counter(z);print(*(c[x]for x in z)) |
s604130553 | p02415 | u682153677 | 1,000 | 131,072 | Wrong Answer | 20 | 5,540 | 70 | Write a program which converts uppercase/lowercase letters to lowercase/uppercase for a given string. | # -*- coding: utf-8 -*-
char = input()
char.swapcase()
print(char)
| s027995764 | Accepted | 20 | 5,544 | 63 | # -*- coding: utf-8 -*-
char = input()
print(char.swapcase())
|
s582697607 | p03698 | u405660020 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 57 | You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different. | s=input()
print('Yes' if len(s)==len(set(s)) else 'No')
| s748847930 | Accepted | 17 | 2,940 | 61 | s = input()
print('yes' if len(s) == len(set(s)) else 'no')
|
s471740765 | p03162 | u359971344 | 2,000 | 1,048,576 | Wrong Answer | 877 | 47,332 | 446 | Taro's summer vacation starts tomorrow, and he has decided to make plans for it now. The vacation consists of N days. For each i (1 \leq i \leq N), Taro will choose one of the following activities and do it on the i-th day: * A: Swim in the sea. Gain a_i points of happiness. * B: Catch bugs in the mountains. Gain b_i points of happiness. * C: Do homework at home. Gain c_i points of happiness. As Taro gets bored easily, he cannot do the same activities for two or more consecutive days. Find the maximum possible total points of happiness that Taro gains. | N = int(input())
matrix = []
for _ in range(N):
matrix.append(list(map(int, input().split())))
dp_matrix = [[0, 0, 0]]
for i in range(N):
dp_matrix_list = [0, 0, 0]
for j in range(3):
for k in range(3):
if (j != k) and (dp_matrix_list[j] < dp_matrix[i][j] + matrix[i][k]):
dp_matrix_list[j] = dp_matrix[i][j] + matrix[i][k]
dp_matrix.append(dp_matrix_list)
print(max(dp_matrix[-1])) | s458304712 | Accepted | 894 | 48,468 | 446 | N = int(input())
matrix = []
for _ in range(N):
matrix.append(list(map(int, input().split())))
dp_matrix = [[0, 0, 0]]
for i in range(N):
dp_matrix_list = [0, 0, 0]
for j in range(3):
for k in range(3):
if (j != k) and (dp_matrix_list[j] < dp_matrix[i][k] + matrix[i][j]):
dp_matrix_list[j] = dp_matrix[i][k] + matrix[i][j]
dp_matrix.append(dp_matrix_list)
print(max(dp_matrix[-1])) |
s253476083 | p03380 | u203748490 | 2,000 | 262,144 | Wrong Answer | 89 | 14,052 | 124 | Let {\rm comb}(n,r) be the number of ways to choose r objects from among n objects, disregarding order. From n non-negative integers a_1, a_2, ..., a_n, select two numbers a_i > a_j so that {\rm comb}(a_i,a_j) is maximized. If there are multiple pairs that maximize the value, any of them is accepted. | n = int(input())
an = [int(i) for i in input().split()]
ai = max(an)
print("%d %d"%(ai, min(an, key=lambda i:abs(ai-i**2)))) | s229683811 | Accepted | 102 | 14,052 | 132 | n = int(input())
an = [int(i) for i in input().split()]
an.sort()
ai = an[-1]
print("%d %d"%(ai, min(an, key=lambda i:abs(ai-i*2)))) |
s669432486 | p02557 | u717792745 | 2,000 | 1,048,576 | Wrong Answer | 222 | 43,128 | 755 | Given are two sequences A and B, both of length N. A and B are each sorted in the ascending order. Check if it is possible to reorder the terms of B so that for each i (1 \leq i \leq N) A_i \neq B_i holds, and if it is possible, output any of the reorderings that achieve it. | n=int(input())
import sys
a=list(map(int,input().split()))
b=list(map(int,input().split()))
d={}
if(n==1):
if(a==b): print('No')
else: print('Yes')
sys.exit()
possible=True
for x in a:
if(d.get(x)==None):
d[x]=1
else:
d[x]+=1
if(d[x]>(n)//2):
possible=False
break
if(not possible):
print('No')
else:
b.sort(reverse=True)
k=0
j=n-1
turn=True
for i in range(0,n):
# print(turn)
if(b[i]==a[i]):
if(turn):
# swap(a[k],a[i])
b[k],b[i]=b[i],b[k]
k+=1
else:
# swap(a[j],a[i])
b[j], b[i] = b[i], b[j]
j-=1
turn=not turn
print(b)
| s248861626 | Accepted | 362 | 69,032 | 404 | import collections
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
A=collections.Counter(a)
B=collections.Counter(b)
inf=10e18
cc,cd=0,0
R=0
for i in range(n+1):
if(A[i]+B[i]>n):
R=inf
else:
cc+=A[i]
R=max(R,cc-cd)
cd+=B[i]
if(R==inf):
print("No")
else:
print("Yes")
ans=" ".join(map(str,b[-R:]+b[:-R]))
print(ans) |
s095407755 | p04035 | u373958718 | 2,000 | 262,144 | Wrong Answer | 260 | 23,116 | 244 | We have N pieces of ropes, numbered 1 through N. The length of piece i is a_i. At first, for each i (1≤i≤N-1), piece i and piece i+1 are tied at the ends, forming one long rope with N-1 knots. Snuke will try to untie all of the knots by performing the following operation repeatedly: * Choose a (connected) rope with a total length of at least L, then untie one of its knots. Is it possible to untie all of the N-1 knots by properly applying this operation? If the answer is positive, find one possible order to untie the knots. | import numpy as np
n,l=map(int,input().split())
a=list(map(int,input().split()))
index=np.argsort(a)
s=sum(a)
for x in index:
if s-a[x]>=l:
s-=a[x]
else:
print("Impossible")
exit(0)
print("Possible")
for x in index:
print(x+1) | s468860905 | Accepted | 277 | 23,104 | 309 | import numpy as np
n,l=map(int,input().split())
a=list(map(int,input().split()))
flag=False;idx=0
for i in range(n-1):
if a[i]+a[i+1]>=l:
flag=True
idx=i
if not flag:
print("Impossible")
exit(0)
print("Possible")
for x in range(idx):
print(x+1)
for x in reversed(range(idx,n-1)):
print(x+1)
|
s803540474 | p03455 | u077019541 | 2,000 | 262,144 | Wrong Answer | 21 | 3,060 | 86 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | a,b = map(int,input().split())
if (a*b)%2==0:
print("Odd")
else:
print("Even") | s168333699 | Accepted | 18 | 2,940 | 86 | a,b = map(int,input().split())
if (a*b)%2==0:
print("Even")
else:
print("Odd") |
s798072921 | p02392 | u626266743 | 1,000 | 131,072 | Wrong Answer | 20 | 7,612 | 88 | Write a program which reads three integers a, b and c, and prints "Yes" if a < b < c, otherwise "No". | a, b, c = map(int, input().split())
if a > b > c:
print("Yes")
else:
print("No") | s903511116 | Accepted | 30 | 7,692 | 88 | a, b, c = map(int, input().split())
if a < b < c:
print("Yes")
else:
print("No") |
s055951159 | p03377 | u536325690 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 95 | 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())
if a > x or a+b < x:
print('No')
else:
print('Yes') | s973289615 | Accepted | 17 | 2,940 | 95 | a, b, x = map(int, input().split())
if a > x or a+b < x:
print('NO')
else:
print('YES') |
s255329657 | p02389 | u482227082 | 1,000 | 131,072 | Wrong Answer | 20 | 5,572 | 47 | Write a program which calculates the area and perimeter of a given rectangle. | a, b = input().split()
print (int(a) * int(b))
| s857219763 | Accepted | 20 | 5,592 | 124 | #
# 1c
#
def main():
a, b = map(int, input().split())
print(a*b, 2*(a+b))
if __name__ == '__main__':
main()
|
s408407693 | p04044 | u779728630 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 142 | 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. | N, L = map(int, input().split())
S = [""]
for i in range(N):
S.append(input())
S.sort()
res = ""
for i in range(N):
res += S[i]
print(res) | s026354095 | Accepted | 17 | 3,060 | 140 | N, L = map(int, input().split())
S = []
for i in range(N):
S.append(input())
S.sort()
res = ""
for i in range(N):
res += S[i]
print(res) |
s352774439 | p03626 | u006657459 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 659 | We have a board with a 2 \times N grid. Snuke covered the board with N dominoes without overlaps. Here, a domino can cover a 1 \times 2 or 2 \times 1 square. Then, Snuke decided to paint these dominoes using three colors: red, cyan and green. Two dominoes that are adjacent by side should be painted by different colors. Here, it is not always necessary to use all three colors. Find the number of such ways to paint the dominoes, modulo 1000000007. The arrangement of the dominoes is given to you as two strings S_1 and S_2 in the following manner: * Each domino is represented by a different English letter (lowercase or uppercase). * The j-th character in S_i represents the domino that occupies the square at the i-th row from the top and j-th column from the left. | N = int(input())
S1 = input()
S2 = input()
patterns = []
flag = False
for i in range(N):
if S1[i] == S2[i]:
patterns.append('row')
elif flag is False:
patterns.append('column')
flag = True
else:
flag = False
print(patterns)
if patterns[0] == 'row':
count = 3
else:
count = 6
for i in range(1, len(patterns)):
prev = patterns[i-1]
current = patterns[i]
if prev == 'row':
if current == 'row':
count *= 2
else:
count *= 2
else: # column
if current == 'row':
count *= 1
else:
count *= 3
print(count % (10**9 + 7)) | s348923116 | Accepted | 18 | 3,064 | 617 | N = int(input())
S1 = input()
S2 = input()
if S1[0] == S2[0]:
count = 3
start_idx = 1
prev = 'v'
else:
count = 6
start_idx = 2
prev = 'h'
skip = False
for i in range(start_idx, len(S1)):
if skip:
skip = False
continue
if S1[i] == S2[i]: # current = v
if prev == 'v':
count *= 2
prev = 'v'
else:
count *= 1
prev = 'v'
else: # current = h
if prev == 'v':
count *= 2
else:
count *= 3
prev = 'h'
skip = True
print(count % (10**9 + 7))
|
s652118341 | p03472 | u497046426 | 2,000 | 262,144 | Wrong Answer | 2,104 | 11,804 | 378 | You are going out for a walk, when you suddenly encounter a monster. Fortunately, you have N katana (swords), Katana 1, Katana 2, …, Katana N, and can perform the following two kinds of attacks in any order: * Wield one of the katana you have. When you wield Katana i (1 ≤ i ≤ N), the monster receives a_i points of damage. The same katana can be wielded any number of times. * Throw one of the katana you have. When you throw Katana i (1 ≤ i ≤ N) at the monster, it receives b_i points of damage, and you lose the katana. That is, you can no longer wield or throw that katana. The monster will vanish when the total damage it has received is H points or more. At least how many attacks do you need in order to vanish it in total? | N, H = map(int, input().split())
A = []
B = []
for i in range(N):
a, b = map(int, input().split())
A.append(a)
B.append(b)
A.sort()
A = A[::-1]
a_max = A[0]
B.sort()
B = B[::-1]
attack = 0
idx = 0
while B[idx] > a_max:
H -= B[idx]
attack += 1
if H <= 0:
print(attack)
break
if H > 0:
attack += (H // a_max) + 1
print(attack) | s946858099 | Accepted | 354 | 12,860 | 373 | import math
N, H = map(int, input().split())
A = []
B = []
for _ in range(N):
a, b = map(int, input().split())
A.append(a)
B.append(b)
a_max = max(A)
B = sorted([x for x in B if x > a_max], reverse = True)
attack = 0
for damage in B:
H -= damage
attack += 1
if H <= 0:
break
if H > 0:
attack += math.ceil(H / a_max)
print(attack) |
s906914943 | p03469 | u647767910 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 37 | On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it. | s = list(input())
s[3] = '8'
print(s) | s009690457 | Accepted | 17 | 2,940 | 31 | s = input()
print('2018'+s[4:]) |
s004273927 | p03449 | u635974378 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 154 | 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())
A1, A2 = list(map(int, input().split())), list(map(int, input().split()))
print(max([A1[0] + sum(A1[:i] + A2[i:]) for i in range(N)]))
| s398081897 | Accepted | 17 | 2,940 | 150 | N = int(input())
A1, A2 = list(map(int, input().split())), list(map(int, input().split()))
print(max([sum(A1[:i + 1] + A2[i:]) for i in range(N)]))
|
s253629869 | p03053 | u945228737 | 1,000 | 1,048,576 | Wrong Answer | 1,057 | 18,036 | 848 | You are given a grid of squares with H horizontal rows and W vertical columns, where each square is painted white or black. HW characters from A_{11} to A_{HW} represent the colors of the squares. A_{ij} is `#` if the square at the i-th row from the top and the j-th column from the left is black, and A_{ij} is `.` if that square is white. We will repeatedly perform the following operation until all the squares are black: * Every white square that shares a side with a black square, becomes black. Find the number of operations that will be performed. The initial grid has at least one black square. | def solve():
H, W = map(int, input().split())
mass = [[0] * W for _ in range(H)]
import pprint
pprint.pprint(mass)
for Hi in range(H):
massHi = [0 if s == '.' else 1 for s in list(input())]
print(massHi)
for Wi in range(W):
mass[Hi][Wi] = massHi[Wi]
import pprint
pprint.pprint(mass)
result = 0
for Hi in range(H):
for Wi in range(W):
if mass[Hi][Wi] == 1:
break
sub_result = 100000
for Hi2 in range(H):
for Wi2 in range(W):
if mass[Hi2][Wi2] == 1:
sub_result = min(sub_result,
abs(Hi2 - Hi) + abs(Wi2 - Wi))
result = max(result, sub_result)
print(result)
if __name__ == '__main__':
solve()
| s049672884 | Accepted | 752 | 106,608 | 1,493 |
from collections import deque
def solve():
H, W = map(int, input().split())
# mass = [[False] * W for _ in range(H)]
queue1 = []
queue2 = []
mass = []
mass.append([False] * (W + 2))
for Hi in range(1, H + 1):
mass.append([False] + [s == '.' for s in list(input())] + [False])
for Wi in range(1, W + 1):
if not mass[Hi][Wi]:
queue1.append((Hi, Wi))
mass.append([False] * (W + 2))
import datetime
d = datetime.datetime.now()
count = -1
while True:
if len(queue1) == 0:
break
for h, w in queue1:
# left
if mass[h][w - 1]:
mass[h][w - 1] = False
queue2.append((h, w - 1))
# right
if mass[h][w + 1]:
mass[h][w + 1] = False
queue2.append((h, w + 1))
# top
if mass[h + 1][w]:
mass[h + 1][w] = False
queue2.append((h + 1, w))
# button
if mass[h - 1][w]:
mass[h - 1][w] = False
queue2.append((h - 1, w))
count += 1
queue1 = queue2
queue2 = []
print(count)
# print(datetime.datetime.now() - d)
if __name__ == '__main__':
solve()
|
s279230445 | p02390 | u966110132 | 1,000 | 131,072 | Wrong Answer | 20 | 5,568 | 33 | Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively. | a = input()
print(a*3600,60*a,a)
| s886127065 | Accepted | 20 | 5,576 | 66 | t=int(input())
print("{0}:{1}:{2}".format(t//3600,t//60%60,t%60))
|
s514685710 | p03696 | u863442865 | 2,000 | 262,144 | Wrong Answer | 22 | 3,316 | 808 | You are given a string S of length N consisting of `(` and `)`. Your task is to insert some number of `(` and `)` into S to obtain a _correct bracket sequence_. Here, a correct bracket sequence is defined as follows: * `()` is a correct bracket sequence. * If X is a correct bracket sequence, the concatenation of `(`, X and `)` in this order is also a correct bracket sequence. * If X and Y are correct bracket sequences, the concatenation of X and Y in this order is also a correct bracket sequence. * Every correct bracket sequence can be derived from the rules above. Find the shortest correct bracket sequence that can be obtained. If there is more than one such sequence, find the lexicographically smallest one. | def main():
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10000000)
from collections import Counter, deque
#from collections import defaultdict
from itertools import combinations, permutations
#from itertools import accumulate, product
from bisect import bisect_left,bisect_right
from math import floor, ceil
#from operator import itemgetter
#mod = 1000000007
N = int(input())
S = list(input())
res = 0
for i in S:
if i == ')':
res += 1
else:
break
add = S[res:].count('(')-S[res:].count(')')
if add>=0:
for _ in range(add):
S.append(')')
print('('*res+''.join(S))
else:
print('('*(res-add)+''.join(S))
if __name__ == '__main__':
main() | s121262742 | Accepted | 21 | 3,444 | 1,116 | def main():
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10000000)
from collections import Counter, deque
#from collections import defaultdict
from itertools import combinations, permutations
#from itertools import accumulate, product
from bisect import bisect_left,bisect_right
from math import floor, ceil
#from operator import itemgetter
#mod = 1000000007
N = int(input())
S = input().rstrip()
t = S
while 1:
a = len(t)
s = t.split('()')
s = ''.join(s)
if len(s)==a:
break
t = s
left = s.count(')')
right = s.count('(')
print('('*left + S + ')'*right)
# N = int(input())
# S = list(input())
# res = 0
# res += 1
# else:
# break
# add = S[res:].count('(')-S[res:].count(')')
# if add>=0:
# for _ in range(add):
# S.append(')')
# print('('*res+''.join(S))
# else:
# print('('*(res-add)+''.join(S))
if __name__ == '__main__':
main() |
s025736036 | p02401 | u175111751 | 1,000 | 131,072 | Wrong Answer | 40 | 7,376 | 274 | 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. | def proc(a, b, op):
if b == '+':
return a + b
if b == '-':
return a - b
if b == '*':
return a * b
if b == '/':
return a // b
while True:
a, op, b = input().split()
if op == '?':
break
print(proc(a, b, op)) | s540074019 | Accepted | 30 | 7,708 | 290 | def proc(a, b, op):
if op == '+':
return a + b
if op == '-':
return a - b
if op == '*':
return a * b
if op == '/':
return a // b
while True:
a, op, b = input().split()
if op == '?':
break
print(proc(int(a), int(b), op)) |
s322606907 | p03449 | u851706118 | 2,000 | 262,144 | Wrong Answer | 33 | 3,752 | 282 | 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)]
res = 0
for i in range(N):
s = 0
for j in range(i+1):
s += A[0][j]
print('j:', s)
for k in range(i, N):
s += A[1][k]
print('i:', s)
res = max(res, s)
print(res) | s584687290 | Accepted | 18 | 3,060 | 239 | N = int(input())
A = [list(map(int, input().split())) for i in range(2)]
res = 0
for i in range(N):
s = 0
for j in range(i + 1):
s += A[0][j]
for k in range(i, N):
s += A[1][k]
res = max(res, s)
print(res)
|
s469695673 | p03564 | u703890795 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 81 | Square1001 has seen an electric bulletin board displaying the integer 1. He can perform the following operations A and B to change this value: * Operation A: The displayed value is doubled. * Operation B: The displayed value increases by K. Square1001 needs to perform these operations N times in total. Find the minimum possible value displayed in the board after N operations. | N = int(input())
K = int(input())
for i in range(N):
N = min(N+K, N*2)
print(N) | s629273311 | Accepted | 17 | 2,940 | 87 | N = int(input())
K = int(input())
m = 1
for i in range(N):
m = min(m+K, m*2)
print(m) |
s197271996 | p03587 | u074201886 | 2,000 | 262,144 | Wrong Answer | 2,104 | 3,060 | 291 | 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? | import sys
N=int(input())
if N%4==0:
print(int(3*N/4),int(3*N/4),int(3*N/4))
else:
for h in range(1,3501):
for n in range(1,3501):
for w in range(1,3501):
if N*(h*n+n*w+w*h)==4*h*n*w:
print(h,n,w)
sys.exit() | s256788227 | Accepted | 17 | 2,940 | 26 | print(input().count("1"))
|
s018422839 | p03351 | u263824932 | 2,000 | 1,048,576 | Wrong Answer | 18 | 2,940 | 98 | 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())
if a+b<=d and b+c<=d:
print('Yes')
else:
print('No')
| s839287721 | Accepted | 19 | 2,940 | 143 | a,b,c,d=map(int,input().split())
if abs(c-a)<=d:
print('Yes')
elif abs(a-b)<=d and abs(b-c)<=d:
print('Yes')
else:
print('No')
|
s629746487 | p02613 | u114365796 | 2,000 | 1,048,576 | Wrong Answer | 31 | 9,404 | 261 | Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format. | from collections import Counter
def c():
n = int(input())
inp = []
for i in range(n):
inp.append(input())
result = Counter(inp)
print(f"AC x {result['AC']}")
print(f"WA x {result['WA']}")
print(f"TLE x {result['TLE']}")
print(f"RE x {result['RE']}") | s034361766 | Accepted | 152 | 9,068 | 378 | n = int(input())
inp = {}
for i in range(n):
a = input()
if inp.get(a) is None:
inp[a] = 0
inp[a] += 1
print(f"AC x {0 if inp.get('AC') is None else inp.get('AC')}")
print(f"WA x {0 if inp.get('WA') is None else inp.get('WA')}")
print(f"TLE x {0 if inp.get('TLE') is None else inp.get('TLE')}")
print(f"RE x {0 if inp.get('RE') is None else inp.get('RE')}")
|
s087507587 | p03778 | u703890795 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 86 | AtCoDeer the deer found two rectangles lying on the table, each with height 1 and width W. If we consider the surface of the desk as a two-dimensional plane, the first rectangle covers the vertical range of AtCoDeer will move the second rectangle horizontally so that it connects with the first rectangle. Find the minimum distance it needs to be moved. | W, a, b = map(int, input().split())
m = min(a, b)
M = max(a, b)
print(max(M-m-2*W, 0)) | s266804638 | Accepted | 17 | 2,940 | 84 | W, a, b = map(int, input().split())
m = min(a, b)
M = max(a, b)
print(max(M-m-W, 0)) |
s538022040 | p04043 | u628965061 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 99 | 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. | list=list(map(int,input().split()))
print("YES" if list.count(7)==2 and list.count(5)==1 else "No") | s721906697 | Accepted | 17 | 2,940 | 99 | list=list(map(int,input().split()))
print("YES" if list.count(7)==1 and list.count(5)==2 else "NO") |
s518650585 | p04031 | u993161647 | 2,000 | 262,144 | Wrong Answer | 25 | 2,940 | 175 | Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal **integers** by transforming some of them. He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2). Find the minimum total cost to achieve his objective. | N = input()
a = list(map(int,input().split()))
for a_i in range(min(a), max(a)+1):
cost_sum = 0
for a_j in a:
cost_sum += ((a_j - a_i) ** 2)
print(cost_sum)
| s420694911 | Accepted | 24 | 2,940 | 200 | N = int(input())
a = list(map(int,input().split()))
cost = []
for i in range(-100, 101):
cost_all = 0
for j in a:
cost_all += (j - i) ** 2
cost.append(cost_all)
print(min(cost))
|
s256147425 | p02392 | u800997102 | 1,000 | 131,072 | Wrong Answer | 20 | 5,584 | 81 | Write a program which reads three integers a, b and c, and prints "Yes" if a < b < c, otherwise "No". | a,b,c=map(int,input().split())
if a<b<c:
print("yes")
else:
print("no")
| s470252975 | Accepted | 20 | 5,588 | 81 | a,b,c=map(int,input().split())
if a<b<c:
print("Yes")
else:
print("No")
|
s650326286 | p04043 | u672882146 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 159 | Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order. | a,b,c=map(int, input().split())
box = [a,b,c]
count1 = box.count(5)
count2 = box.count(7)
if count1 == 2 and count2 ==1:
print('Yes')
else:
print('No') | s032273170 | Accepted | 17 | 2,940 | 159 | a,b,c=map(int, input().split())
box = [a,b,c]
count1 = box.count(5)
count2 = box.count(7)
if count1 == 2 and count2 ==1:
print('YES')
else:
print('NO') |
s091629550 | p00002 | u877201735 | 1,000 | 131,072 | Wrong Answer | 30 | 7,544 | 153 | Write a program which computes the digit number of sum of two integers a and b. | while True:
try:
x = list(map(int, input().split(' ')))
n = x[0] * x[1]
print(len(str(n)))
except EOFError:
break | s172369125 | Accepted | 20 | 7,604 | 153 | while True:
try:
x = list(map(int, input().split(' ')))
n = x[0] + x[1]
print(len(str(n)))
except EOFError:
break |
s202394992 | p03470 | u902917675 | 2,000 | 262,144 | Wrong Answer | 26 | 9,164 | 236 | An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have? | n = int(input())
strlist = [input() for i in range(n)]
intlist = [int(x) for x in strlist]
sortlist = sorted(intlist)
print(sortlist)
temp = 0
ans = 0
for d in sortlist:
if int(d) > temp:
temp = int(d)
ans += 1
print(ans) | s300325907 | Accepted | 29 | 9,072 | 219 | n = int(input())
strlist = [input() for i in range(n)]
intlist = [int(x) for x in strlist]
sortlist = sorted(intlist)
temp = 0
ans = 0
for d in sortlist:
if int(d) > temp:
temp = int(d)
ans += 1
print(ans) |
s056982552 | p03567 | u077019541 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 76 | Snuke built an online judge to hold a programming contest. When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.) Determine whether the judge can return the string `AC` as the verdict to a program. | s = [i for i in input()]
if "AC"in s:
print("Yes")
else:
print("No") | s916053636 | Accepted | 17 | 2,940 | 63 | s = input()
if "AC"in s:
print("Yes")
else:
print("No") |
s348377999 | p03251 | u935984175 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,060 | 313 | 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. | def solve():
N, M, X, Y = map(int, input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
max_x = max(X, max(x))
min_y = min(Y, min(y))
if max_x < min_y:
print('NO War')
else:
print('War')
if __name__ == '__main__':
solve()
| s460418945 | Accepted | 18 | 3,064 | 313 | def solve():
N, M, X, Y = map(int, input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
max_x = max(X, max(x))
min_y = min(Y, min(y))
if max_x < min_y:
print('No War')
else:
print('War')
if __name__ == '__main__':
solve()
|
s004498682 | p02619 | u981418135 | 2,000 | 1,048,576 | Wrong Answer | 42 | 9,224 | 494 | Let's first write a program to calculate the score from a pair of input and output. You can know the total score by submitting your solution, or an official program to calculate a score is often provided for local evaluation as in this contest. Nevertheless, writing a score calculator by yourself is still useful to check your understanding of the problem specification. Moreover, the source code of the score calculator can often be reused for solving the problem or debugging your solution. So it is worthwhile to write a score calculator unless it is very complicated. | d =int(input())
c = list(map(int, input().split()))
s = []
t = []
d_l = []
score = 0
soft = 0
for i in range(d):
s.append(0)
s[i] = list(map(int, input().split()))
for i in range(26):
d_l.append(0)
for i in range(d):
t = int(input())
soft = 0
for j in range(26):
if t-1 == j:
d_l[j] = 0
score = s[i][j]
else:
d_l[j] += 1
soft += c[j] * d_l[j]
score -= soft
print(score)
| s233388940 | Accepted | 34 | 9,380 | 495 | d =int(input())
c = list(map(int, input().split()))
s = []
t = []
d_l = []
score = 0
soft = 0
for i in range(d):
s.append(0)
s[i] = list(map(int, input().split()))
for i in range(26):
d_l.append(0)
for i in range(d):
t = int(input())
soft = 0
for j in range(26):
if t-1 == j:
d_l[j] = 0
score += s[i][j]
else:
d_l[j] += 1
soft += c[j] * d_l[j]
score -= soft
print(score)
|
s774107349 | p03962 | u079699418 | 2,000 | 262,144 | Wrong Answer | 25 | 9,052 | 46 | AtCoDeer the deer recently bought three paint cans. The color of the one he bought two days ago is a, the color of the one he bought yesterday is b, and the color of the one he bought today is c. Here, the color of each paint can is represented by an integer between 1 and 100, inclusive. Since he is forgetful, he might have bought more than one paint can in the same color. Count the number of different kinds of colors of these paint cans and tell him. | x = list(map(int,input().split()))
len(set(x)) | s266391396 | Accepted | 28 | 9,040 | 53 | x = list(map(int,input().split()))
print(len(set(x))) |
s948420493 | p04044 | u475018333 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 268 | 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. | n, l = map(int, input().split())
l = []
i = 0
# while i <= n-1:
# list[i] = list(input())
l = [input() for i in range(n)]
result = sorted(l)
# print('--------------')
# print(result)
# print('--------------')
for i in range(n):
print(result[i])
| s466027394 | Accepted | 17 | 3,060 | 149 | n, l = map(int, input().split())
l = []
i = 0
l = [input() for i in range(n)]
result = sorted(l)
s = ''.join(result)
print(s)
|
s411139899 | p02601 | u552116325 | 2,000 | 1,048,576 | Wrong Answer | 30 | 9,184 | 257 | M-kun has the following three cards: * A red card with the integer A. * A green card with the integer B. * A blue card with the integer C. He is a genius magician who can do the following operation at most K times: * Choose one of the three cards and multiply the written integer by 2. His magic is successful if both of the following conditions are satisfied after the operations: * The integer on the green card is **strictly** greater than the integer on the red card. * The integer on the blue card is **strictly** greater than the integer on the green card. Determine whether the magic can be successful. | A, B, C = map(int, input().split())
K = int(input())
for i in range(1, K):
for ii in range(1, K - i):
iii = K - i - ii
A = A * (2 ** i)
B = B * (2 ** ii)
C = C * (2 ** iii)
if A < B < C:
print('Yes')
exit(0)
print('No')
| s760079776 | Accepted | 25 | 8,780 | 251 | A, B, C = map(int, input().split())
K = int(input())
for i in range(K):
for ii in range(K - i):
iii = K - i - ii
a = A * (2 ** i)
b = B * (2 ** ii)
c = C * (2 ** iii)
if a < b < c:
print('Yes')
exit(0)
print('No')
|
s289680103 | p03471 | u360258922 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 363 | 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(' '))
s_n = [0, 0, 0]
s_l = [10000, 5000, 1000]
for i, s in enumerate(s_l):
if n > -1:
s_n[i] = int(y/s)
n = n - s_n[i]
y = y - s * s_n[i]
else:
print('-1 -1 -1')
if n > -1:
if y == 0:
print(' '.join(map(str, s_n)))
else:
print('-1 -1 -1')
else:
print('-1 -1 -1') | s166574741 | Accepted | 897 | 3,064 | 346 | N, Y = map(int, input().split(' '))
res10000 = -1
res5000 = -1
res1000 = -1
for a in range(N+1):
for b in range(N-a+1):
c = N - a - b
total = 10000*a + 5000*b + 1000*c
if total == Y:
res10000 = a
res5000 = b
res1000 = c
print(str(' '.join(map(str, [res10000, res5000, res1000])))) |
s792114792 | p02613 | u607114738 | 2,000 | 1,048,576 | Wrong Answer | 166 | 16,176 | 209 | 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 = str(input())
S.append(s)
result = ['AC', 'WA', 'TLE', 'RE']
C = [S.count(r) for r in result]
for (re, c) in zip(result, C):
print(re + 'x' + str(c)) | s963150043 | Accepted | 164 | 16,128 | 211 | N = int(input())
S = []
for i in range(N):
s = str(input())
S.append(s)
result = ['AC', 'WA', 'TLE', 'RE']
C = [S.count(r) for r in result]
for (re, c) in zip(result, C):
print(re + ' x ' + str(c)) |
s768306166 | p03360 | u595893956 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 83 | 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())
print(max(a,b,c)*(2**int(input())+a+b+c-max(a,b,c))) | s280016961 | Accepted | 17 | 2,940 | 84 | a,b,c=map(int,input().split())
print(max(a,b,c)*(2**int(input()))+a+b+c-max(a,b,c))
|
s628626227 | p02237 | u067677727 | 1,000 | 131,072 | Wrong Answer | 20 | 7,508 | 433 | There are two standard ways to represent a graph $G = (V, E)$, where $V$ is a set of vertices and $E$ is a set of edges; Adjacency list representation and Adjacency matrix representation. An adjacency-list representation consists of an array $Adj[|V|]$ of $|V|$ lists, one for each vertex in $V$. For each $u \in V$, the adjacency list $Adj[u]$ contains all vertices $v$ such that there is an edge $(u, v) \in E$. That is, $Adj[u]$ consists of all vertices adjacent to $u$ in $G$. An adjacency-matrix representation consists of $|V| \times |V|$ matrix $A = a_{ij}$ such that $a_{ij} = 1$ if $(i, j) \in E$, $a_{ij} = 0$ otherwise. Write a program which reads a directed graph $G$ represented by the adjacency list, and prints its adjacency-matrix representation. $G$ consists of $n\; (=|V|)$ vertices identified by their IDs $1, 2,.., n$ respectively. | def main():
n = int(input())
adj = [[0 for i in range(n)] for j in range(n)]
for i in range(n):
tmp = list(map(int, input().split()))
u = tmp[0]
u = u -1
k = tmp[1]
v = tmp[2:]
for j in range(k):
adj[u][v[j]-1] = 1
for i in range(n):
print(" ".join(list(map(str, adj[i]))))
for i in adj:
print(*i)
if __name__ == "main":
main() | s820746948 | Accepted | 30 | 8,528 | 365 | def main():
n = int(input())
adj = [[0 for i in range(n)] for j in range(n)]
for i in range(n):
tmp = list(map(int, input().split()))
u = tmp[0]
u = u -1
k = tmp[1]
v = tmp[2:]
for j in range(k):
adj[u][v[j]-1] = 1
for i in adj:
print(*i)
if __name__ == "__main__":
main() |
s366739912 | p04043 | u679817762 | 2,000 | 262,144 | Wrong Answer | 25 | 8,964 | 159 | Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order. |
lst = input().split()
for i in lst:
i = int(i)
lst.sort()
if lst == [5, 5, 7]:
print('YES')
else:
print('NO') | s423017142 | Accepted | 27 | 8,996 | 184 |
lst = input().split()
for i in range(len(lst)):
lst[i] = int(lst[i])
lst.sort()
if lst == [5, 5, 7]:
print('YES')
else:
print('NO') |
s010751876 | p02613 | u934609868 | 2,000 | 1,048,576 | Wrong Answer | 157 | 9,108 | 319 | 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())
ac = 0
wa = 0
tle = 0
re = 0
for i in range(1,N+1):
S = str(input())
if S == "AC":
ac += 1
elif S == "WA":
wa += 1
elif S == "TLE":
tle += 1
elif S == "RE":
re += 1
print("AC × " + str(ac))
print("WA × " + str(wa))
print("TLE × " + str(tle))
print("AC × " + str(re))
| s527442766 | Accepted | 155 | 9,204 | 305 | N = int(input())
ac = 0
wa = 0
tle = 0
re = 0
for i in range(1,N+1):
S = str(input())
if S == "AC":
ac += 1
elif S == "WA":
wa += 1
elif S == "TLE":
tle += 1
else:
re += 1
print("AC x " + str(ac))
print("WA x " + str(wa))
print("TLE x " + str(tle))
print("RE x " + str(re))
|
s225758895 | p02608 | u383450070 | 2,000 | 1,048,576 | Wrong Answer | 2,206 | 8,972 | 260 | 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). | import math
n=int(input())
for p in range(1, n):
right=math.floor(p*0.5)
cnt=0
for i in range(right):
for j in range(right):
for k in range(right):
if i**2+j**2+k**2+i*j+j*k+k*i==p and i>0 and j>0 and k>0:
cnt+=1
print(cnt) | s644877205 | Accepted | 326 | 87,396 | 445 | n=int(input())
lst=[0]*10000000
for i in range(1, 101):
for j in range(1, i+1):
for k in range(1, j+1):
temp=i**2+j**2+k**2+i*j+j*k+k*i
if i==j and j==k:
lst[temp+1]+=1
elif i==j and j!=k:
lst[temp+1]+=3
elif j==k and k!=i:
lst[temp+1]+=3
elif k==i and i!=j:
lst[temp+1]+=3
elif i!=j and j!=k and k!=i:
lst[temp+1]+=6
for p in range(1, n+1):
print(lst[p+1]) |
s958124187 | p03251 | u013408661 | 2,000 | 1,048,576 | Wrong Answer | 18 | 3,064 | 289 | 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()
Y.sort()
Z=[i for i in range(x+1,y)]
if X[-1]<Y[0]:
z=[i for i in range(X[-1]+1,Y[0])]
if len(set(z)&set(Z))>0:
print("No War")
else:
print("War")
else:
print("War") | s757074647 | Accepted | 17 | 3,064 | 293 | n,m,x,y=map(int,input().split())
X=list(map(int,input().split()))
Y=list(map(int,input().split()))
X.sort()
Y.sort()
Z=[i for i in range(x+1,y+1)]
if X[-1]<Y[0]:
z=[i for i in range(X[-1]+1,Y[0]+1)]
if len(set(z)&set(Z))>0:
print("No War")
else:
print("War")
else:
print("War") |
s207202617 | p03672 | u766566560 | 2,000 | 262,144 | Time Limit Exceeded | 2,104 | 7,920 | 132 | We will call a string that can be obtained by concatenating two equal strings an _even_ string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not. You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input. | S = input()
while True:
if S[0:len(S)//2] == S[len(S)//2:len(S)-1]:
print(len(S))
else:
tmp = S[0:len(S)-2]
S = tmp | s959416781 | Accepted | 18 | 3,060 | 179 | S = input()
cnt = 0
for i in range(len(S)):
if S[0:len(S)//2] == S[len(S)//2:len(S) + 1] and cnt != 0:
print(len(S))
exit()
else:
cnt += 1
S = S[0:len(S) - 1] |
s965636195 | p03854 | u912652535 | 2,000 | 262,144 | Wrong Answer | 18 | 3,188 | 133 | 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`. | s = input().replace("eracer",'').replace('erace','').replace('dreamer','').replace('dreame','')
print('YES' if len(s) == 0 else 'NO') | s792240276 | Accepted | 18 | 3,188 | 132 | s = input().replace("eraser",'').replace('erase','').replace('dreamer','').replace('dream','')
print('YES' if len(s) == 0 else 'NO') |
s447431732 | p02612 | u542541293 | 2,000 | 1,048,576 | Wrong Answer | 26 | 9,148 | 54 | 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())
q, mod = divmod(N, 1000)
print(mod) | s301950579 | Accepted | 28 | 9,156 | 91 | N = int(input())
q, mod = divmod(N, 1000)
if mod == 0:
print(0)
else:
print(1000-mod) |
s643980636 | p02678 | u970267139 | 2,000 | 1,048,576 | Wrong Answer | 739 | 60,204 | 433 | 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. | n, m = map(int, input().split())
edge = [list(map(int, input().split())) for _ in range(m)]
next_node = [[] for _ in range(n)]
for a, b in edge:
next_node[a-1].append(b)
next_node[b-1].append(a)
q = [1]
visit = [1] + [0] * (n-1)
prev = [0]*n
while q:
now = q.pop()
for i in next_node[now-1]:
if visit[i-1] == 0:
q.append(i)
visit[i-1] = 1
prev[i-1] = now
print(prev[1:]) | s508278624 | Accepted | 1,375 | 59,588 | 460 | n, m = map(int, input().split())
edge = [list(map(int, input().split())) for _ in range(m)]
next_node = [[] for _ in range(n)]
for a, b in edge:
next_node[a-1].append(b)
next_node[b-1].append(a)
q = [1]
visit = [1] + [0] * (n-1)
prev = [0] * n
while q:
now = q.pop(0)
for i in next_node[now-1]:
if visit[i-1] == 0:
q.append(i)
visit[i-1] = 1
prev[i-1] = now
print('Yes')
print(*prev[1:], sep='\n') |
s512210561 | p02613 | u426250125 | 2,000 | 1,048,576 | Wrong Answer | 154 | 16,324 | 243 | 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())
status={'AC':0, 'WA':0, 'TLE':0, 'RE':0}
results = [input() for n in range(N)]
for result in results:
if result in status.keys():
status[result] += 1
for k, v in status.items():
print('{} × {}'.format(k, v)) | s664742060 | Accepted | 155 | 16,332 | 242 | N = int(input())
status={'AC':0, 'WA':0, 'TLE':0, 'RE':0}
results = [input() for n in range(N)]
for result in results:
if result in status.keys():
status[result] += 1
for k, v in status.items():
print('{} x {}'.format(k, v)) |
s781785263 | p03494 | u044916426 | 2,000 | 262,144 | Wrong Answer | 18 | 3,188 | 230 | 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. | li = list(map(int, input().split()))
a = 0
br = False
while True:
for i in li:
if i % 2 != 0:
br = True
break
if br:
break
a += 1
li = list(map(lambda x: x / 2, li))
print(a) | s070153938 | Accepted | 19 | 2,940 | 247 | n = int(input())
li = list(map(int, input().split()))
a = 0
br = False
while True:
for i in li:
if i % 2 != 0:
br = True
break
if br:
break
a += 1
li = list(map(lambda x: x / 2, li))
print(a) |
s023220865 | p00033 | u945345165 | 1,000 | 131,072 | Wrong Answer | 20 | 7,628 | 1,053 | 図のように二股に分かれている容器があります。1 から 10 までの番号が付けられた10 個の玉を容器の開口部 A から落とし、左の筒 B か右の筒 C に玉を入れます。板 D は支点 E を中心に左右に回転できるので、板 D を動かすことで筒 B と筒 C のどちらに入れるか決めることができます。 開口部 A から落とす玉の並びを与えます。それらを順番に筒 B 又は筒 Cに入れていきます。このとき、筒 B と筒 C のおのおのが両方とも番号の小さい玉の上に大きい玉を並べられる場合は YES、並べられない場合は NO と出力するプログラムを作成してください。ただし、容器の中で玉の順序を入れ替えることはできないものとします。また、続けて同じ筒に入れることができるものとし、筒 B, C ともに 10 個の玉がすべて入るだけの余裕があるものとします。 | import sys
def solve(balls):
ans = distribute(balls, [], [])
if ans is True: print('YES')
else: print("NO")
def distribute(balls, R, L):
if len(balls) != 0:
next = balls[0]
ans = False
#case R
if isMutch(next, R):
neoR = R
neoR.append(next)
ans = distribute(balls[1:], neoR, L)
if isMutch(next, L):
neoL = L
neoL.append(next)
ans =distribute(balls[1:], R, neoL)
if ans: return True
else:
return isOrdered(R) and isOrdered(L)
def isMutch(next, lis):
if len(lis) != 0:
return next > lis[len(lis)-1]
return True
def isOrdered(lis):
#check both R and L are ordered
checker = sorted(lis)
return checker == lis
limit = 2**11
sys.setrecursionlimit(limit)
line = input()
size = -1;
while True:
if size == -1:
size = int(line)
else:
solve(line.split(' '))
size -= 1
print(size)
if size == 0: break
line = input() | s034515552 | Accepted | 20 | 7,616 | 568 | import sys
def solve(balls):
if distribute(balls, 0, 0): print('YES')
else: print('NO')
def distribute(balls, lastR, lastL):
if len(balls) != 0:
next = balls[0]
if next>lastR:
if distribute(balls[1:], next, lastL): return True
if next>lastL:
if distribute(balls[1:], lastR, next): return True
else:
return True
sys.setrecursionlimit(2**11)
size = int(input());
for i in range(size):
balls = []
for a in input().split(' '):
balls.append(int(a))
solve(balls) |
s591735661 | p03386 | u167681750 | 2,000 | 262,144 | Wrong Answer | 2,103 | 3,064 | 98 | 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())
print([i for i in range(A, B + 1) if A + K > i or B - K < i]) | s718187739 | Accepted | 17 | 3,060 | 148 | A, B, K = map(int, input().split())
for i in range(A, min(A + K, B + 1)):
print(i)
for i in range(max(B - K + 1, A + K), B + 1):
print(i)
|
s983747858 | p02612 | u346812984 | 2,000 | 1,048,576 | Wrong Answer | 29 | 9,156 | 230 | 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. | import sys
sys.setrecursionlimit(10 ** 6)
INF = float("inf")
MOD = 10 ** 9 + 7
def input():
return sys.stdin.readline().strip()
def main():
N = int(input())
print(N % 1000)
if __name__ == "__main__":
main()
| s576769307 | Accepted | 34 | 9,160 | 290 | import sys
sys.setrecursionlimit(10 ** 6)
INF = float("inf")
MOD = 10 ** 9 + 7
def input():
return sys.stdin.readline().strip()
def main():
N = int(input())
if N % 1000 == 0:
print(0)
else:
print(1000 - N % 1000)
if __name__ == "__main__":
main()
|
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