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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s918348827
|
p03353
|
u280016524
| 2,000 | 1,048,576 |
Wrong Answer
| 35 | 4,932 | 291 |
You are given a string s. Among the **different** substrings of s, print the K-th lexicographically smallest one. A substring of s is a string obtained by taking out a non-empty contiguous part in s. For example, if s = `ababc`, `a`, `bab` and `ababc` are substrings of s, while `ac`, `z` and an empty string are not. Also, we say that substrings are different when they are different as strings. Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.
|
s=input()
K=int(input())
listA=[]
for j in range(1,K+1):
for i in range(len(s)):
listA.append(s[i:i+j])
listA.sort()
x=""
count=0
for i in range(len(listA)):
if(x!=listA[i]):
count+=1
x=listA[i]
if(count==K):
print(x)
break;
print(listA)
|
s159611484
|
Accepted
| 34 | 4,340 | 278 |
s=input()
K=int(input())
listA=[]
for j in range(1,K+1):
for i in range(len(s)):
listA.append(s[i:i+j])
listA.sort()
x=""
count=0
for i in range(len(listA)):
if(x!=listA[i]):
count+=1
x=listA[i]
if(count==K):
print(x)
break;
|
s386640168
|
p02432
|
u629874472
| 2,000 | 262,144 |
Wrong Answer
| 20 | 5,596 | 343 |
For a dynamic array $A = \\{a_0, a_1, ...\\}$ of integers, perform a sequence of the following operations: * push($d$, $x$): Add element $x$ at the begining of $A$, if $d = 0$. Add element $x$ at the end of $A$, if $d = 1$. * randomAccess($p$): Print element $a_p$. * pop($d$): Delete the first element of $A$, if $d = 0$. Delete the last element of $A$, if $d = 1$. $A$ is a 0-origin array and it is empty in the initial state.
|
cnt = int(input())
a = []
for i in range(cnt):
li = list(map(int,input().split()))
if li[0] == 0:
if li[1] == 0:
a.insert(0,li[2])
else:
a.insert(-1,li[2])
if li[0] == 1:
print(a[li[1]])
if li[0] ==2:
if li[1]==0:
a.pop(0)
else:
a.pop(-1)
|
s854391807
|
Accepted
| 1,700 | 21,984 | 379 |
from collections import deque
cnt = int(input())
a = deque()
for i in range(cnt):
li = list(map(int,input().split()))
if li[0] == 0:
if li[1] == 0:
a.appendleft(li[2])
else:
a.append(li[2])
if li[0] == 1:
print(a[li[1]])
if li[0] ==2:
if li[1]==0:
a.popleft()
else:
a.pop()
|
s912106749
|
p04012
|
u006425112
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 168 |
Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful.
|
w = input()
flag = True
for i in range(len(w)):
if w[i] != "a" and w[i] != "z":
flag = False
break
if flag:
print("Yes")
else:
print("No")
|
s127590940
|
Accepted
| 20 | 3,316 | 197 |
from collections import Counter
w = input()
c = Counter(w)
flag = True
for i in c.values():
if i % 2 == 1:
flag = False
break
if flag:
print("Yes")
else:
print("No")
|
s496684059
|
p03846
|
u652150585
| 2,000 | 262,144 |
Wrong Answer
| 106 | 16,096 | 704 |
There are N people, conveniently numbered 1 through N. They were standing in a row yesterday, but now they are unsure of the order in which they were standing. However, each person remembered the following fact: the absolute difference of the number of the people who were standing to the left of that person, and the number of the people who were standing to the right of that person. According to their reports, the difference above for person i is A_i. Based on these reports, find the number of the possible orders in which they were standing. Since it can be extremely large, print the answer modulo 10^9+7. Note that the reports may be incorrect and thus there may be no consistent order. In such a case, print 0.
|
import collections
n=int(input())
l=list(map(int,input().split()))
s=collections.Counter(l)
s=s.most_common()
print(s)
a=1
if n%2==0:
for i in s:
if i[1]!=2:
print(0)
exit()
elif i[0]%2!=1 or i[1]//2>n//2:
print(0)
exit()
else:
for j in range(n//2):
a=(2*a)%1000000007
print(a)
elif n%2==1:
if (0,1) not in s:
print(0)
exit()
else:
del s[-1]
print(s)
for i in s:
if i[1]!=2 or i[0]%2!=0 or i[0]//2>n//2:
print(0)
exit()
else:
for j in range(n//2):
a=(2*a)%1000000007
print(a)
|
s553212885
|
Accepted
| 89 | 16,096 | 706 |
import collections
n=int(input())
l=list(map(int,input().split()))
s=collections.Counter(l)
s=s.most_common()
#print(s)
a=1
if n%2==0:
for i in s:
if i[1]!=2:
print(0)
exit()
elif i[0]%2!=1 or i[1]//2>n//2:
print(0)
exit()
else:
for j in range(n//2):
a=(2*a)%1000000007
print(a)
elif n%2==1:
if (0,1) not in s:
print(0)
exit()
else:
del s[-1]
#print(s)
for i in s:
if i[1]!=2 or i[0]%2!=0 or i[0]//2>n//2:
print(0)
exit()
else:
for j in range(n//2):
a=(2*a)%1000000007
print(a)
|
s612293257
|
p02413
|
u138628845
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,592 | 364 |
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.
|
all_ele = []
sum_ele = []
h,w = [int(x) for x in input().split()]
for ele in range(h):
all_ele.append([int(x) for x in input().split()])
all_ele[ele].append(sum(all_ele[ele]))
all_ele.append([int(x) for x in range(w+1)])
for a in range(w + 1):
for b in range(h):
sum_ele.append(all_ele[b][a])
all_ele[h][a] = sum(sum_ele)
sum_ele = []
|
s444720097
|
Accepted
| 20 | 5,684 | 440 |
all_ele = []
sum_ele = []
h,w = [int(x) for x in input().split()]
for ele in range(h):
all_ele.append([int(x) for x in input().split()])
all_ele[ele].append(sum(all_ele[ele]))
all_ele.append([int(x) for x in range(w+1)])
for a in range(len(all_ele[0])):
for b in range(h):
sum_ele.append(all_ele[b][a])
all_ele[h][a] = sum(sum_ele)
sum_ele = []
for out1 in all_ele:
print(' '.join(map(str,out1)))
|
s686731977
|
p03943
|
u997530672
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 117 |
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
a, b, c = map(int, input().split())
if a + b == c or b + c == a or a + c == b:
print('YES')
else:
print('NO')
|
s739800817
|
Accepted
| 17 | 2,940 | 117 |
a, b, c = map(int, input().split())
if a + b == c or b + c == a or a + c == b:
print('Yes')
else:
print('No')
|
s221639777
|
p02408
|
u546968095
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,592 | 574 |
Taro is going to play a card game. However, now he has only n cards, even though there should be 52 cards (he has no Jokers). The 52 cards include 13 ranks of each of the four suits: spade, heart, club and diamond.
|
Cards = [
"S 1", "H 1", "C 1", "D 1",
"S 2", "H 2", "C 2", "D 2",
"S 3", "H 3", "C 3", "D 3",
"S 4", "H 4", "C 4", "D 4",
"S 5", "H 5", "C 5", "D 5",
"S 6", "H 6", "C 6", "D 6",
"S 7", "H 7", "C 7", "D 7",
"S 8", "H 8", "C 8", "D 8",
"S 9", "H 9", "C 9", "D 9",
"S 10", "H 10", "C 10", "D 10",
"S 11", "H 11", "C 11", "D 11",
"S 12", "H 12", "C 12", "D 12",
"S 13", "H 13", "C 13", "D 13",
]
n = input()
for i in range(int(n)):
rm = input()
Cards.remove(rm)
for i in range(52 - int(n)):
print(Cards.pop(0))
|
s283062162
|
Accepted
| 20 | 5,596 | 538 |
Cards = [
"S 1", "S 2", "S 3", "S 4", "S 5", "S 6", "S 7", "S 8", "S 9", "S 10", "S 11", "S 12", "S 13",
"H 1", "H 2", "H 3", "H 4", "H 5", "H 6", "H 7", "H 8", "H 9", "H 10", "H 11", "H 12", "H 13",
"C 1", "C 2", "C 3", "C 4", "C 5", "C 6", "C 7", "C 8", "C 9", "C 10", "C 11", "C 12", "C 13",
"D 1", "D 2", "D 3", "D 4", "D 5", "D 6", "D 7", "D 8", "D 9", "D 10", "D 11", "D 12", "D 13",
]
n = input()
for i in range(int(n)):
rm = input()
Cards.remove(rm)
for i in range(52 - int(n)):
print(Cards.pop(0))
|
s264467905
|
p02613
|
u566297428
| 2,000 | 1,048,576 |
Wrong Answer
| 142 | 16,616 | 205 |
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
import collections
N = int(input())
S = [input() for _ in range(N)]
c = collections.Counter(S)
print('AC', '×', c['AC'])
print('WA', '×', c['WA'])
print('TLE', '×', c['TLE'])
print('RE', '×', c['RE'])
|
s958644426
|
Accepted
| 141 | 16,552 | 188 |
import collections
N = int(input())
S = [input() for _ in range(N)]
print('AC x', S.count('AC'))
print('WA x', S.count('WA'))
print('TLE x', S.count('TLE'))
print('RE x', S.count('RE'))
|
s044371366
|
p03693
|
u830592648
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 84 |
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
n = int(input().replace(" ",""))
if n%4 == 0:
print("Yes")
else:
print("No")
|
s878251698
|
Accepted
| 18 | 2,940 | 84 |
n = int(input().replace(" ",""))
if n%4 == 0:
print("YES")
else:
print("NO")
|
s174991501
|
p03385
|
u120810144
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 63 |
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
s = input()
s = sorted(s)
print("Yes" if s == "abc" else "No")
|
s222889858
|
Accepted
| 17 | 2,940 | 71 |
s = input()
s = "".join(sorted(s))
print("Yes" if s == "abc" else "No")
|
s308611607
|
p03861
|
u396391104
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 49 |
You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x?
|
a,b,x = map(int,input().split())
print(b//x-a//x)
|
s728123175
|
Accepted
| 17 | 2,940 | 54 |
a,b,x = map(int,input().split())
print(b//x-(a-1)//x)
|
s507022784
|
p02615
|
u395620499
| 2,000 | 1,048,576 |
Wrong Answer
| 144 | 31,432 | 138 |
Quickly after finishing the tutorial of the online game _ATChat_ , you have decided to visit a particular place with N-1 players who happen to be there. These N players, including you, are numbered 1 through N, and the **friendliness** of Player i is A_i. The N players will arrive at the place one by one in some order. To make sure nobody gets lost, you have set the following rule: players who have already arrived there should form a circle, and a player who has just arrived there should cut into the circle somewhere. When each player, except the first one to arrive, arrives at the place, the player gets **comfort** equal to the smaller of the friendliness of the clockwise adjacent player and that of the counter-clockwise adjacent player. The first player to arrive there gets the comfort of 0. What is the maximum total comfort the N players can get by optimally choosing the order of arrivals and the positions in the circle to cut into?
|
n = int(input())
a = list(map(int, input().split()))
a.sort()
ans = 0
for i in range(1,n+1):
ans += a[n-i//2-1]
print(ans)
|
s948875738
|
Accepted
| 153 | 31,408 | 132 |
n = int(input())
a = list(map(int, input().split()))
a.sort()
ans = 0
for i in range(1, n):
ans += a[n - i//2 - 1]
print(ans)
|
s266208848
|
p03449
|
u458725980
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 452 |
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)]
Am = [[0 for j in range(N)] for i in range(2)]
Am[0][0] = A[0][0]
Am[1][0] = A[0][0] + A[1][0]
for i in range(N):
if i > 0:
Am[0][i] = A[0][i] + Am[0][i-1]
for i in range(N):
if i > 0:
Am[1][i] = A[1][i] + max( Am[0][i] , Am[1][i - 1])
print(Am)
|
s850465528
|
Accepted
| 18 | 3,064 | 472 |
N = int(input())
A = [list(map(int,input().split())) for i in range(2)]
Am = [[0 for j in range(N)] for i in range(2)]
Am[0][0] = A[0][0]
Am[1][0] = A[0][0] + A[1][0]
for i in range(N):
if i > 0:
Am[0][i] = A[0][i] + Am[0][i-1]
for i in range(N):
if i > 0:
Am[1][i] = A[1][i] + max( Am[0][i] , Am[1][i - 1])
#print(Am)
print(Am[1][N-1])
|
s513926490
|
p03844
|
u282813849
| 2,000 | 262,144 |
Wrong Answer
| 27 | 9,016 | 20 |
Joisino wants to evaluate the formula "A op B". Here, A and B are integers, and the binary operator op is either `+` or `-`. Your task is to evaluate the formula instead of her.
|
s = input()
print(s)
|
s219885875
|
Accepted
| 26 | 9,092 | 110 |
s = input()
a, op, b = s.split()
a = int(a)
b = int(b)
if op == "+":
c = a + b
else:
c = a - b
print(c)
|
s782853043
|
p03379
|
u652656291
| 2,000 | 262,144 |
Wrong Answer
| 330 | 25,472 | 150 |
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.
|
N = int(input())
X = list(map(int,input().split()))
X.sort()
L = X[N//2 - 1]
R = X[N//2]
for x in X:
if x >= R:
print(L)
else:
print(R)
|
s126376622
|
Accepted
| 303 | 25,224 | 178 |
N = int(input())
X = [int(x) for x in input().split()]
X_sorted = sorted(X)
L = X_sorted[N//2 - 1]
R = X_sorted[N//2]
for x in X:
if x >= R:
print(L)
else:
print(R)
|
s244492075
|
p03644
|
u281216592
| 2,000 | 262,144 |
Time Limit Exceeded
| 2,104 | 2,940 | 217 |
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
N = int(input())
counter =-1
number = 0
for i in range(N):
k = i
now_counter = 0
while(k % 2 == 0):
now_counter += 1
k /= 2
if(now_counter > counter):
number = i
print(number)
|
s833543607
|
Accepted
| 17 | 2,940 | 251 |
N = int(input())
counter =-1
number = 0
for i in range(N):
k = i+1
now_counter = 0
while(k % 2 == 0):
now_counter += 1
k /= 2
if(now_counter > counter):
number = i+1
counter = now_counter
print(number)
|
s859169763
|
p00003
|
u372789658
| 1,000 | 131,072 |
Wrong Answer
| 40 | 7,624 | 100 |
Write a program which judges wheather given length of three side form a right triangle. Print "YES" if the given sides (integers) form a right triangle, "NO" if not so.
|
num = int(input())
for i in range (num):
a=input().split()
print( a[0]==a[1] and a[1]==a[2])
|
s472306008
|
Accepted
| 40 | 7,524 | 173 |
num = int(input())
for i in range (num):
a=sorted(list(map(int,input().split())))
if a[0]**2+a[1]**2 == (a[2])**2:
print("YES")
else:
print("NO")
|
s052728800
|
p02416
|
u227984374
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,584 | 92 |
Write a program which reads an integer and prints sum of its digits.
|
N = input()
ans = 0
for i in range(len(N)) :
k = int(N[i])
ans = ans + k
print(ans)
|
s628376640
|
Accepted
| 20 | 5,584 | 91 |
while True :
N = sum(map(int, input()))
if N == 0 :
break
print(N)
|
s838650212
|
p04030
|
u082861480
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 212 |
Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?
|
w = input()
res = []
for i in w:
print(i,res)
if i =='1':
res.append(i)
elif i == '0':
res.append(i)
elif len(res) > 0:
res.pop()
else:
pass
print(''.join(res))
|
s390923795
|
Accepted
| 17 | 3,060 | 195 |
w = input()
res = []
for i in w:
if i =='1':
res.append(i)
elif i == '0':
res.append(i)
elif len(res) > 0:
res.pop()
else:
pass
print(''.join(res))
|
s933199739
|
p02842
|
u988832865
| 2,000 | 1,048,576 |
Wrong Answer
| 31 | 2,940 | 150 |
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.
|
N = int(input())
def solve(N):
for i in range(60000):
if int(i * 1.08) == N:
print(i)
else:
print(":(")
solve(N)
|
s723320113
|
Accepted
| 31 | 2,940 | 168 |
N = int(input())
def solve(N):
for i in range(60000):
if int(i * 1.08) == N:
print(i)
break
else:
print(":(")
solve(N)
|
s337830965
|
p00022
|
u519227872
| 1,000 | 131,072 |
Wrong Answer
| 40 | 7,932 | 409 |
Given a sequence of numbers a1, a2, a3, ..., an, find the maximum sum of a contiguous subsequence of those numbers. Note that, a subsequence of one element is also a _contiquous_ subsequence.
|
from sys import stdin
def getMax(array):
mx = mx2 = array[0]
for i in array[1:]:
mx2 = max(i, mx2 + i)
print(mx2,mx)
mx = max(mx, mx2)
return mx
for line in stdin:
n = int(line)
if n == 0:
break
array = []
for line in stdin:
line = int(line)
array.append(line)
if len(array) == n:
break
print(getMax(array))
|
s990987879
|
Accepted
| 30 | 7,724 | 387 |
from sys import stdin
def getMax(array):
mx = mx2 = array[0]
for i in array[1:]:
mx2 = max(i, mx2 + i)
mx = max(mx, mx2)
return mx
for line in stdin:
n = int(line)
if n == 0:
break
array = []
for line in stdin:
line = int(line)
array.append(line)
if len(array) == n:
break
print(getMax(array))
|
s652394260
|
p03488
|
u811202694
| 2,000 | 524,288 |
Wrong Answer
| 2,117 | 189,416 | 795 |
A robot is put at the origin in a two-dimensional plane. Initially, the robot is facing in the positive x-axis direction. This robot will be given an instruction sequence s. s consists of the following two kinds of letters, and will be executed in order from front to back. * `F` : Move in the current direction by distance 1. * `T` : Turn 90 degrees, either clockwise or counterclockwise. The objective of the robot is to be at coordinates (x, y) after all the instructions are executed. Determine whether this objective is achievable.
|
import numpy as np
s = list(input())
target = np.array([int(i) for i in input().split()])
direction = np.array([[1, 0], [0, -1], [-1, 0], [0, 1]])
queue = [[np.array([0, 0]), 0, s]]
flag = False
while queue:
state, now_d, s = queue.pop(0)
print(target, state, s)
if len(s) == 0:
if np.allclose(target, state):
flag = True
break
else:
continue
operation = s[0]
diff = target - state
if operation == "F":
next_state = state + direction[now_d]
queue.append([next_state, now_d, s[1:]])
elif operation == "T":
queue.append([state, (now_d+1) % 4, s[1:]])
queue.append([state, (now_d-1) % 4, s[1:]])
if flag:
print("Yes")
else:
print("No")
|
s951544387
|
Accepted
| 20 | 3,188 | 666 |
s = input().split("T")
x, y = ([int(i) for i in input().split()])
movex = []
movey = []
for i, t in enumerate(s):
if i % 2 == 0:
movex.append(len(t))
else:
movey.append(len(t))
def t_dfs(move, target, is_x):
if is_x:
target -= move.pop(0)
move.sort(reverse=True)
for dx in move:
if target > 0:
target -= dx
else:
target += dx
return target == 0
if t_dfs(movex, x, True) and t_dfs(movey, y, False):
print("Yes")
else:
print("No")
|
s980345161
|
p03380
|
u497952650
| 2,000 | 262,144 |
Wrong Answer
| 113 | 14,428 | 196 |
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())
a = list(map(int,input().split()))
a.sort()
M = max(a)
m = M//2
tmp = 1e9
for i in range(n-1):
if tmp > abs(a[i]-m):
tmp = abs(a[i]-m)
ans = i
print(M,a[ans])
|
s825680020
|
Accepted
| 69 | 14,428 | 203 |
n = int(input())
a = list(map(int,input().split()))
m = max(a)
res = 1e10
ans = 0
for i in a:
if i != m:
if abs(m//2-i) < res:
res = abs(m//2-i)
ans = i
print(m,ans)
|
s951919024
|
p03386
|
u844789719
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 206 |
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
A, B, K = [int(_) for _ in input().split()]
if A+K < B-K:
print("\n".join([str(_) for _ in list(range(A,A+K+1)) + list(range(B-K,B+1))]))
else:
print("\n".join([str(_) for _ in list(range(A,B+1))]))
|
s622831927
|
Accepted
| 17 | 2,940 | 208 |
A, B, K = [int(_) for _ in input().split()]
if 2*K <= B-A:
print("\n".join([str(_) for _ in list(range(A,A+K)) + list(range(B-K+1,B+1))]))
else:
print("\n".join([str(_) for _ in list(range(A,B+1))]))
|
s555865999
|
p03693
|
u154297942
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 109 |
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
a,b,c = map(int, input().split())
if (100 * a + 10 * b + c) % 4 == 0:
print("Yes")
else:
print("No")
|
s407439563
|
Accepted
| 18 | 2,940 | 109 |
a,b,c = map(int, input().split())
if (100 * a + 10 * b + c) % 4 == 0:
print("YES")
else:
print("NO")
|
s873114489
|
p03679
|
u729133443
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 88 |
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though. Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious. Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it. Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
|
x,a,b=map(int,input().split());print([[['safe','dangerous'][b-a>x]],'delicious'][a-b<0])
|
s774245911
|
Accepted
| 27 | 9,148 | 84 |
x,a,b=map(int,input().split())
print(('delicious',('safe','dangerous')[b-a>x])[a<b])
|
s847559461
|
p02612
|
u286754585
| 2,000 | 1,048,576 |
Wrong Answer
| 32 | 8,996 | 82 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n=int(input())
if (n%1000==0):
print(0)
else:
print(int(n-(n//1000)*1000))
|
s177422077
|
Accepted
| 27 | 9,028 | 83 |
n=int(input())
a=n//1000
if (n%1000==0):
print(0)
else:
print((a+1)*1000-n)
|
s467102968
|
p02865
|
u364061715
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 77 |
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
N = int(input())
if N%2==1:
print(int((N-1)/2))
else:
print(int(N/2))
|
s418219845
|
Accepted
| 17 | 2,940 | 81 |
N = int(input())
if N%2==1:
print(int((N-1)/2))
else:
print(int((N-2)/2))
|
s042861721
|
p03485
|
u636822224
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 97 |
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
a, b = map(int, input().split())
if type(a/b) == int:
print(int(a/b))
else:
print(int(a/b+1))
|
s310761979
|
Accepted
| 25 | 3,064 | 54 |
a,b = map(int,input().split())
print(int((a+b+2-1)/2))
|
s631329918
|
p03645
|
u712187387
| 2,000 | 262,144 |
Wrong Answer
| 2,107 | 61,600 | 304 |
In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands. For convenience, we will call them Island 1, Island 2, ..., Island N. There are M kinds of regular boat services between these islands. Each service connects two islands. The i-th service connects Island a_i and Island b_i. Cat Snuke is on Island 1 now, and wants to go to Island N. However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services. Help him.
|
N, M = map(int,input().split())
ships = []
for m in range(M):
ships.append(list(map(int,input().split())))
print(ships)
flag = 0
for n in range(2,N+1):
print(n)
if [1,n] in ships and [n,N] in ships:
print("POSSIBLE")
flag=1
break
if flag==0:
print("IMPOSSIBLE")
|
s760085849
|
Accepted
| 669 | 21,812 | 339 |
N, M = map(int,input().split())
ships_1 = []
ships_2 = []
for m in range(M):
tmp = list(map(int,input().split()))
if tmp[0]==1:
ships_1.append(tmp[1])
elif tmp[1] == N:
ships_2.append(tmp[0])
matched_list = list(set(ships_1) & set(ships_2))
if matched_list:
print("POSSIBLE")
else:
print("IMPOSSIBLE")
|
s172730918
|
p03720
|
u396391104
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 123 |
There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
|
n,m = map(int,input().split())
list = [list(input().split()) for i in range(m)]
for i in range(n):
print(list.count(i+1))
|
s539349229
|
Accepted
| 17 | 2,940 | 154 |
n,m = map(int,input().split())
list = [0]*n
for i in range(m):
a,b = map(int,input().split())
list[a-1] += 1
list[b-1] += 1
[print(c) for c in list]
|
s651001019
|
p03623
|
u768896740
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 66 |
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
n, a, b = map(int, input().split())
print(min(abs(n-a), abs(n-b)))
|
s053365492
|
Accepted
| 17 | 2,940 | 111 |
n, a, b = map(int, input().split())
if min(abs(n-a), abs(n-b)) == abs(n-a):
print('A')
else:
print('B')
|
s073516529
|
p03836
|
u006657459
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 199 |
Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him.
|
sx, sy, tx, ty = map(int, input().split())
tx -= sx
ty -= sy
r = 'U' * ty + 'R' * tx + 'D' * ty + 'L' * (tx+1) + \
'U' * (ty+1) + 'U' * (tx+1) + 'DR' + 'D' * (ty+1) + 'L' * (tx+1) + 'U'
print(r)
|
s356264327
|
Accepted
| 17 | 3,060 | 199 |
sx, sy, tx, ty = map(int, input().split())
tx -= sx
ty -= sy
r = 'U' * ty + 'R' * tx + 'D' * ty + 'L' * (tx+1) + \
'U' * (ty+1) + 'R' * (tx+1) + 'DR' + 'D' * (ty+1) + 'L' * (tx+1) + 'U'
print(r)
|
s374904319
|
p03024
|
u681150536
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 189 |
Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility.
|
from sys import exit
S = input()
min_win = len(S) - 7
win_count = 0
for s in S:
if s == 'o':
win_count += 1
if win_count >= min_win:
print('Yes')
else:
print('No')
|
s028128623
|
Accepted
| 17 | 2,940 | 189 |
from sys import exit
S = input()
min_win = len(S) - 7
win_count = 0
for s in S:
if s == 'o':
win_count += 1
if win_count >= min_win:
print('YES')
else:
print('NO')
|
s207767322
|
p03719
|
u663710122
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 75 |
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
A, B, C = map(int, input().split())
print("YES" if A <= C <= B else "NO")
|
s625874743
|
Accepted
| 21 | 3,316 | 84 |
print((lambda A, B, C: "Yes" if A <= C <= B else "No")(*map(int, input().split())))
|
s015658014
|
p02747
|
u072606168
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 117 |
A Hitachi string is a concatenation of one or more copies of the string `hi`. For example, `hi` and `hihi` are Hitachi strings, while `ha` and `hii` are not. Given a string S, determine whether S is a Hitachi string.
|
s = list(input())
if len(s)%2 == 1:
print("No")
elif s.count("hi") == len(s)//2:
print("Yes")
else:
print("No")
|
s039344564
|
Accepted
| 17 | 2,940 | 111 |
s = input()
if len(s)%2 == 1:
print("No")
elif s.count("hi") == len(s)//2:
print("Yes")
else:
print("No")
|
s416982790
|
p03448
|
u165318982
| 2,000 | 262,144 |
Wrong Answer
| 60 | 9,148 | 244 |
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
A = int(input())
B = int(input())
C = int(input())
X = int(input())
answer = 0
for i in range(A):
for j in range(B):
for k in range(C):
total = 500 * i + 100 * j + 50 * k
if total == X:
answer += 1
print(str(answer))
|
s893226044
|
Accepted
| 58 | 9,172 | 249 |
A = int(input())
B = int(input())
C = int(input())
X = int(input())
answer = 0
for i in range(A+1):
for j in range(B+1):
for k in range(C+1):
total = 500 * i + 100 * j + 50 * k
if total == X:
answer += 1
print(str(answer))
|
s809434231
|
p03448
|
u627283301
| 2,000 | 262,144 |
Wrong Answer
| 51 | 3,060 | 239 |
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.
|
#ABC087 B
A=int(input())
B=int(input())
C=int(input())
X=int(input())
able_number=0
for a in range(A):
for b in range(B):
for c in range(C):
if X==500*a+100*b+50*c:
able_number+=1
print(able_number)
|
s301087210
|
Accepted
| 51 | 3,064 | 245 |
#ABC087 B
A=int(input())
B=int(input())
C=int(input())
X=int(input())
able_number=0
for a in range(A+1):
for b in range(B+1):
for c in range(C+1):
if X==500*a+100*b+50*c:
able_number+=1
print(able_number)
|
s891512431
|
p03699
|
u702582248
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 176 |
You are taking a computer-based examination. The examination consists of N questions, and the score allocated to the i-th question is s_i. Your answer to each question will be judged as either "correct" or "incorrect", and your grade will be the sum of the points allocated to questions that are answered correctly. When you finish answering the questions, your answers will be immediately judged and your grade will be displayed... if everything goes well. However, the examination system is actually flawed, and if your grade is a multiple of 10, the system displays 0 as your grade. Otherwise, your grade is displayed correctly. In this situation, what is the maximum value that can be displayed as your grade?
|
n = int(input())
a = [int(input()) for i in range(n)]
b = min([i for i in a if i % 10] + [0])
if sum(a) % 10:
print(sum(a))
elif b:
print(sum(a) - b)
else:
print(0)
|
s666560538
|
Accepted
| 17 | 3,060 | 160 |
n = int(input())
a = [int(input()) for i in range(n)]
b = min([i for i in a if i % 10] + [sum(a)])
if sum(a) % 10:
print(sum(a))
else:
print(sum(a) - b)
|
s332903739
|
p04029
|
u912115033
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 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)
|
s417808695
|
Accepted
| 17 | 2,940 | 34 |
n = int(input())
print(n*(n+1)//2)
|
s953988534
|
p03494
|
u309018392
| 2,000 | 262,144 |
Time Limit Exceeded
| 2,104 | 3,060 | 108 |
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.
|
input()
A = list(map(int,input().split()))
count=0
while all(a%2==0 for a in A):
count+=1
print(count)
|
s114777705
|
Accepted
| 18 | 3,060 | 129 |
input()
A = list(map(int,input().split()))
count=0
while all(a%2==0 for a in A):
A=[a/2 for a in A]
count+=1
print(count)
|
s921280711
|
p03386
|
u127856129
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 219 |
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,c=map(int,input().split())
d=c+1
e=b-a
f=0
g=0
if d==e:
while e>f:
print(a)
a+=1
f+=1
else:
while c>f:
print(a)
a+=1
f+=1
while c>g:
print(b-1)
b+=1
g+=1
|
s880815984
|
Accepted
| 18 | 3,060 | 169 |
a,b,k=map(int,input().split())
if b-a+1<=k*2:
for i in range(a,1+b):
print(i)
else:
for i in range(a,a+k):
print(i)
for y in range(b-k+1,b+1):
print(y)
|
s305382088
|
p03448
|
u922769680
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 236 |
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
a=int(input())
b=int(input())
c=int(input())
x=int(input())
x=500*a+100*b+50*c
0<= a, b, c <=50
1<= a+b+c <=50
50<= x <=20000
list_a=[0]
for i in range(0,a):
for j in range(0,b):
list_a.append(str("i""j"))
print(len(list_a))
|
s805193485
|
Accepted
| 49 | 3,064 | 244 |
a=int(input())
b=int(input())
c=int(input())
X=int(input())
0<=a,b,c<=50
a+b+c>=1
count=0
for i in range(a+1):
for j in range(b+1):
for k in range(c+1):
if(500*i)+(100*j)+(50*k)==X:
count+=1
print(count)
|
s431214808
|
p02369
|
u067972379
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,008 | 1,091 |
Find a cycle in a directed graph G(V, E).
|
from collections import deque, defaultdict
def topological_sort(V, E):
'''
Kahn's Algorithm (O(|V| + |E|) time)
Input:
V = [0, 1, ..., N-1]: a list of vertices of the digraph
E: the adjacency list of the digraph (dict)
Output:
If the input digraph is acyclic, then return a topological sorting of the digraph.
Else, return None.
'''
indeg = {v: 0 for v in V}
for ends in E.values():
for v in ends:
indeg[v] += 1
q = deque([v for v in V if indeg[v] == 0])
top_sorted = []
while q:
v = q.popleft()
top_sorted.append(v)
for u in E[v]:
indeg[u] -= 1
if indeg[u] == 0:
q.append(u)
if len(top_sorted) == len(V): # The input digraph is acyclic.
return top_sorted
else: # There is a directed cycle in the digraph.
return None
N, M = map(int, input().split())
V = range(N)
E = defaultdict(list)
for _ in range(M):
s, t = map(int, input().split())
E[s].append(t)
print(0 if topological_sort(V, E) is None else 1)
|
s648181220
|
Accepted
| 30 | 6,012 | 1,091 |
from collections import deque, defaultdict
def topological_sort(V, E):
'''
Kahn's Algorithm (O(|V| + |E|) time)
Input:
V = [0, 1, ..., N-1]: a list of vertices of the digraph
E: the adjacency list of the digraph (dict)
Output:
If the input digraph is acyclic, then return a topological sorting of the digraph.
Else, return None.
'''
indeg = {v: 0 for v in V}
for ends in E.values():
for v in ends:
indeg[v] += 1
q = deque([v for v in V if indeg[v] == 0])
top_sorted = []
while q:
v = q.popleft()
top_sorted.append(v)
for u in E[v]:
indeg[u] -= 1
if indeg[u] == 0:
q.append(u)
if len(top_sorted) == len(V): # The input digraph is acyclic.
return top_sorted
else: # There is a directed cycle in the digraph.
return None
N, M = map(int, input().split())
V = range(N)
E = defaultdict(list)
for _ in range(M):
s, t = map(int, input().split())
E[s].append(t)
print(1 if topological_sort(V, E) is None else 0)
|
s711639942
|
p03813
|
u619728370
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 78 |
Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise. You are given Smeke's current rating, x. Print `ABC` if Smeke will participate in ABC, and print `ARC` otherwise.
|
s = input()
a = s.find("A")
z = s.rfind("Z")
print(a)
print(z)
print(z-a)
|
s168079115
|
Accepted
| 17 | 2,940 | 183 |
# -*- coding: utf-8 -*-
x = int(input())
if x<1200:
print("ABC")
else:
print("ARC")
|
s583211350
|
p02646
|
u181644161
| 2,000 | 1,048,576 |
Wrong Answer
| 20 | 9,176 | 187 |
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
a,v=map(int,input().split())
b,w=map(int,input().split())
t=int(input())
flag=0
if w-v<0:
if abs(a-b) <= t*abs(v-w):
flag=1
if flag==1:
print("Yes")
else:
print("No")
|
s458068525
|
Accepted
| 25 | 9,176 | 187 |
a,v=map(int,input().split())
b,w=map(int,input().split())
t=int(input())
flag=0
if w-v<0:
if abs(a-b) <= t*abs(v-w):
flag=1
if flag==1:
print("YES")
else:
print("NO")
|
s714734030
|
p02613
|
u954954040
| 2,000 | 1,048,576 |
Wrong Answer
| 28 | 9,164 | 33 |
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())
print(N+N*N+N*N*N)
|
s418358806
|
Accepted
| 145 | 16,248 | 260 |
list=[]
count=int(input())
for i in range(count):
list.append(input())
print("AC"+" "+"x"+" "+str(list.count("AC")))
print("WA"+" "+"x"+" "+str(list.count("WA")))
print("TLE"+" "+"x"+" "+str(list.count("TLE")))
print("RE"+" "+"x"+" "+str(list.count("RE")))
|
s625290020
|
p02645
|
u081365646
| 2,000 | 1,048,576 |
Wrong Answer
| 20 | 9,020 | 24 |
When you asked some guy in your class his name, he called himself S, where S is a string of length between 3 and 20 (inclusive) consisting of lowercase English letters. You have decided to choose some three consecutive characters from S and make it his nickname. Print a string that is a valid nickname for him.
|
s = input()
print(s[:2])
|
s602585775
|
Accepted
| 22 | 9,024 | 24 |
s = input()
print(s[:3])
|
s928769825
|
p02613
|
u966542724
| 2,000 | 1,048,576 |
Wrong Answer
| 150 | 9,200 | 294 |
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(n):
s = input()
if s == 'AC':
ac += 1
elif s == 'WA':
wa += 1
elif s == 'TLE':
tle += 1
else:
re += 1
print('AC ×', ac)
print('WA ×', wa)
print('TLE ×', tle)
print('RE ×', re)
|
s081404524
|
Accepted
| 151 | 9,004 | 290 |
n = int(input())
ac = 0
wa = 0
tle = 0
re = 0
for i in range(n):
s = input()
if s == 'AC':
ac += 1
elif s == 'WA':
wa += 1
elif s == 'TLE':
tle += 1
else:
re += 1
print('AC x', ac)
print('WA x', wa)
print('TLE x', tle)
print('RE x', re)
|
s631983588
|
p02612
|
u733866054
| 2,000 | 1,048,576 |
Wrong Answer
| 30 | 9,156 | 137 |
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())
otsuri=10000
for i in range(10) :
if N-1000*i>=0 :
coin=N-1000*i
otsuri=min(otsuri,coin)
print(otsuri)
|
s057293660
|
Accepted
| 26 | 9,080 | 137 |
N=int(input())
otsuri=10000
for i in range(11) :
if 1000*i-N>=0 :
coin=1000*i-N
otsuri=min(otsuri,coin)
print(otsuri)
|
s933750682
|
p03814
|
u546440137
| 2,000 | 262,144 |
Wrong Answer
| 60 | 10,580 | 189 |
Snuke has decided to construct a string that starts with `A` and ends with `Z`, by taking out a substring of a string s (that is, a consecutive part of s). Find the greatest length of the string Snuke can construct. Here, the test set guarantees that there always exists a substring of s that starts with `A` and ends with `Z`.
|
s=list(input())
a=0
z=0
for i in range(len(s)):
if i=="A":
a=i
break
ss=s.reverse()
for i in range(len(s)):
if i=="Z":
z=i
break
print(len(s)-(a+z))
|
s946612618
|
Accepted
| 48 | 11,948 | 194 |
s=list(input())
a=0
z=0
for i in range(len(s)):
if s[i]=="A":
a=i
break
S=list(reversed(s))
for j in range(len(S)):
if S[j]=="Z":
z=j
break
print(len(s)-a-z)
|
s016854446
|
p03471
|
u123648284
| 2,000 | 262,144 |
Wrong Answer
| 759 | 3,060 | 289 |
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 = list(map(int, input().split()))
l = [0, 0, 0]
flg = False
for i in range(N+1):
if flg:
break
for j in range(N+1-i):
k = N-(i+j)
if Y == 1000*i + 5000*j + 10000*k:
print('{} {} {}'.format(i, j, k))
flg = True
break
if not flg:
print('-1 -1 -1')
|
s066927841
|
Accepted
| 1,097 | 3,060 | 312 |
N, Y = map(int, input().split())
res = ["-1", "-1", "-1"]
for i in range(N+1):
if i * 10000 > Y:
break
for j in range(N-i+1):
if i * 10000 + j * 5000 > Y:
break
k = N-(i+j)
if i * 10000 + j * 5000 + k * 1000 == Y:
res = [str(i), str(j), str(k)]
break
print(" ".join(res))
|
s461282452
|
p03606
|
u088552457
| 2,000 | 262,144 |
Wrong Answer
| 23 | 3,060 | 156 |
Joisino is working as a receptionist at a theater. The theater has 100000 seats, numbered from 1 to 100000. According to her memo, N groups of audiences have come so far, and the i-th group occupies the consecutive seats from Seat l_i to Seat r_i (inclusive). How many people are sitting at the theater now?
|
n = int(input())
answer = 0
for i in range(n):
a, b = map(int, input().split())
if a == b:
answer += 1
else:
answer += b - a
print(answer)
|
s896253253
|
Accepted
| 20 | 3,060 | 123 |
n = int(input())
answer = 0
for i in range(n):
a, b = map(int, input().split())
answer += (b + 1) - a
print(answer)
|
s397992513
|
p03457
|
u558782626
| 2,000 | 262,144 |
Wrong Answer
| 378 | 3,060 | 213 |
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())
t0, x0, y0 = (0, 0, 0)
for _ in range(N):
t1, x1, y1 = map(int, input().split())
if abs(x1-x0)+abs(y1-y0) > t1-t0 or (abs(x1-x0)+abs(y1-y0)) % (t1-t0):
print('No')
break
else:
print('Yes')
|
s147245896
|
Accepted
| 392 | 3,064 | 243 |
N = int(input())
t0, x0, y0 = (0, 0, 0)
for _ in range(N):
t1, x1, y1 = map(int, input().split())
if abs(x1-x0)+abs(y1-y0) > t1-t0 or (abs(x1-x0)+abs(y1-y0))%2 != (t1-t0)%2:
print('No')
break
t0, x0, y0 = t1, x1, y1
else:
print('Yes')
|
s092792313
|
p04043
|
u490553751
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 164 |
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.
|
import sys
input = sys.stdin.readline
li = [int(i) for i in input().split()]
li.sort()
a = "No"
if li[0] == 5 and li[1] == 5 and li[2] == 7 :
a = "Yes"
print(a)
|
s167319169
|
Accepted
| 16 | 2,940 | 164 |
import sys
input = sys.stdin.readline
li = [int(i) for i in input().split()]
li.sort()
a = "NO"
if li[0] == 5 and li[1] == 5 and li[2] == 7 :
a = "YES"
print(a)
|
s850391597
|
p03067
|
u929813319
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 190 |
There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise.
|
a,b,c = list(map(int,input().split()))
if b-a> 0:
if c-b > 0:
print("no")
else:
print("yes")
else:
if b-c > 0:
print("no")
else:
print("yes")
|
s074912732
|
Accepted
| 17 | 3,060 | 145 |
a,b,c = list(map(int,input().split()))
if b-c > 0 and c-a > 0:
print("Yes")
elif a-c > 0 and c-b > 0:
print("Yes")
else:
print("No")
|
s416575159
|
p03624
|
u315868385
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,188 | 141 |
You are given a string S consisting of lowercase English letters. Find the lexicographically (alphabetically) smallest lowercase English letter that does not occur in S. If every lowercase English letter occurs in S, print `None` instead.
|
S = input()
for i in range(ord('a'), ord('z')+1):
c = chr(i)
if c not in S:
print(c)
if c == 'z':
print("None")
|
s026005819
|
Accepted
| 17 | 3,188 | 137 |
S = input()
l = [chr(i) for i in range(ord('a'), ord('z')+1) if chr(i) not in S]
if len(l) == 0:
print("None")
else:
print(l[0])
|
s001989615
|
p03160
|
u683956577
| 2,000 | 1,048,576 |
Wrong Answer
| 141 | 14,660 | 343 |
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
import sys
stdin = sys.stdin
ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
ns = lambda: stdin.readline().rstrip() # ignore trailing spaces
n = ni()
h = na()
dp = [0]*n
dp[0] = 0
dp[1] = abs(h[1]-h[0])
count = 1
for i in range(2,n):
dp[i] = min(abs(h[i-1]-h[i])+dp[i-1], abs(h[i-2]-h[i])+dp[i-2])
print(dp)
|
s179442893
|
Accepted
| 153 | 14,012 | 393 |
import sys
stdin = sys.stdin
ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
ns = lambda: stdin.readline().rstrip() # ignore trailing spaces
n = ni()
h = na()
A = [0 for _ in range(n)]
for i in range(n):
if i == 1:
A[i] = abs(h[i]-h[i-1])
elif i>1:
A[i] = min(A[i-1]+abs(h[i]-h[i-1]),A[i-2]+abs(h[i]-h[i-2]))
else:
A[i] = A[i]
print(A[-1])
|
s822375857
|
p02397
|
u002193969
| 1,000 | 131,072 |
Wrong Answer
| 50 | 5,612 | 114 |
Write a program which reads two integers x and y, and prints them in ascending order.
|
while True:
a, b = [int(x) for x in input().split()]
if a == 0 and b == 0:
break
print(a, b)
|
s432988419
|
Accepted
| 60 | 5,616 | 162 |
while True:
a, b = [int(x) for x in input().split()]
if a == 0 and b == 0:
break
if a < b:
print(a, b)
else:
print(b, a)
|
s930562745
|
p03672
|
u859897687
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 121 |
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(len(s)-1,0,-1):
if i%2>0:
continue
if s[:i//2]==s[i//2:i]:
print(s[:i//2])
break
|
s753317549
|
Accepted
| 18 | 2,940 | 114 |
s=input()
for i in range(len(s)-1,0,-1):
if i%2>0:
continue
if s[:i//2]==s[i//2:i]:
print(i)
break
|
s027310739
|
p02393
|
u152353734
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,724 | 104 |
Write a program which reads three integers, and prints them in ascending order.
|
(a, b, c) = [int(i) for i in input().split()]
if a < b < c:
print('1 3 8')
else:
print('3 1 8')
|
s550593200
|
Accepted
| 30 | 6,728 | 282 |
nums = input().split()
a = int(nums[0])
b = int(nums[1])
c = int(nums[2])
if a <= b <= c:
print(a, b, c)
elif a <= c <= b:
print(a, c, b)
elif b <= a <= c:
print(b, a, c)
elif b <= c <= a:
print(b, c, a)
elif c <= a <= b:
print(c, a, b)
else:
print(c, b, a)
|
s065479205
|
p02394
|
u727538672
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,528 | 147 |
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
|
W, H, x, y, r = map(int, input().split())
if (x + r) > W and (x - r) > 0 and (y + r) > H and (y - r) > 0 :
print("Yes")
else :
print("No")
|
s609701180
|
Accepted
| 20 | 7,656 | 151 |
W, H, x, y, r = map(int, input().split())
if (x + r) <= W and (x - r) >= 0 and (y + r) <= H and (y - r) >= 0 :
print("Yes")
else :
print("No")
|
s449735868
|
p03455
|
u684084063
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 164 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
# -*- coding: utf-8 -*-
import math
a,b =map(int,input().split())
N =str(a)+str(b)
ans= math.sqrt(int(N))
if int(N)==ans*ans:
print("Yes")
else:
print("No")
|
s497306239
|
Accepted
| 17 | 2,940 | 112 |
# -*- coding: utf-8 -*-
a,b =map(int,input().split())
N=a*b
if 0==N%2:
print("Even")
else:
print("Odd")
|
s740173202
|
p03574
|
u798316285
| 2,000 | 262,144 |
Wrong Answer
| 22 | 3,444 | 446 |
You are given an H × W grid. The squares in the grid are described by H strings, S_1,...,S_H. The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W). `.` stands for an empty square, and `#` stands for a square containing a bomb. Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square. (Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.) He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square. Print the strings after the process.
|
h,w=map(int,input().split())
S=[["."]*(w+2)]+[["."]+list(input())+["."] for _ in range(h)]+[["."]*(w+2)]
ans=[[0]*w]*h
for i in range(1,h+1):
for j in range(1,w+1):
if S[i][j]=="#":
ans[i-1][j-1]="#"
else:
count=(S[i-1][j-1]=="#")+(S[i-1][j]=="#")+(S[i-1][j+1]=="#")+(S[i][j-1]=="#")+(S[i][j+1]=="#")+(S[i+1][j-1]=="#")+(S[i+1][j]=="#")+(S[i+1][j+1]=="#")
ans[i-1][j-1]=count
for i in range(h):
print(*ans[i],sep="")
|
s197306352
|
Accepted
| 21 | 3,444 | 478 |
h,w=map(int,input().split())
S=[["."]*(w+2)]+[["."]+list(input())+["."] for _ in range(h)]+[["."]*(w+2)]
ans=[[0 for i in range(w)] for j in range(h)]
for i in range(1,h+1):
for j in range(1,w+1):
if S[i][j]=="#":
ans[i-1][j-1]="#"
else:
count=(S[i-1][j-1]=="#")+(S[i-1][j]=="#")+(S[i-1][j+1]=="#")+(S[i][j-1]=="#")+(S[i][j+1]=="#")+(S[i+1][j-1]=="#")+(S[i+1][j]=="#")+(S[i+1][j+1]=="#")
ans[i-1][j-1]=count
for i in range(h):
print(*ans[i],sep="")
|
s000992162
|
p03005
|
u267222408
| 2,000 | 1,048,576 |
Wrong Answer
| 22 | 2,940 | 79 |
Takahashi is distributing N balls to K persons. If each person has to receive at least one ball, what is the maximum possible difference in the number of balls received between the person with the most balls and the person with the fewest balls?
|
N, K = map(int, input().split())
if K == 1:
print(0)
else:
print(N-K-1)
|
s180229798
|
Accepted
| 17 | 2,940 | 78 |
N, K = map(int, input().split())
if K == 1:
print(0)
else:
print(N-K)
|
s391190439
|
p03387
|
u588558668
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 217 |
You are given three integers A, B and C. Find the minimum number of operations required to make A, B and C all equal by repeatedly performing the following two kinds of operations in any order: * Choose two among A, B and C, then increase both by 1. * Choose one among A, B and C, then increase it by 2. It can be proved that we can always make A, B and C all equal by repeatedly performing these operations.
|
l=list(map(int,input().split()))
l.sort()
a=l[0]
b=l[1]
c=l[2]
ans=0
while a<c-1:
a+=2
ans+=1
while b<c-1:
b+=2
ans+=1
if a==c and b==c:
print(ans)
elif a!=c and b!=c:
print(ans+2)
else:
print(ans+1)
|
s283516449
|
Accepted
| 17 | 3,060 | 218 |
l=list(map(int,input().split()))
l.sort()
a=l[0]
b=l[1]
c=l[2]
ans=0
while a<c-1:
a+=2
ans+=1
while b<c-1:
b+=2
ans+=1
if a==c and b==c:
print(ans)
elif a!=c and b!=c:
print(ans+1)
else:
print(ans+2)
|
s734067568
|
p02743
|
u933341648
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 93 |
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
|
import math
a, b, c = map(int, input().split())
ans = a + b < c
print('Yes' if ans else 'No')
|
s445802785
|
Accepted
| 17 | 2,940 | 114 |
a, b, c = map(int, input().split())
x = c - a - b
ans = x > 0 and 4 * a * b < x ** 2
print('Yes' if ans else 'No')
|
s281697288
|
p03388
|
u263830634
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 347 |
10^{10^{10}} participants, including Takahashi, competed in two programming contests. In each contest, all participants had distinct ranks from first through 10^{10^{10}}-th. The _score_ of a participant is the product of his/her ranks in the two contests. Process the following Q queries: * In the i-th query, you are given two positive integers A_i and B_i. Assuming that Takahashi was ranked A_i-th in the first contest and B_i-th in the second contest, find the maximum possible number of participants whose scores are smaller than Takahashi's.
|
import math
Q = int(input())
for _ in range(Q):
A, B = map(int, input().split())
score = A*B
if A > B:
A, B = B, A
if A == B:
print (2 * A -2)
elif A+1 == B:
print (2 * A - 2)
else:
c = int(math.sqrt(A*B))
print (2 * c -1)
|
s844915096
|
Accepted
| 18 | 3,064 | 464 |
import math
Q = int(input())
for _ in range(Q):
A, B = map(int, input().split())
score = A*B
if A > B:
A, B = B, A
if A == B:
print (2 * A -2)
elif A+1 == B:
print (2 * A - 2)
else:
c = int(math.sqrt(A*B))
if c ** 2 == A * B:
c -= 1
if c * (c+1) >= A * B:
print (2 * c -2)
else:
print (2 * c -1)
|
s192502481
|
p00027
|
u560214129
| 1,000 | 131,072 |
Wrong Answer
| 30 | 8,064 | 550 |
Your task is to write a program which reads a date (from 2004/1/1 to 2004/12/31) and prints the day of the date. Jan. 1, 2004, is Thursday. Note that 2004 is a leap year and we have Feb. 29.
|
from datetime import date
import sys
for line in sys.stdin.readlines():
m , d = map(int,line.split())
if m != 0 and d != 0 :
val = date(2004,m,d)
day = val.isoweekday()
if day == 0:
print("Sunday")
elif day == 1:
print("Monday")
elif day == 2:
print("Tuesday")
elif day == 3:
print("Wednesday")
elif day == 4:
print("Thursday")
elif day == 5:
print("Friday")
else:
print("Saturday")
else:
break
|
s929465513
|
Accepted
| 30 | 8,048 | 550 |
from datetime import date
import sys
for line in sys.stdin.readlines():
m , d = map(int,line.split())
if m != 0 and d != 0 :
val = date(2004,m,d)
day = val.isoweekday()
if day == 7:
print("Sunday")
elif day == 1:
print("Monday")
elif day == 2:
print("Tuesday")
elif day == 3:
print("Wednesday")
elif day == 4:
print("Thursday")
elif day == 5:
print("Friday")
else:
print("Saturday")
else:
break
|
s474894075
|
p02390
|
u830563109
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,536 | 44 |
Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively.
|
s = 2000
m = s / 60
h = m / 60
print(s,m,h)
|
s598239501
|
Accepted
| 20 | 5,588 | 92 |
a = int(input())
h = int(a/3600)
m=int((a%3600)/60)
s=int((a%3600)%60)
print(h,m,s,sep=":")
|
s270022212
|
p02393
|
u050103511
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,444 | 57 |
Write a program which reads three integers, and prints them in ascending order.
|
list = [i for i in input().split()]
print(" ".join(list))
|
s403381218
|
Accepted
| 20 | 7,672 | 82 |
L = [int(i) for i in input().split()]
print(' '.join([str(i) for i in sorted(L)]))
|
s418538757
|
p03854
|
u697559326
| 2,000 | 262,144 |
Wrong Answer
| 87 | 3,572 | 694 |
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()
word = ["dream","dreamer","erase","eraser"]
S = S[::-1]
for i in range(len(word)): word[i] = word[i][::-1]
print(S, S[5:], S[11:] == "")
while True:
for i in range(len(word)):
flag2 = False
if word[i] == S[:len(word[i])]:
flag2 = True
S = S[len(word[i]):]
break
if S == "":
print("YES")
break
elif flag2 == False:
print("NO")
break
|
s710761112
|
Accepted
| 92 | 3,188 | 629 |
S = input()
word = ["dream","dreamer","erase","eraser"]
S = S[::-1]
for i in range(len(word)): word[i] = word[i][::-1]
while True:
for i in range(len(word)):
flag2 = False
if word[i] == S[:len(word[i])]:
flag2 = True
S = S[len(word[i]):]
break
if S == "":
print("YES")
break
elif flag2 == False:
print("NO")
break
|
s839981874
|
p00042
|
u078042885
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,680 | 327 |
宝物がたくさん収蔵されている博物館に、泥棒が大きな風呂敷を一つだけ持って忍び込みました。盗み出したいものはたくさんありますが、風呂敷が耐えられる重さが限られており、これを超えると風呂敷が破れてしまいます。そこで泥棒は、用意した風呂敷を破らず且つ最も価値が高くなるようなお宝の組み合わせを考えなくてはなりません。 風呂敷が耐えられる重さ W、および博物館にある個々のお宝の価値と重さを読み込んで、重さの総和が W を超えない範囲で価値の総和が最大になるときの、お宝の価値総和と重さの総和を出力するプログラムを作成してください。ただし、価値の総和が最大になる組み合わせが複数あるときは、重さの総和が小さいものを出力することとします。
|
c=1
while 1:
W=int(input())
if W==0:break
n=int(input())
dp=[0]*(W+1)
for i in range(n):
v,w = map(int, input().split(','))
for j in range(W,w-1,-1):
dp[j]=max(dp[j-w]+v,dp[j])
for i in range(W):
if dp[W]==dp[i]:break
print('Case %d\n%d\n%d'%(c,dp[W],i))
c+=1
|
s763289547
|
Accepted
| 370 | 7,652 | 328 |
c=1
while 1:
W=int(input())
if W==0:break
dp=[0]*(W+1)
for i in range(int(input())):
v,w=map(int, input().split(','))
for j in range(W,w-1,-1):
if dp[j]<dp[j-w]+v:dp[j]=dp[j-w]+v
for i in range(W+1):
if dp[W]==dp[i]:break
print('Case %d:\n%d\n%d'%(c,dp[W],i))
c+=1
|
s811424017
|
p03644
|
u982139831
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 80 |
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
N = int(input())
if N%2==0:
print(N)
elif N==1:
print(N)
else:
print(N-1)
|
s367998253
|
Accepted
| 17 | 2,940 | 223 |
N = int(input())
num_list=[]
for num in range(1,N+1):
counter=0
while True:
if int(num/2)==num/2:
num/=2
counter+=1
else:
break
num_list.append(counter)
print(num_list.index(max(num_list))+1)
|
s709506157
|
p03024
|
u296518383
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 47 |
Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility.
|
print("No" if input().count("x")>=8 else "Yes")
|
s337032455
|
Accepted
| 17 | 2,940 | 47 |
print("NO" if input().count("x")>=8 else "YES")
|
s724345343
|
p03050
|
u063896676
| 2,000 | 1,048,576 |
Wrong Answer
| 160 | 2,940 | 173 |
Snuke received a positive integer N from Takahashi. A positive integer m is called a _favorite number_ when the following condition is satisfied: * The quotient and remainder of N divided by m are equal, that is, \lfloor \frac{N}{m} \rfloor = N \bmod m holds. Find all favorite numbers and print the sum of those.
|
# coding: utf-8
n = int(input())
output = 0
# n = (m+1)*x = y*x (y>=2)
for x in range(1, int(pow(n, 0.5))):
if n % x == 0:
output += n // x - 1
print(output)
|
s772846437
|
Accepted
| 311 | 3,060 | 199 |
# coding: utf-8
n = int(input())
output = 0
# n = (m+1)*x = y*x (y>=2)
limit = pow(n+0.25, 0.5)-0.5
x = 1
while x < limit:
if n % x == 0:
output += n // x - 1
x += 1
print(output)
|
s574569700
|
p02314
|
u724548524
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,600 | 409 |
Find the minimum number of coins to make change for n cents using coins of denominations d1, d2,.., dm. The coins can be used any number of times.
|
def cs(n, p, c):
global count, m
if p >= count:
return m
if n % c[0] == 0:
count = min(count, p + n // c[0])
return count
elif c[0] < n:
return min(cs(n - c[0], p + 1, c), cs(n, p, c[1:]))
else:
return cs(n, p, c[1:])
n, m = map(int, input().split())
c = list(map(int, input().split()))
c.sort(reverse = True)
print(c)
count = n
print(cs(n, 0, c))
|
s194791978
|
Accepted
| 60 | 5,604 | 414 |
def cs(n, p, c):
global count, m
if n % c[0] == 0:
count = min(count, p + n // c[0])
return count
elif p + n // c[0] >= count:
return n
elif c[0] < n:
return min(cs(n - c[0], p + 1, c), cs(n, p, c[1:]))
else:
return cs(n, p, c[1:])
n, m = map(int, input().split())
c = list(map(int, input().split()))
c.sort(reverse = True)
count = n
print(cs(n, 0, c))
|
s427879074
|
p03852
|
u246572032
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 96 |
Given a lowercase English letter c, determine whether it is a vowel. Here, there are five vowels in the English alphabet: `a`, `e`, `i`, `o` and `u`.
|
x = input()
print('vowel' if x=='a' and x=='e' and x=='i' and x=='o' and x=='u' else'consonant')
|
s619727228
|
Accepted
| 17 | 2,940 | 59 |
c = input()
print('vowel' if c in 'aeiou' else 'consonant')
|
s341539371
|
p03353
|
u227082700
| 2,000 | 1,048,576 |
Wrong Answer
| 35 | 4,340 | 119 |
You are given a string s. Among the **different** substrings of s, print the K-th lexicographically smallest one. A substring of s is a string obtained by taking out a non-empty contiguous part in s. For example, if s = `ababc`, `a`, `bab` and `ababc` are substrings of s, while `ac`, `z` and an empty string are not. Also, we say that substrings are different when they are different as strings. Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.
|
s=input()
k=int(input())
a=[]
for i in range(k):
for j in range(len(s)-i):a.append(s[j:j+i+1])
a.sort()
print(a[k-1])
|
s418422407
|
Accepted
| 34 | 4,592 | 122 |
s=input()
k=int(input())
a={s}
for i in range(k):
for j in range(len(s)-i):a.add(s[j:j+i+1])
print(sorted(list(a))[k-1])
|
s477957747
|
p03455
|
u023229441
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 79 |
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("Even")
else:
print("Odd")
|
s021793515
|
Accepted
| 17 | 2,940 | 59 |
a,b=map(int,input().split())
print("EOvdedn"[a*b%2==1::2])
|
s731322656
|
p03796
|
u589730912
| 2,000 | 262,144 |
Wrong Answer
| 2,104 | 10,136 | 99 |
Snuke loves working out. He is now exercising N times. Before he starts exercising, his _power_ is 1. After he exercises for the i-th time, his power gets multiplied by i. Find Snuke's power after he exercises N times. Since the answer can be extremely large, print the answer modulo 10^{9}+7.
|
N = int(input())
power = 1
for i in range(1, N+1):
power *= i
print(power **N % (10 ** 9 + 7))
|
s698699050
|
Accepted
| 34 | 2,940 | 111 |
N = int(input())
power = 1
div = 10 ** 9 + 7
for i in range(1, N+1):
power = power * i % div
print(power)
|
s053721163
|
p03486
|
u379424722
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 161 |
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.
|
n = input()
#print(n)
n = ''.join(sorted(n))
m = input()
#print(m)
m = ''.join(sorted(m))
#print(n)
#print(m)
if n < m:
print("Yes")
else:
print("No")
|
s107949640
|
Accepted
| 20 | 3,060 | 177 |
n = input()
#print(n)
n = ''.join(sorted(n))
m = input()
#print(m)
m = ''.join(sorted(m, reverse = True))
#print(n)
#print(m)
if n < m:
print("Yes")
else:
print("No")
|
s830438502
|
p02747
|
u661977789
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 68 |
A Hitachi string is a concatenation of one or more copies of the string `hi`. For example, `hi` and `hihi` are Hitachi strings, while `ha` and `hii` are not. Given a string S, determine whether S is a Hitachi string.
|
h = input().replace("hi", "")
o = "Yes" if len(h) else "No"
print(o)
|
s324612004
|
Accepted
| 17 | 2,940 | 51 |
print("No" if input().replace("hi", "") else "Yes")
|
s969583103
|
p03474
|
u220345792
| 2,000 | 262,144 |
Wrong Answer
| 19 | 2,940 | 127 |
The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.
|
l = input().split()
a=input()
b=a.split("-")
if len(b[0]) == l[0] and len(b[1])==l[1]:
print('YES')
else:
print('NO')
|
s634358400
|
Accepted
| 17 | 2,940 | 119 |
A, B = map(int, input().split())
S = input()
if S[A] == "-" and S.count("-") == 1:
print("Yes")
else:
print("No")
|
s402871765
|
p04029
|
u475675023
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 31 |
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))
|
s241167526
|
Accepted
| 17 | 2,940 | 32 |
N=int(input());print(N*(N+1)//2)
|
s901940380
|
p02972
|
u186967328
| 2,000 | 1,048,576 |
Wrong Answer
| 556 | 20,540 | 327 |
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.
|
N = int(input())
a = list(map(int,input().split()))
b = [0 for i in range(N)]
def check(i):
cnt = 0
n = i
while n <= N:
cnt += b[n-1]
n += i
return cnt
for i in range(N-1,-1,-1):
if a[i] + check(i+1) == 1:
b[i] = 1
print(b.count(1))
L = [str(int(i)) for i in b]
print(' '.join(L))
|
s550455592
|
Accepted
| 569 | 22,324 | 423 |
N = int(input())
a = list(map(int,input().split()))
b = [0 for i in range(N)]
def check(i):
cnt = 0
n = i
while n <= N:
cnt += b[n-1]
n += i
return cnt
for i in range(N-1,-1,-1):
if (a[i] + check(i+1))%2 == 1:
b[i] = 1
ans = []
cnt = 0
for i in range(N):
if b[i] == 1:
cnt += 1
ans.append(i+1)
print(cnt)
L = [str(int(i)) for i in ans]
print(' '.join(L))
|
s733622852
|
p03681
|
u216392490
| 2,000 | 262,144 |
Wrong Answer
| 36 | 3,060 | 350 |
Snuke has N dogs and M monkeys. He wants them to line up in a row. As a Japanese saying goes, these dogs and monkeys are on bad terms. _("ken'en no naka", literally "the relationship of dogs and monkeys", means a relationship of mutual hatred.)_ Snuke is trying to reconsile them, by arranging the animals so that there are neither two adjacent dogs nor two adjacent monkeys. How many such arrangements there are? Find the count modulo 10^9+7 (since animals cannot understand numbers larger than that). Here, dogs and monkeys are both distinguishable. Also, two arrangements that result from reversing each other are distinguished.
|
import sys
sys.setrecursionlimit(int(1e6))
N, M = list(map(int, input().split()))
base = int(1e9 + 7)
ans = 0
if abs(N -M) > 1:
pass
else:
fact = 1
for i in range(1, min(N+1, M+1)):
fact = (fact * i) % (base)
if N == M:
ans = 2 * fact * fact
else:
ans = fact * fact * max(N, M)
print(ans)
ans = ans % base
print(ans)
|
s330959585
|
Accepted
| 36 | 3,060 | 338 |
import sys
sys.setrecursionlimit(int(1e6))
N, M = list(map(int, input().split()))
base = int(1e9 + 7)
ans = 0
if abs(N -M) > 1:
pass
else:
fact = 1
for i in range(1, min(N+1, M+1)):
fact = (fact * i) % (base)
if N == M:
ans = 2 * fact * fact
else:
ans = fact * fact * max(N, M)
ans = ans % base
print(ans)
|
s023780876
|
p03693
|
u835924161
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 95 |
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
a,b,c=map(str,input().split())
a+=b
a+=c
if int(a)%4==0:
print("Yes")
else:
print("No")
|
s175203201
|
Accepted
| 17 | 2,940 | 95 |
a,b,c=map(str,input().split())
a+=b
a+=c
if int(a)%4==0:
print("YES")
else:
print("NO")
|
s725896869
|
p02401
|
u940150266
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,604 | 271 |
Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part.
|
while(True):
a,op,b = input().split()
if op == '+':
print(int(a)+int(b))
elif op == '-':
print(int(a)-int(b))
elif op == '*':
print(int(a)*int(b))
elif op == '/':
print(int(a)/int(b))
elif op == '?':
break
|
s731156631
|
Accepted
| 20 | 5,600 | 276 |
while(True):
a,op,b = input().split()
if op == '+':
print(int(a)+int(b))
elif op == '-':
print(int(a)-int(b))
elif op == '*':
print(int(a)*int(b))
elif op == '/':
print(int(int(a)/int(b)))
elif op == '?':
break
|
s875081265
|
p03163
|
u763115743
| 2,000 | 1,048,576 |
Wrong Answer
| 2,110 | 109,428 | 585 |
There are N items, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), Item i has a weight of w_i and a value of v_i. Taro has decided to choose some of the N items and carry them home in a knapsack. The capacity of the knapsack is W, which means that the sum of the weights of items taken must be at most W. Find the maximum possible sum of the values of items that Taro takes home.
|
def solve(n, w, items):
dp = [[0] * (w + 1) for i in range(n + 1)]
for i in range(1, n + 1):
for j in range(1, w + 1):
prev = i - 1
dp[i][j] = max(dp[i][j], dp[prev][j])
weight, value = items[i - 1][0], items[i - 1][1]
if j - weight >= 0:
dp[i][j] = max(dp[i][j], dp[i][j - weight] + value)
return dp[n][w]
def main():
n, w = map(int, input().split())
items = []
for i in range(n):
items.append(list(map(int, input().split())))
ans = solve(n, w, items)
print(ans)
main()
|
s195418154
|
Accepted
| 303 | 20,116 | 370 |
import numpy as np
def solve(w, items):
dp = np.zeros(w + 1, dtype=int)
for weight, value in items:
dp[weight:] = np.maximum(dp[weight:], dp[:w-weight+1] + value)
return dp[w]
def main():
n, w = map(int, input().split())
items = list(list(map(int, input().split())) for _ in range(n))
ans = solve(w, items)
print(ans)
main()
|
s356490979
|
p02401
|
u922112509
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,612 | 332 |
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.
|
# Simple Calculator
end = 0
while end == 0:
n = [i for i in input().rstrip().split()]
a = int(n[0])
op = n[1]
b = int(n[2])
if op == '+':
print(a + b)
elif op == '-':
print(a - b)
elif op == '*':
print(a * b)
elif op == '/':
print(a / b)
else:
end += 1
|
s538306254
|
Accepted
| 20 | 5,604 | 333 |
# Simple Calculator
end = 0
while end == 0:
n = [i for i in input().rstrip().split()]
a = int(n[0])
op = n[1]
b = int(n[2])
if op == '+':
print(a + b)
elif op == '-':
print(a - b)
elif op == '*':
print(a * b)
elif op == '/':
print(a // b)
else:
end += 1
|
s754922559
|
p02612
|
u848097937
| 2,000 | 1,048,576 |
Wrong Answer
| 32 | 9,024 | 78 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N = int(input())
if N % 1000:
print(1000)
else:
print(1000 - N % 1000)
|
s949182830
|
Accepted
| 28 | 9,028 | 77 |
n = int(input())
if n % 1000 == 0:
print(0)
else:
print(1000 - n % 1000)
|
s120435557
|
p03860
|
u508141157
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 24 |
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%sC",input()[0])
|
s573230800
|
Accepted
| 20 | 2,940 | 25 |
print("A"+input()[8]+"C")
|
s134726885
|
p03548
|
u497046426
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 67 |
We have a long seat of width X centimeters. There are many people who wants to sit here. A person sitting on the seat will always occupy an interval of length Y centimeters. We would like to seat as many people as possible, but they are all very shy, and there must be a gap of length at least Z centimeters between two people, and between the end of the seat and a person. At most how many people can sit on the seat?
|
X, Y, Z = map(int, input().split())
ans = X // (Y + 2*Z)
print(ans)
|
s868307765
|
Accepted
| 17 | 3,064 | 71 |
X, Y, Z = map(int, input().split())
ans = (X - Z) // (Y + Z)
print(ans)
|
s834090465
|
p03672
|
u397563544
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 115 |
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()
n = len(s)//2
for i in range(n):
if s[:n-i] == s[n-i:2*(n-i)]:
print(2*(n-i))
break
|
s520033011
|
Accepted
| 18 | 2,940 | 117 |
s = input()
n = len(s)//2
for i in range(1,n):
if s[:n-i] == s[n-i:2*(n-i)]:
print(2*(n-i))
break
|
s321164929
|
p02678
|
u029785897
| 2,000 | 1,048,576 |
Wrong Answer
| 881 | 48,888 | 661 |
There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.
|
from collections import deque
N, M = map(int, input().split())
Gs = [[] for _ in range(N+1)]
for _ in range(M):
a, b = map(int, input().split())
Gs[a].append(b)
Gs[b].append(a)
ms = [0] * (N+1)
def bfs(c):
queue = deque()
ns = Gs[c]
for n in ns:
queue.append((n, c))
while queue:
c, p = queue.popleft()
if ms[c] != 0:
continue
ms[c] = p
ns = Gs[c]
for n in ns:
if n != p:
queue.append((n, c))
bfs(1)
print(ms)
for m in ms[2:]:
if m == 0:
print('No')
break
else:
print('Yes')
for m in ms[2:]:
print(m)
|
s562498203
|
Accepted
| 800 | 48,808 | 651 |
from collections import deque
N, M = map(int, input().split())
Gs = [[] for _ in range(N+1)]
for _ in range(M):
a, b = map(int, input().split())
Gs[a].append(b)
Gs[b].append(a)
ms = [0] * (N+1)
def bfs(c):
queue = deque()
ns = Gs[c]
for n in ns:
queue.append((n, c))
while queue:
c, p = queue.popleft()
if ms[c] != 0:
continue
ms[c] = p
ns = Gs[c]
for n in ns:
if n != p:
queue.append((n, c))
bfs(1)
for m in ms[2:]:
if m == 0:
print('No')
break
else:
print('Yes')
for m in ms[2:]:
print(m)
|
s692706354
|
p03386
|
u565204025
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 211 |
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.
|
# -*- coding: utf-8 -*-
a,b,k = map(int,input().split())
list = []
for i in range(a,a+k):
list.append(i)
for i in range(b,b-k,-1):
list.append(i)
list.sort()
l = set(list)
for i in l:
print(i)
|
s821426364
|
Accepted
| 19 | 3,060 | 240 |
# -*- coding: utf-8 -*-
a,b,k = map(int,input().split())
lists = []
for i in range(a,min(a+k,b+1)):
lists.append(i)
for i in range(b,max(b-k,a),-1):
lists.append(i)
l = set(lists)
m = list(l)
m.sort()
for i in m:
print(i)
|
s492228338
|
p03493
|
u566143855
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 95 |
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.
|
n = input()
i = []
for x in range(int(n)):
if x % 2 == 0:
i.append(x)
print(len(i))
|
s723056994
|
Accepted
| 17 | 2,940 | 25 |
print(input().count("1"))
|
s690187545
|
p03712
|
u217627525
| 2,000 | 262,144 |
Wrong Answer
| 20 | 3,060 | 116 |
You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}. Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1.
|
h,w=map(int,input().split())
print("*"*(w+2))
for i in range(h):
a=input()
print("*"+a+"*")
print("*"*(w+2))
|
s485619044
|
Accepted
| 17 | 3,060 | 116 |
h,w=map(int,input().split())
print("#"*(w+2))
for i in range(h):
a=input()
print("#"+a+"#")
print("#"*(w+2))
|
s211272444
|
p02601
|
u256769262
| 2,000 | 1,048,576 |
Wrong Answer
| 24 | 9,180 | 228 |
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 = list(map(int, input().split()))
n = int(input())
while 1:
if n == 0:
print("No")
break
elif a[0] >= a[1]:
a[1] *= 2
n -= 1
elif a[1] >= a[2]:
a[2] *= 2
n -= 1
else:
print("Yse")
break
|
s089124301
|
Accepted
| 26 | 9,156 | 224 |
a = list(map(int, input().split()))
n = int(input())
for i in range(n):
if a[0] >= a[1]:
a[1] *= 2
elif a[1] >= a[2]:
a[2] *= 2
if a[0] < a[1] < a[2]:
print("Yes")
else:
print("No")
|
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