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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s523657649
|
p03352
|
u943015560
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,060 | 327 |
You are given a positive integer X. Find the largest _perfect power_ that is at most X. Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2.
|
def Exponential():
X = int(input())
if X<1 or X>1000:
return ("Oops! X is out of the range!")
if X ==1: return(X)
res=[]
for b in range(X, 0, -1):
for p in range(2, X+1, 1):
if b**p <=X:
res.append(b**p)
else: break
return(max(res))
Exponential()
|
s138910416
|
Accepted
| 18 | 3,060 | 306 |
def Exponential():
X = int(input())
if X<1 or X>1000:
return ("Oops! X is out of the range!")
res=[1]
for b in range(1, X+1):
for p in range(2, X+1, 1):
if b**p <=X:
res.append(b**p)
else: break
print(max(res))
Exponential()
|
s716269564
|
p03543
|
u209918867
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 66 |
We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**?
|
s=int(input())
print(any([str(i)*3 in str(s) for i in range(10)]))
|
s683875823
|
Accepted
| 17 | 2,940 | 88 |
s=int(input())
print("Yes" if any([str(i)*3 in str(s) for i in range(10)]) else "No" )
|
s535847188
|
p03501
|
u232852711
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 55 |
You are parking at a parking lot. You can choose from the following two fee plans: * Plan 1: The fee will be A×T yen (the currency of Japan) when you park for T hours. * Plan 2: The fee will be B yen, regardless of the duration. Find the minimum fee when you park for N hours.
|
n, a, b = list(map(int, input().split()))
min(n*a, b)
|
s409528904
|
Accepted
| 17 | 2,940 | 62 |
n, a, b = list(map(int, input().split()))
print(min(n*a, b))
|
s835847206
|
p03699
|
u773686010
| 2,000 | 262,144 |
Wrong Answer
| 29 | 9,204 | 255 |
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())
dp = [[0]*2 for i in range(N + 1)]
for i in range(1,N + 1):
P = int(input())
dp[i][False] = max(dp[i-1])
dp[i][True] = max(dp[i-1][False] + P,dp[i-1][True] + P)
print(max(list(map(lambda x:x if x % 10 != 0 else 0,dp[-1]))))
|
s144385532
|
Accepted
| 66 | 10,208 | 201 |
N = int(input())
dp = [0]
for i in range(N):
P = int(input())
dp_p = list(map(lambda x:x + P,dp))
dp = list(set(dp + dp_p))
P_List = [0] + [i for i in dp if i % 10 != 0]
print(max(P_List))
|
s889886246
|
p03110
|
u151005508
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 230 |
Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins. If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?
|
N = int(input())
xv=[]
for _ in range(N):
xv.append(tuple(input().split()))
profit=0
for val in xv:
if val[1]=='JPY':
profit+=int(val[0])
else:
profit+=float(val[0])*380000.0
print(N, xv)
print(profit)
|
s904750251
|
Accepted
| 18 | 3,060 | 231 |
N = int(input())
xv=[]
for _ in range(N):
xv.append(tuple(input().split()))
profit=0
for val in xv:
if val[1]=='JPY':
profit+=int(val[0])
else:
profit+=float(val[0])*380000.0
#print(N, xv)
print(profit)
|
s241901848
|
p03379
|
u883048396
| 2,000 | 262,144 |
Wrong Answer
| 247 | 29,616 | 274 |
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.
|
iN = int(input())
aX = [int(_) for _ in input().split()]
aXs = sorted(aX)
iMu = iN//2
iMd = iN//2 -1
iMuV = aX[iMu]
iMdV = aX[iMd]
if iMuV == iMdV:
print("\n".join([str(iMuV)]*iN))
else:
print("\n".join(map(str,map(lambda x: iMuV if x < iMuV else iMdV ,aX))))
|
s244971577
|
Accepted
| 241 | 29,808 | 275 |
iN = int(input())
aX = [int(_) for _ in input().split()]
aXs = sorted(aX)
iMu = iN//2
iMd = iN//2 -1
iMuV = aXs[iMu]
iMdV = aXs[iMd]
if iMuV == iMdV:
print("\n".join([str(iMuV)]*iN))
else:
print("\n".join(map(str,map(lambda x: iMuV if x < iMuV else iMdV ,aX))))
|
s748738608
|
p03139
|
u008582165
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 86 |
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())
a = min(A,B)
b = A + B - N
print("{} {}".format(a,b))
|
s962861354
|
Accepted
| 17 | 3,060 | 207 |
N,A,B = map(int,input().split())
if A == 0 or B == 0:
print("0 0")
elif N >= A + B:
a = min(A,B)
print("{} {}".format(a,0))
else:
a = min(A,B)
b = A + B - N
print("{} {}".format(a,b))
|
s451548827
|
p03740
|
u600402037
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 201 |
Alice and Brown loves games. Today, they will play the following game. In this game, there are two piles initially consisting of X and Y stones, respectively. Alice and Bob alternately perform the following operation, starting from Alice: * Take 2i stones from one of the piles. Then, throw away i of them, and put the remaining i in the other pile. Here, the integer i (1≤i) can be freely chosen as long as there is a sufficient number of stones in the pile. The player who becomes unable to perform the operation, loses the game. Given X and Y, determine the winner of the game, assuming that both players play optimally.
|
import sys
sr = lambda: sys.stdin.readline().rstrip()
ir = lambda: int(sr())
lr = lambda: list(map(int, sr().split()))
X, Y = lr()
if abs(Y - X) >= 2:
print('Alice')
else:
print('Bob')
# 24
|
s403235842
|
Accepted
| 18 | 2,940 | 203 |
import sys
sr = lambda: sys.stdin.readline().rstrip()
ir = lambda: int(sr())
lr = lambda: list(map(int, sr().split()))
X, Y = lr()
if abs(Y - X) >= 2:
print('Alice')
else:
print('Brown')
# 24
|
s568295428
|
p03130
|
u572012241
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,064 | 592 |
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads. Determine if we can visit all the towns by traversing each of the roads exactly once.
|
import heapq
from sys import stdin
input = stdin.readline
# n = int(input())
# a = list(map(int,input().split()))
# ab=[]
# a,b = map(int, input().split())
# ab.append([a,b])
def main():
count =[0]*4
for i in range(3):
a,b = map(int, input().split())
count[a-1]+=1
count[b-1]+=1
flag = True
for i in count:
if i ==3:
flag=False
if flag:
print("Yes")
else:
print("No")
if __name__ == '__main__':
main()
|
s672767802
|
Accepted
| 18 | 3,060 | 592 |
import heapq
from sys import stdin
input = stdin.readline
# n = int(input())
# a = list(map(int,input().split()))
# ab=[]
# a,b = map(int, input().split())
# ab.append([a,b])
def main():
count =[0]*4
for i in range(3):
a,b = map(int, input().split())
count[a-1]+=1
count[b-1]+=1
flag = True
for i in count:
if i ==3:
flag=False
if flag:
print("YES")
else:
print("NO")
if __name__ == '__main__':
main()
|
s558913931
|
p03693
|
u708211626
| 2,000 | 262,144 |
Wrong Answer
| 25 | 9,020 | 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?
|
a=input().split()
b=''.join(a)
if int(b)%4==0:
print('Yes')
else:
print('No')
|
s748311911
|
Accepted
| 24 | 9,080 | 84 |
a=input().split()
b=''.join(a)
if int(b)%4==0:
print('YES')
else:
print('NO')
|
s817124800
|
p03416
|
u790301364
| 2,000 | 262,144 |
Wrong Answer
| 77 | 3,064 | 464 |
Find the number of _palindromic numbers_ among the integers between A and B (inclusive). Here, a palindromic number is a positive integer whose string representation in base 10 (without leading zeros) reads the same forward and backward.
|
def main6():
str = input('');
buf = str.split();
num = [];
result = 0;
for i in range(2):
num.insert(i,int(buf[i]));
for i in range(num[1] - num[0] + 1):
k = num[0] + i;
k1 = int(k/10000);
k2 = int(k/1000)%10;
k4 = int(k/10)%10;
k5 = k%10;
if k1 == k5 & k2 == k4:
result = result + 1;
print(result);
if __name__ == '__main__':
#main4()
#main5()
main6()
|
s739181324
|
Accepted
| 71 | 3,064 | 469 |
def main6():
str = input('');
buf = str.split();
num = [];
result = 0;
for i in range(2):
num.insert(i,int(buf[i]));
for i in range(num[1] - num[0] + 1):
k = num[0] + i;
k1 = int(k/10000)%10;
k2 = int(k/1000)%10;
k4 = int(k/10)%10;
k5 = k%10;
if k1 == k5 and k2 == k4:
result = result + 1;
print(result);
if __name__ == '__main__':
#main4()
#main5()
main6()
|
s009628349
|
p02972
|
u668785999
| 2,000 | 1,048,576 |
Wrong Answer
| 985 | 15,332 | 488 |
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()))
ans = [0 for i in range(N+1)]
flag = True
for i in range(N,0,-1):
for j in range(1,N+1):
if i*j>N:
break
else:
if ans[i*j] %2 == a[i*j -1 ]%2:
pass
else:
if j == 1:
ans[i*j] +=1
else:
flag = False
break
if flag:
print(sum(ans))
print(*a[1:])
else:
print(-1)
|
s411833640
|
Accepted
| 678 | 17,652 | 403 |
N = int(input())
a = list(map(int,input().split()))
ans = [0 for i in range(N+1)]
for i in range(N,0,-1):
res = 0
for j in range(1,N+1):
if i*j>N:
break
else:
res += ans[i*j]
if res%2 != a[i-1]:
ans[i] += 1
print(sum(ans))
A = []
for i,v in enumerate(ans):
if v>0:
A.append(i)
print(*A)
|
s474929009
|
p02831
|
u201928947
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 175 |
Takahashi is organizing a party. At the party, each guest will receive one or more snack pieces. Takahashi predicts that the number of guests at this party will be A or B. Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted. We assume that a piece cannot be divided and distributed to multiple guests.
|
a,b = map(int,input().split())
def gcd(x,y):
if x % y == 0:
return y
elif y % x == 0:
return x
else:
return gcd(y,abs(y-x))
print(gcd(a,b))
|
s992875543
|
Accepted
| 17 | 3,060 | 182 |
a,b = map(int,input().split())
def gcd(x,y):
if x % y == 0:
return y
elif y % x == 0:
return x
else:
return gcd(y,abs(y-x))
print(a*b//(gcd(a,b)))
|
s150367933
|
p03067
|
u386249594
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 131 |
There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise.
|
A,B,C = map(int,input().split())
if A < B and B < C:
print('Yes')
elif C < B and B < A:
print('Yes')
else:
print('No')
|
s168028421
|
Accepted
| 17 | 2,940 | 131 |
A,B,C = map(int,input().split())
if A < C and C < B:
print('Yes')
elif B < C and C < A:
print('Yes')
else:
print('No')
|
s346172495
|
p03477
|
u130900604
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 126 |
A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R. Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan. Print `Left` if the balance scale tips to the left; print `Balanced` if it balances; print `Right` if it tips to the right.
|
a,b,c,d=map(int,input().split())
l=a+b
r=c+d
if l==r:
print("Balanced")
elif r<l:
print("Right")
else:
print("Left")
|
s344282259
|
Accepted
| 17 | 2,940 | 127 |
a,b,c,d=map(int,input().split())
l=a+b
r=c+d
if l==r:
print("Balanced")
elif r>l:
print("Right")
else:
print("Left")
|
s864271468
|
p02612
|
u601393594
| 2,000 | 1,048,576 |
Wrong Answer
| 29 | 9,040 | 53 |
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())
p = n - (n // 1000) * 1000
print(p)
|
s873601937
|
Accepted
| 27 | 9,092 | 81 |
n = int(input())
if n % 1000 == 0:
p=0
else:
p=((n//1000)+1)*1000-n
print(p)
|
s160605272
|
p03997
|
u337820403
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 76 |
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a = int(input())
b = int(input())
h = int(input())
print(((a + b) * h) / 2)
|
s866406193
|
Accepted
| 17 | 2,940 | 81 |
a = int(input())
b = int(input())
h = int(input())
print(int(((a + b) * h) / 2))
|
s991551176
|
p02831
|
u037190233
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 134 |
Takahashi is organizing a party. At the party, each guest will receive one or more snack pieces. Takahashi predicts that the number of guests at this party will be A or B. Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted. We assume that a piece cannot be divided and distributed to multiple guests.
|
x,y = input().split()
x = int(x)
y = int(y)
if x < y:
a = y
b = x
else:
a = x
b = y
while b:
a, b = b, a % b
print(x*y/a)
|
s487198152
|
Accepted
| 17 | 2,940 | 140 |
x,y = input().split()
x = int(x)
y = int(y)
if x < y:
a = y
b = x
else:
a = x
b = y
while b:
a, b = b, a % b
print(x * y // a)
|
s716266007
|
p03730
|
u341926204
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 168 |
We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`.
|
def b():
a, b, c = map(int, input().split())
for i in range(c+1):
if a*i % b == c:
print('YES')
break
else: print('NO')
b()
|
s656944476
|
Accepted
| 17 | 2,940 | 169 |
def b():
a, b, c = map(int, input().split())
for i in range(2, b):
if a*i % b == c:
print('YES')
break
else: print('NO')
b()
|
s163429361
|
p03434
|
u845937249
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 403 |
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 = input()
n = int(n)
nums = input().split()
nums = list(map(int,nums))
print(nums)
nums.sort(reverse=True)
print(nums)
i = 0
Alis = 0
Bob = 0
for i in range(0,n):
if i % 2 == 0:
Alis = Alis + nums[i]
else:
Bob = Bob + nums[i]
print(Alis-Bob)
#print(Alis)
#print(Bob)
|
s334697332
|
Accepted
| 18 | 3,060 | 405 |
n = input()
n = int(n)
nums = input().split()
nums = list(map(int,nums))
#print(nums)
nums.sort(reverse=True)
#print(nums)
i = 0
Alis = 0
Bob = 0
for i in range(0,n):
if i % 2 == 0:
Alis = Alis + nums[i]
else:
Bob = Bob + nums[i]
print(Alis-Bob)
#print(Alis)
#print(Bob)
|
s498780202
|
p03494
|
u396495667
| 2,000 | 262,144 |
Time Limit Exceeded
| 2,104 | 2,940 | 140 |
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
n = int(input())
a = [int() for _ in input().split()]
cnt =0
while all(i%2==0 for i in a):
a = [int(i//2) for i in a]
cnt +=1
print(cnt)
|
s550597140
|
Accepted
| 19 | 2,940 | 152 |
n = int(input())
A = list(map(int,input().split()))
c = 0
while all(a%2 == 0 for a in A) :
for i in range(len(A)): A[i] = A[i]/2
c += 1
print(c)
|
s066094299
|
p02390
|
u335508235
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,708 | 79 |
Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively.
|
s=int(input())
hour=s/3600
min=s/600%60
sec=s%60
print(hour, min,sec,sep=":")
|
s839441212
|
Accepted
| 70 | 7,704 | 91 |
a = int(input())
h=int(a/3600)
m=int((a%3600)/60)
s=int((a%3600)%60)
print (h,m,s,sep=":")
|
s452043721
|
p02603
|
u997641430
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 9,064 | 375 |
To become a millionaire, M-kun has decided to make money by trading in the next N days. Currently, he has 1000 yen and no stocks - only one kind of stock is issued in the country where he lives. He is famous across the country for his ability to foresee the future. He already knows that the price of one stock in the next N days will be as follows: * A_1 yen on the 1-st day, A_2 yen on the 2-nd day, ..., A_N yen on the N-th day. In the i-th day, M-kun can make the following trade **any number of times** (possibly zero), **within the amount of money and stocks that he has at the time**. * Buy stock: Pay A_i yen and receive one stock. * Sell stock: Sell one stock for A_i yen. What is the maximum possible amount of money that M-kun can have in the end by trading optimally?
|
n = int(input())
*A, = map(int, input().split())
n = 10
A = [200] * n
A = [200] + A + [0]
money = 1000
stock = 0
for i in range(1, n + 1):
if (A[i - 1] > A[i]) and (A[i] <= A[i + 1]):
cnt = money // A[i]
money -= cnt * A[i]
stock += cnt
if (A[i - 1] <= A[i]) and (A[i] > A[i + 1]):
money += stock * A[i]
stock = 0
print(money)
|
s909477544
|
Accepted
| 27 | 9,136 | 145 |
n = int(input())
*A, = map(int, input().split())
ans = 1000
for i in range(n - 1):
ans += (ans // A[i]) * max(A[i + 1] - A[i], 0)
print(ans)
|
s661663505
|
p03456
|
u127535085
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 140 |
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.
|
from math import sqrt
a,b = input().split()
ans = sqrt(int(a+b))
print(ans)
if ans / int(ans) == 1:
print("Yes")
else:
print("No")
|
s513324165
|
Accepted
| 17 | 2,940 | 129 |
from math import sqrt
a,b = input().split()
ans = sqrt(int(a+b))
if ans / int(ans) == 1:
print("Yes")
else:
print("No")
|
s657983422
|
p03473
|
u037485167
| 2,000 | 262,144 |
Time Limit Exceeded
| 2,104 | 5,612 | 620 |
How many hours do we have until New Year at M o'clock (24-hour notation) on 30th, December?
|
candidates = [2] + list(range(3,100001,2))
i = 1
got_one = True
while got_one:
p = candidates[i]
new_candidates = candidates[:i+1]
for j in candidates[i+1:]:
if j%p != 0:
new_candidates.append(j)
if len(new_candidates) == len(candidates):
got_one = False
candidates = new_candidates
i = i + 1
similar = []
for p in candidates:
if (p+1)/2 in candidates:
similar.append(p)
answers = []
n = int(input())
for q in range(n):
[l, r] = [int(i) for i in input().split()]
answers.append(sum([l <= p <= r for p in similar]))
for a in answers:
print(a)
|
s804803225
|
Accepted
| 17 | 2,940 | 33 |
s = int(input())
print(24-s + 24)
|
s976714094
|
p03433
|
u088974156
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 106 |
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
n=int(input())
a=int(input())
if(a==0):
print("YES")
elif((a%500)>n):
print("NO")
else:
print("YES")
|
s816988670
|
Accepted
| 17 | 2,940 | 80 |
n=int(input())
a=int(input())
if((n%500)<=a):
print("Yes")
else:
print("No")
|
s534282350
|
p02406
|
u217069758
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,592 | 145 |
In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops. Note that use of goto statement is highly discouraged, because it is difficult to trace the control flow of a program which includes goto. Write a program which does precisely the same thing as the following program (this example is wrtten in C++). Let's try to write the program without goto statements. void call(int n){ int i = 1; CHECK_NUM: int x = i; if ( x % 3 == 0 ){ cout << " " << i; goto END_CHECK_NUM; } INCLUDE3: if ( x % 10 == 3 ){ cout << " " << i; goto END_CHECK_NUM; } x /= 10; if ( x ) goto INCLUDE3; END_CHECK_NUM: if ( ++i <= n ) goto CHECK_NUM; cout << endl; }
|
n = int(input())
result = []
for i in range(1, n+1):
if (i % 3 == 0) or (i // 10 == 3):
result.append(str(i))
print(' '.join(result))
|
s030035490
|
Accepted
| 20 | 5,636 | 156 |
n = int(input())
result = ''
for i in range(3, n+1):
if (i % 3 == 0) or (i % 10 == 3) or ('3' in str(i)):
result += (' ' + str(i))
print(result)
|
s060247417
|
p03386
|
u098982053
| 2,000 | 262,144 |
Wrong Answer
| 2,226 | 2,076,568 | 138 |
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(" "))
List = [i for i in range(A,B+1)]
la = List[:K]
lb = List[-K:]
L = la+lb
for l in set(L):
print(l)
|
s631621710
|
Accepted
| 18 | 3,060 | 195 |
A,B,K = map(int,input().split(" "))
if 2*K >= B-A+1:
for i in range(A,B+1):
print(i)
else:
for i in range(A,A+K):
print(i)
for i in range(B-K+1,B+1):
print(i)
|
s531065190
|
p03161
|
u365375535
| 2,000 | 1,048,576 |
Wrong Answer
| 2,104 | 14,888 | 412 |
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 one of the following: Stone i + 1, i + 2, \ldots, i + K. 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.
|
def run():
N, K = map(int, input().split())
h_li = [int(h) for h in input().split()]
cost_li = [0] * (N)
cost_li[1] = abs(h_li[0] - h_li[1])
for n in range(2, N):
if n < K: k = n
else: k = K
cost = float('inf')
h = h_li[n]
for _k in range(1, k+1):
cost = min(cost, cost_li[n-_k] + abs(h_li[n-_k]-h))
cost_li[n] = cost
print(cost_li)
if __name__ == '__main__':
run()
|
s579416342
|
Accepted
| 1,866 | 13,980 | 359 |
def run():
N, K = map(int, input().split())
h_li = [int(h) for h in input().split()]
cost_li = [0] * N
cost_li[1] = abs(h_li[0] - h_li[1])
for n in range(2, N):
k = max(0, n - K)
h = h_li[n]
cost = min([cost_li[i] + abs(h_li[i] - h) for i in range(k, n)])
cost_li[n] = cost
print(cost_li[N-1])
if __name__ == '__main__':
run()
|
s128618651
|
p03069
|
u248670337
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 9,036 | 26 |
There are N stones arranged in a row. Every stone is painted white or black. A string S represents the color of the stones. The i-th stone from the left is white if the i-th character of S is `.`, and the stone is black if the character is `#`. Takahashi wants to change the colors of some stones to black or white so that there will be no white stone immediately to the right of a black stone. Find the minimum number of stones that needs to be recolored.
|
print(input().count('#.'))
|
s905591866
|
Accepted
| 110 | 9,252 | 87 |
_,s=open(b:=0);w=a=s.count('.')
for c in s:b+=(x:=c<'.');w-=(x<1);a=min(a,b+w)
print(a)
|
s427591508
|
p03139
|
u703890795
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 65 |
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())
print(max(A,B), max(A+B-N,0))
|
s797409990
|
Accepted
| 17 | 2,940 | 65 |
N, A, B = map(int, input().split())
print(min(A,B), max(A+B-N,0))
|
s905662789
|
p02601
|
u540698208
| 2,000 | 1,048,576 |
Wrong Answer
| 31 | 9,052 | 296 |
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.
|
def main():
A, B, C = (int(x) for x in input().split())
K = int(input())
while not A < B < C and K > 0:
if A >= B: B *= 2
else: C *= 2
K -= 1
print(A, B, C, K)
if A < B < C: print('Yes')
else: print('No')
if __name__ == '__main__':
main()
|
s168757330
|
Accepted
| 31 | 9,052 | 298 |
def main():
A, B, C = (int(x) for x in input().split())
K = int(input())
while not A < B < C and K > 0:
if A >= B: B *= 2
else: C *= 2
K -= 1
if A < B < C: print('Yes')
else: print('No')
if __name__ == '__main__':
main()
|
s817617243
|
p03486
|
u440161695
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 78 |
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=sorted(input())
t=sorted(input())
if s<t:
print("Yes")
else:
print("No")
|
s735995730
|
Accepted
| 17 | 2,940 | 63 |
print("Yes" if sorted(input())<sorted(input())[::-1] else "No")
|
s505766460
|
p03964
|
u239981649
| 2,000 | 262,144 |
Wrong Answer
| 22 | 3,188 | 248 |
AtCoDeer the deer is seeing a quick report of election results on TV. Two candidates are standing for the election: Takahashi and Aoki. The report shows the ratio of the current numbers of votes the two candidates have obtained, but not the actual numbers of votes. AtCoDeer has checked the report N times, and when he checked it for the i-th (1≦i≦N) time, the ratio was T_i:A_i. It is known that each candidate had at least one vote when he checked the report for the first time. Find the minimum possible total number of votes obtained by the two candidates when he checked the report for the N-th time. It can be assumed that the number of votes obtained by each candidate never decreases.
|
import math
n = int(input())
a = [list(map(int, input().split())) for _ in range(n)]
for k in range(n):
if k:
m = min(math.ceil(a[k-1][0]/a[k][0]), math.ceil(a[k-1][1]/a[k][1]))
a[k] = [m*a[k][0], m*a[k][1]]
print(sum(a[n-1]))
|
s842300532
|
Accepted
| 84 | 5,332 | 340 |
from math import ceil
from decimal import Decimal as Deci
n = int(input())
a = [list(map(int, input().split())) for _ in range(n)]
for k in range(n):
if k:
m0 = ceil(Deci(a[k-1][0])/Deci(a[k][0]))
m1 = ceil(Deci(a[k-1][1])/Deci(a[k][1]))
m = max(m0, m1)
a[k] = [m*a[k][0], m*a[k][1]]
print(sum(a[n-1]))
|
s114374542
|
p03407
|
u140480594
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 89 |
An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.
|
A, B, C = list(map(int, input().split()))
if A + B <= C : print("Yes")
else : print("No")
|
s394132121
|
Accepted
| 17 | 2,940 | 89 |
A, B, C = list(map(int, input().split()))
if A + B >= C : print("Yes")
else : print("No")
|
s924622018
|
p03711
|
u136090046
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 260 |
Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group.
|
val = [int(x) for x in input().split()]
if val[0] == 2 or val[1] == 2:
print("NO")
elif val[0] in [4,6,9,11] and val[1] in [4,6,9,11]:
print("YES")
elif val[0] in [1,3,5,7,8,10,12] and val[1] in [1,3,5,7,8,10,12]:
print("YES")
else:
print("NO")
|
s002300578
|
Accepted
| 19 | 3,064 | 251 |
x,y = map(int,input().split())
list1 = [1,3,5,7,8,10,12]
list2 = [4,6,9,11]
list3 = [2]
if x in list1 and y in list1:
print('Yes')
elif x in list2 and y in list2:
print('Yes')
elif x in list3 and y in list3:
print('Yes')
else:
print('No')
|
s954780482
|
p02603
|
u242196904
| 2,000 | 1,048,576 |
Wrong Answer
| 31 | 9,160 | 172 |
To become a millionaire, M-kun has decided to make money by trading in the next N days. Currently, he has 1000 yen and no stocks - only one kind of stock is issued in the country where he lives. He is famous across the country for his ability to foresee the future. He already knows that the price of one stock in the next N days will be as follows: * A_1 yen on the 1-st day, A_2 yen on the 2-nd day, ..., A_N yen on the N-th day. In the i-th day, M-kun can make the following trade **any number of times** (possibly zero), **within the amount of money and stocks that he has at the time**. * Buy stock: Pay A_i yen and receive one stock. * Sell stock: Sell one stock for A_i yen. What is the maximum possible amount of money that M-kun can have in the end by trading optimally?
|
n = int(input())
a = list(map(int, input().split()))
money = 1000
for i in range(n - 1):
if a[i+1] > a[i]:
money = money % a[i] + money // a[i] * a[i+1]
money
|
s357632667
|
Accepted
| 31 | 9,108 | 179 |
n = int(input())
a = list(map(int, input().split()))
money = 1000
for i in range(n - 1):
if a[i+1] > a[i]:
money = money % a[i] + money // a[i] * a[i+1]
print(money)
|
s986733894
|
p03563
|
u663438907
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 49 |
Takahashi is a user of a site that hosts programming contests. When a user competes in a contest, the _rating_ of the user (not necessarily an integer) changes according to the _performance_ of the user, as follows: * Let the current rating of the user be a. * Suppose that the performance of the user in the contest is b. * Then, the new rating of the user will be the avarage of a and b. For example, if a user with rating 1 competes in a contest and gives performance 1000, his/her new rating will be 500.5, the average of 1 and 1000. Takahashi's current rating is R, and he wants his rating to be exactly G after the next contest. Find the performance required to achieve it.
|
R = int(input())
G = int(input())
print(G - 2*R)
|
s629748508
|
Accepted
| 17 | 2,940 | 49 |
R = int(input())
G = int(input())
print(2*G - R)
|
s711923642
|
p03339
|
u233437481
| 2,000 | 1,048,576 |
Wrong Answer
| 463 | 21,064 | 258 |
There are N people standing in a row from west to east. Each person is facing east or west. The directions of the people is given as a string S of length N. The i-th person from the west is facing east if S_i = `E`, and west if S_i = `W`. You will appoint one of the N people as the leader, then command the rest of them to face in the direction of the leader. Here, we do not care which direction the leader is facing. The people in the row hate to change their directions, so you would like to select the leader so that the number of people who have to change their directions is minimized. Find the minimum number of people who have to change their directions.
|
N = int(input())
S = list(input())
E = [0]*(N+1)
W = [0]*(N+1)
ans = 10000*10000
for i in range(N-1):
if S[i] == 'E':
E[i+1] += E[i] + 1
else:
W[i+1] += W[i] + 1
for i in range(N):
ans = min(ans,E[i]+W[i])
print(ans)
|
s230103803
|
Accepted
| 329 | 19,916 | 336 |
N = int(input())
S = list(input())
E = [0]*N
W = [0]*N
ans = 10000*10000
for i in range(N-1):
if S[i] == 'W':
W[i+1] += W[i] + 1
else:
W[i+1] = W[i]
for i in range(N)[:0:-1]:
if S[i] == 'E':
E[i-1] += E[i] + 1
else:
E[i-1] = E[i]
for i in range(N):
ans = min(ans,E[i]+W[i])
print(ans)
|
s397076318
|
p02647
|
u469550773
| 2,000 | 1,048,576 |
Wrong Answer
| 2,206 | 51,460 | 525 |
We have N bulbs arranged on a number line, numbered 1 to N from left to right. Bulb i is at coordinate i. Each bulb has a non-negative integer parameter called intensity. When there is a bulb of intensity d at coordinate x, the bulb illuminates the segment from coordinate x-d-0.5 to x+d+0.5. Initially, the intensity of Bulb i is A_i. We will now do the following operation K times in a row: * For each integer i between 1 and N (inclusive), let B_i be the number of bulbs illuminating coordinate i. Then, change the intensity of each bulb i to B_i. Find the intensity of each bulb after the K operations.
|
import numpy as np
Debug = False
# Calculate intensity
N, K = [int(v) for v in input().split(" ")]
A_arr = [int(v) for v in input().split(" ")]
def Luminate(A_arr, i):
step = np.arange(len(A_arr))
distance = abs(step - i)
isLuminated = distance < np.array(A_arr)+1
if Debug:
print(step, distance,isLuminated)
return isLuminated
def get_Intensity(A_arr):
return [np.sum(Luminate(A_arr, i)) for i in range(len(A_arr))]
Intensity = A_arr
for i in range(K):
Intensity = get_Intensity(Intensity)
print(Intensity)
|
s350664820
|
Accepted
| 1,434 | 51,568 | 396 |
import numpy as np
N, K = map(int, input().split(' '))
A = tuple(map(int, input().split(' ')))
indexes = np.arange(0, N)
dp = np.zeros(shape=N + 1, dtype=np.int64)
for k in range(K):
np.add.at(dp, np.maximum(0, indexes - A), 1)
np.add.at(dp, np.minimum(N, indexes + A + 1), -1)
A = dp.cumsum()[:-1]
if np.all(A == N):
break
dp *= 0
print(' '.join(map(str, A)))
|
s253859640
|
p03813
|
u312078744
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 79 |
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()
num = len(s)
A = s.find("A")
Z = s.rfind("Z")
ans = Z - A
print(Z)
|
s278494947
|
Accepted
| 18 | 3,064 | 77 |
score = int(input())
if (score < 1200):
print("ABC")
else:
print("ARC")
|
s834930045
|
p02646
|
u433836112
| 2,000 | 1,048,576 |
Wrong Answer
| 23 | 9,180 | 182 |
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 = list(map(int, input().strip().split()))
b, w = list(map(int, input().strip().split()))
t = int(input().strip())
if abs(a-b) <= (v-w)*t:
print('Yes')
else:
print('No')
|
s711581398
|
Accepted
| 22 | 9,176 | 182 |
a, v = list(map(int, input().strip().split()))
b, w = list(map(int, input().strip().split()))
t = int(input().strip())
if abs(a-b) <= (v-w)*t:
print('YES')
else:
print('NO')
|
s513525674
|
p03494
|
u600402037
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 324 |
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
n = int(input())
a_list = list(map(int, input().split()))
min_a = min(a_list)
count = 0
while min_a % 2 == 0:
min_a /= 2
count += 1
#print(count)
while True:
for i in range(n):
if a_list[i] / 2 ** count % 2 !=0:
break
else:
count += 1
continue
break
print(count)
|
s464117639
|
Accepted
| 151 | 12,504 | 149 |
import numpy as np
input()
A = np.array(list(map(int, input().split())))
count = 0
while all(A%2==0):
count += 1
A //= 2
print(count)
|
s347843082
|
p03623
|
u952130512
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 71 |
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x,a,b=input().split()
print(min(abs(int(x)-int(a)),abs(int(x)-int(b))))
|
s530070726
|
Accepted
| 18 | 2,940 | 95 |
x,a,b=input().split()
if abs(int(x)-int(a))<abs(int(x)-int(b)):
print("A")
else:
print("B")
|
s100434315
|
p03433
|
u824326335
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 57 |
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
print("YES" if int(input())%500 < int(input()) else "NO")
|
s980654369
|
Accepted
| 18 | 2,940 | 58 |
print("Yes" if int(input())%500 <= int(input()) else "No")
|
s668456974
|
p03624
|
u753682919
| 2,000 | 262,144 |
Wrong Answer
| 24 | 3,956 | 123 |
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=list(input())
A=list("abcdefghijklmnopqrstuvwxyz")
if set(S)!=set(A):
print(A[A.index(min(S))-1])
else:
print("None")
|
s058064999
|
Accepted
| 21 | 3,188 | 183 |
S=set(input())
Al="qwertyuiopasdfghjklzxcvbnm"
AL=list(Al)
l=[]
for i in range(len(Al)):
if not AL[i] in S:
l.append(AL[i])
if l!=[]:
print(sorted(l)[0])
else:
print("None")
|
s433156110
|
p03457
|
u925567828
| 2,000 | 262,144 |
Wrong Answer
| 445 | 12,496 | 787 |
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.
|
import sys
N = int(input())
t=[0]*(N+1)
x=[0]*(N+1)
y=[0]*(N+1)
dist = 0
dt = 0
now_x =0
now_y =0
t[0] = 0
x[0] = now_x
y[0] = now_y
for i in range(1,N+1):
t[i],x[i],y[i] = map(int, input().split())
for i in range(1,N+1):
dist = abs(x[i]-x[i-1])+abs(y[i]-y[i-1])
dt = abs(t[i] - t[i-1])
if(dist > dt):
print('No')
sys.exit()
else:
if(dt %2 != dist %2):
print('No')
sys.exit()
else:
print('Yes')
|
s653263563
|
Accepted
| 374 | 11,636 | 755 |
import sys
N = int(input())
t=[0]*(N+1)
x=[0]*(N+1)
y=[0]*(N+1)
dist = 0
dt = 0
now_x =0
now_y =0
t[0] = 0
x[0] = now_x
y[0] = now_y
for i in range(1,N+1):
t[i],x[i],y[i] = map(int, input().split())
for i in range(1,N+1):
dist = abs(x[i]-x[i-1])+abs(y[i]-y[i-1])
dt = abs(t[i] - t[i-1])
if(dist > dt):
print('No')
sys.exit()
else:
if(dt %2 != dist %2):
print('No')
sys.exit()
print('Yes')
|
s943613917
|
p02613
|
u228303592
| 2,000 | 1,048,576 |
Wrong Answer
| 160 | 16,164 | 304 |
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())
k = [input() for i in range(n)]
ac = 0
wa = 0
tle = 0
re = 0
for j in range(n):
if k[j] == 'AC':
ac += 1
elif k[j] == 'WA':
wa += 1
elif k[j] == 'TLE':
tle += 1
elif k[j] == 'RE':
re += 1
print("AC *",ac)
print("WA *",wa)
print("TLE *",tle)
print("RE *",re)
|
s238764478
|
Accepted
| 160 | 16,192 | 304 |
n = int(input())
k = [input() for i in range(n)]
ac = 0
wa = 0
tle = 0
re = 0
for j in range(n):
if k[j] == 'AC':
ac += 1
elif k[j] == 'WA':
wa += 1
elif k[j] == 'TLE':
tle += 1
elif k[j] == 'RE':
re += 1
print("AC x",ac)
print("WA x",wa)
print("TLE x",tle)
print("RE x",re)
|
s030004100
|
p03854
|
u193182854
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,316 | 158 |
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()
words = "eraser", "erase", "dreamer", "dream"
for word in words:
s = s.replace(word, "")
print(s)
print("YES" if len(s) == 0 else "NO")
|
s286106229
|
Accepted
| 19 | 3,188 | 145 |
s = input()
words = "eraser", "erase", "dreamer", "dream"
for word in words:
s = s.replace(word, "")
print("YES" if len(s) == 0 else "NO")
|
s418097994
|
p03438
|
u462329577
| 2,000 | 262,144 |
Wrong Answer
| 22 | 4,600 | 187 |
You are given two integer sequences of length N: a_1,a_2,..,a_N and b_1,b_2,..,b_N. Determine if we can repeat the following operation zero or more times so that the sequences a and b become equal. Operation: Choose two integers i and j (possibly the same) between 1 and N (inclusive), then perform the following two actions **simultaneously** : * Add 2 to a_i. * Add 1 to b_j.
|
#!/usr/bin/env python
N = int(input())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
if N > sum(b)-sum(a):
print("No")
else:
print("Yes")
print(sum(b)-sum(a))
|
s237632658
|
Accepted
| 38 | 10,588 | 407 |
#!/usr/bin/env python3
# input
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
res = 0
for i in range(n):
if a[i] > b[i]:
res -= a[i] - b[i]
else:
res += (b[i] - a[i]) // 2
if res >= 0:
print("Yes")
else:
print("No")
|
s151468390
|
p02389
|
u678690126
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,584 | 88 |
Write a program which calculates the area and perimeter of a given rectangle.
|
a, b = map(int, input().split())
if 1 <= a <= 100 and 1 <= b <= 100:
print(a * b)
|
s486228302
|
Accepted
| 20 | 5,596 | 142 |
a, b = map(int, input().split())
if 1 <= a <= 100 and 1 <= b <= 100:
area = a * b
perimeter = (a + b) * 2
print(area, perimeter)
|
s669011812
|
p03796
|
u032189172
| 2,000 | 262,144 |
Wrong Answer
| 229 | 3,984 | 79 |
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.
|
from math import factorial
N = int(input())
P = factorial(N)
print(N%(10**9+7))
|
s403283897
|
Accepted
| 231 | 4,020 | 79 |
from math import factorial
N = int(input())
P = factorial(N)
print(P%(10**9+7))
|
s625262924
|
p03846
|
u332385682
| 2,000 | 262,144 |
Wrong Answer
| 86 | 14,820 | 348 |
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.
|
from collections import Counter
N = int(input())
A = [int(a) for a in input().split()]
c = Counter(A)
c_v = sorted(c.values())
if N % 2 == 0:
if c_v == [2] * (N // 2):
print(2 ** (N // 2))
else:
print(0)
else:
if c_v == [1].extend([2] * ((N - 1) // 2)):
print(2 ** ((N - 1) // 2))
else:
print(0)
|
s494368645
|
Accepted
| 93 | 18,272 | 330 |
from collections import Counter
N = int(input())
A = [int(a) for a in input().split()]
check = {}
if N % 2 == 0:
for i in range(1, N, 2):
check[i] = 2
else:
check[0] = 1
for i in range(2, N, 2):
check[i] = 2
if check == Counter(A):
print((2 ** (N // 2)) % (10 ** 9 + 7))
else:
print(0)
|
s646020812
|
p03730
|
u281303342
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 128 |
We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`.
|
A,B,C = map(int,input().split())
Ans = "No"
for i in range(1,101):
if (A*i)%B==C:
Ans="Yes"
break
print(Ans)
|
s454595964
|
Accepted
| 17 | 3,060 | 251 |
# atcoder : python3 (3.4.3)
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10**6)
mod = 10**9+7
# main
A,B,C = map(int,input().split())
ans = "NO"
for i in range(1,B+1):
if (A*i)%B == C:
ans = "YES"
break
print(ans)
|
s592729210
|
p03556
|
u385167811
| 2,000 | 262,144 |
Wrong Answer
| 22 | 2,940 | 131 |
Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer.
|
n = int(input())
i = 1
while True:
if i ** 2 > n:
print(i-1)
break
i += 1
if i >= 10000:
break
|
s761635342
|
Accepted
| 31 | 2,940 | 139 |
n = int(input())
i = 1
while True:
if i ** 2 > n:
print((i-1)**2)
break
i += 1
if i >= 10000000:
break
|
s216093813
|
p02678
|
u536034761
| 2,000 | 1,048,576 |
Wrong Answer
| 830 | 38,572 | 500 |
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())
ways = [[] for i in range(N)]
for i in range(M):
a, b = map(int, input().split())
a -= 1
b -= 1
ways[a].append(b)
ways[b].append(a)
print(ways)
searched = [0 for i in range(N)]
searched[0] = 1
ans = [0 for i in range(N)]
d = deque([])
d.append(0)
while d:
tmp = d.popleft()
for w in ways[tmp]:
if searched[w] == 0:
searched[w] = 1
ans[w] = tmp
d.append(w)
print("Yes")
for a in ans[1:]:
print(a + 1)
|
s209396647
|
Accepted
| 723 | 35,804 | 488 |
from collections import deque
N, M = map(int, input().split())
ways = [[] for i in range(N)]
for i in range(M):
a, b = map(int, input().split())
a -= 1
b -= 1
ways[a].append(b)
ways[b].append(a)
searched = [0 for i in range(N)]
searched[0] = 1
ans = [0 for i in range(N)]
d = deque([])
d.append(0)
while d:
tmp = d.popleft()
for w in ways[tmp]:
if searched[w] == 0:
searched[w] = 1
ans[w] = tmp
d.append(w)
print("Yes")
for a in ans[1:]:
print(a + 1)
|
s013763781
|
p03760
|
u785220618
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 240 |
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions. You are given two strings O and E. O contains the characters at the odd- numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
|
# -*- coding: utf-8 -*-
o = list(input()[::2])
e = list(input()[1::2])
ans = []
while len(o) and len(e):
if len(o):
ans.append(o[0])
del o[0]
if len(e):
ans.append(e[0])
del e[0]
print(''.join(ans))
|
s602306656
|
Accepted
| 17 | 2,940 | 228 |
# -*- coding: utf-8 -*-
o = list(input())
e = list(input())
ans = []
while len(o) or len(e):
if len(o):
ans.append(o[0])
del o[0]
if len(e):
ans.append(e[0])
del e[0]
print(''.join(ans))
|
s139289172
|
p03646
|
u923712635
| 2,000 | 262,144 |
Wrong Answer
| 2,104 | 39,412 | 84 |
We have a sequence of length N consisting of non-negative integers. Consider performing the following operation on this sequence until the largest element in this sequence becomes N-1 or smaller. * Determine the largest element in the sequence (if there is more than one, choose one). Decrease the value of this element by N, and increase each of the other elements by 1. It can be proved that the largest element in the sequence becomes N-1 or smaller after a finite number of operations. You are given an integer K. Find an integer sequence a_i such that the number of times we will perform the above operation is exactly K. It can be shown that there is always such a sequence under the constraints on input and output in this problem.
|
K = int(input())
print(K+1)
for x in range(K+1,1,-1):
print(x,end=' ')
print(0)
|
s956672387
|
Accepted
| 18 | 3,064 | 568 |
K = int(input())
if(K==0):
print(4)
print(3,3,3,3)
elif(K==1):
print(3)
print(1,0,3)
elif(K<50):
n = K
li = [i for i in range(n)]
for i in range(n):
for j in range(n):
if(i==j):
li[j] += n
else:
li[j] -= 1
print(n)
print(*li)
else:
x = K%50
y = K//50
li = [y+i for i in range(50)]
for i in range(x):
for j in range(50):
if(i==j):
li[j] += 50
else:
li[j] -= 1
print(50)
print(*li)
|
s517025478
|
p03377
|
u870518235
| 2,000 | 262,144 |
Wrong Answer
| 28 | 9,108 | 93 |
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 0 < X - A <= B:
print("Yes")
else:
print("No")
|
s458643622
|
Accepted
| 27 | 9,112 | 94 |
A, B, X = map(int, input().split())
if 0 <= X - A <= B:
print("YES")
else:
print("NO")
|
s707870654
|
p03493
|
u330314953
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 89 |
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.
|
s = input()
cnt = 0
for i in range(len(s)):
if s[i] == 1:
cnt += 1
print(cnt)
|
s033416070
|
Accepted
| 17 | 2,940 | 91 |
s = input()
cnt = 0
for i in range(len(s)):
if s[i] == "1":
cnt += 1
print(cnt)
|
s599780800
|
p03494
|
u018591138
| 2,000 | 262,144 |
Wrong Answer
| 20 | 3,060 | 281 |
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
n = int(input(">>"))
a = map(int, input(">>").split())
count = 0
flag = True
while(flag):
tmp_a = []
for i in a:
if i % 2 !=0:
flag = False
break
tmp_a.append(i/2)
a = tmp_a
if flag == True:
count += 1
print(count)
|
s061149329
|
Accepted
| 19 | 3,060 | 273 |
n = int(input())
a = map(int, input().split())
count = 0
flag = True
while(flag):
tmp_a = []
for i in a:
if i % 2 !=0:
flag = False
break
tmp_a.append(i/2)
a = tmp_a
if flag == True:
count += 1
print(count)
|
s047422443
|
p02613
|
u798093965
| 2,000 | 1,048,576 |
Wrong Answer
| 142 | 9,136 | 320 |
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())
C = 0
D = 0
E = 0
F = 0
S = ''
for i in range(N):
S = input()
if S == 'AC':
C += 1
elif S == 'WA':
D += 1
elif S == 'TLE':
E += 1
elif S == 'RE':
F += 1
print('AC × '+ str(C))
print('WA × '+ str(D))
print('TLE × '+ str(E))
print('RE × '+ str(F))
|
s403306009
|
Accepted
| 148 | 8,988 | 313 |
N = int(input())
C = 0
D = 0
E = 0
F = 0
S = ''
for i in range(N):
S = input()
if S == 'AC':
C += 1
elif S == 'WA':
D += 1
elif S == 'TLE':
E += 1
elif S == 'RE':
F += 1
print('AC x '+ str(C), 'WA x ' + str(D), 'TLE x ' + str(E), 'RE x ' + str(F), sep='\n')
|
s624821151
|
p03795
|
u365254117
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 79 |
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
n =int(input())
if n >= 15:
print(800*n-(n-15)*200)
else:
print(800*n)
|
s467230611
|
Accepted
| 17 | 2,940 | 39 |
n =int(input())
print(800*n-n//15*200)
|
s550513030
|
p03448
|
u729911693
| 2,000 | 262,144 |
Wrong Answer
| 49 | 3,060 | 235 |
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())
count = 0
for l in range(0, a):
for m in range(0, b):
for n in range(0, c):
if(500*l+100*m+50*n == x):
count+=1
print(count)
|
s191061795
|
Accepted
| 49 | 3,060 | 241 |
a = int(input())
b = int(input())
c = int(input())
x = int(input())
count = 0
for l in range(0, a+1):
for m in range(0, b+1):
for n in range(0, c+1):
if(500*l+100*m+50*n == x):
count+=1
print(count)
|
s079494934
|
p03814
|
u319818856
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,516 | 175 |
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`.
|
def a2z_string(s: str)->int:
a = s.find('a')
z = s.rfind('z')
return z - a + 1
if __name__ == "__main__":
s = input()
ans = a2z_string(s)
print(ans)
|
s169218320
|
Accepted
| 17 | 3,512 | 175 |
def a2z_string(s: str)->int:
a = s.find('A')
z = s.rfind('Z')
return z - a + 1
if __name__ == "__main__":
s = input()
ans = a2z_string(s)
print(ans)
|
s397954512
|
p03712
|
u591779169
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 161 |
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())
a = []
for i in range(h):
a.append(input())
print("#"*(w+2))
for j in range(h):
print("#" + a[0] + "#")
print("#"*(w+2))
|
s297579456
|
Accepted
| 18 | 3,060 | 161 |
h, w = map(int, input().split())
a = []
for i in range(h):
a.append(input())
print("#"*(w+2))
for j in range(h):
print("#" + a[j] + "#")
print("#"*(w+2))
|
s194106733
|
p03456
|
u640753006
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 137 |
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.
|
import math
a,b = map(int, input().split())
if a%2 == 0 or b%2 == 0:
print(math.floor(a*b/2))
else:
print(math.floor(a*b/2+1))
|
s737945277
|
Accepted
| 17 | 2,940 | 140 |
import math
a, b = input().split()
ab = a+b
ab = int(ab)
ans = math.sqrt(ab)
if ans.is_integer():
print('Yes')
else:
print('No')
|
s579818008
|
p03729
|
u735008991
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 84 |
You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`.
|
a, b, c = input().split()
print("YES" if a[0] == b[-1] and b[-1] == c[0] else "NO")
|
s109721702
|
Accepted
| 17 | 2,940 | 84 |
a, b, c = input().split()
print("YES" if a[-1] == b[0] and b[-1] == c[0] else "NO")
|
s134117599
|
p03455
|
u155774139
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 88 |
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 == 0:
print("Even")
else:
print("Odd")
|
s354235930
|
Accepted
| 18 | 2,940 | 90 |
a, b = map(int, input().split())
if a*b % 2 == 0:
print("Even")
else:
print("Odd")
|
s289023543
|
p02612
|
u277312083
| 2,000 | 1,048,576 |
Wrong Answer
| 33 | 9,136 | 27 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
print(int(input()) % 1000)
|
s243142900
|
Accepted
| 31 | 9,140 | 78 |
n = int(input())
if n % 1000:
print(1000 - (n % 1000))
else:
print(0)
|
s583049850
|
p03472
|
u091051505
| 2,000 | 262,144 |
Wrong Answer
| 406 | 12,084 | 550 |
You are going out for a walk, when you suddenly encounter a monster. Fortunately, you have N katana (swords), Katana 1, Katana 2, …, Katana N, and can perform the following two kinds of attacks in any order: * Wield one of the katana you have. When you wield Katana i (1 ≤ i ≤ N), the monster receives a_i points of damage. The same katana can be wielded any number of times. * Throw one of the katana you have. When you throw Katana i (1 ≤ i ≤ N) at the monster, it receives b_i points of damage, and you lose the katana. That is, you can no longer wield or throw that katana. The monster will vanish when the total damage it has received is H points or more. At least how many attacks do you need in order to vanish it in total?
|
from collections import deque
n, h = map(int, input().split())
a_list = []
b_list = []
time = 0
for _ in range(n):
a, b = map(int, input().split())
a_list.append(a)
b_list.append(b)
a_list.sort()
b_list.sort(reverse=True)
b_stack = deque(b_list)
while h > 0:
if len(b_stack) > 0:
if a_list[-1] < b_stack[0]:
h -= b_stack.popleft()
time += 1
else:
time += h // a_list[-1]
h = 0
else:
time += h // a_list[-1]
h = 0
print(time)
|
s573330252
|
Accepted
| 409 | 12,056 | 582 |
from collections import deque
import math
n, h = map(int, input().split())
a_list = []
b_list = []
time = 0
for _ in range(n):
a, b = map(int, input().split())
a_list.append(a)
b_list.append(b)
a_list.sort()
b_list.sort(reverse=True)
b_stack = deque(b_list)
while h > 0:
if len(b_stack) > 0:
if a_list[-1] < b_stack[0]:
h -= b_stack.popleft()
time += 1
else:
time += math.ceil(h / a_list[-1])
h = 0
else:
time += math.ceil(h / a_list[-1])
h = 0
print(time)
|
s569534872
|
p03007
|
u350997995
| 2,000 | 1,048,576 |
Wrong Answer
| 1,833 | 21,524 | 346 |
There are N integers, A_1, A_2, ..., A_N, written on a blackboard. We will repeat the following operation N-1 times so that we have only one integer on the blackboard. * Choose two integers x and y on the blackboard and erase these two integers. Then, write a new integer x-y. Find the maximum possible value of the final integer on the blackboard and a sequence of operations that maximizes the final integer.
|
from heapq import heapify, heappop, heappush
N = int(input())
A = list(map(int,input().split()))
heapify(A)
ans = []
for i in range(N-1,0,-1):
n = i+1
a,b = A.pop(0),A.pop()
if n%2==1:
ans.append([a,b])
heappush(A,a-b)
else:
ans.append([b,a])
heappush(A,b-a)
print(A[0])
for a in ans:
print(*a)
|
s893821718
|
Accepted
| 270 | 21,676 | 312 |
N = int(input())
A = list(map(int,input().split()))
ans = []
A.sort()
a = A.pop(0)
b = A.pop()
N -= 2
k = 1
for i in range(N):
y = A.pop()
if y<0:k=0
if k:
ans.append([a,y])
a -= y
else:
ans.append([b,y])
b -= y
ans.append([b,a])
print(b-a)
for a in ans: print(*a)
|
s247800609
|
p02612
|
u707808519
| 2,000 | 1,048,576 |
Wrong Answer
| 30 | 9,140 | 30 |
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)
|
s655029638
|
Accepted
| 29 | 9,148 | 74 |
N = int(input())
if N%1000 == 0:
print(0)
else:
print(1000-N%1000)
|
s861917379
|
p03457
|
u473633103
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 365 |
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.
|
# coding: utf-8
# Your code here!
n = int(input())
now = [0,0]
time = 0
flag = True
for i in range(n):
t,x,y = map(int,input().split())
if t-time >= abs((now[0]-x)*(now[1]-y)) and (t-time)%2==((now[0]-x)*(now[1]-y))%2:
now = [x,y]
time = t
else:
flag = False
break
if flag:
print("Yes")
else:
print("No")
|
s000926790
|
Accepted
| 379 | 3,064 | 365 |
# coding: utf-8
# Your code here!
n = int(input())
now = [0,0]
time = 0
flag = True
for i in range(n):
t,x,y = map(int,input().split())
if t-time >= abs((now[0]-x)+(now[1]-y)) and (t-time)%2==((now[0]-x)+(now[1]-y))%2:
now = [x,y]
time = t
else:
flag = False
break
if flag:
print("Yes")
else:
print("No")
|
s203160662
|
p03524
|
u024782094
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,188 | 296 |
Snuke has a string S consisting of three kinds of letters: `a`, `b` and `c`. He has a phobia for palindromes, and wants to permute the characters in S so that S will not contain a palindrome of length 2 or more as a substring. Determine whether this is possible.
|
import sys
def input(): return sys.stdin.readline().strip()
def mp(): return map(int,input().split())
def lmp(): return list(map(int,input().split()))
s=input()
l=[]
l.append(s.count("a"))
l.append(s.count("b"))
l.append(s.count("c"))
if max(l)-min(l)<=1:
print("Yes")
else:
print("No")
|
s988325147
|
Accepted
| 18 | 3,188 | 296 |
import sys
def input(): return sys.stdin.readline().strip()
def mp(): return map(int,input().split())
def lmp(): return list(map(int,input().split()))
s=input()
l=[]
l.append(s.count("a"))
l.append(s.count("b"))
l.append(s.count("c"))
if max(l)-min(l)<=1:
print("YES")
else:
print("NO")
|
s267270178
|
p02288
|
u027872723
| 2,000 | 131,072 |
Wrong Answer
| 30 | 7,656 | 561 |
A binary heap which satisfies max-heap property is called max-heap. In a max- heap, for every node $i$ other than the root, $A Write a program which reads an array and constructs a max-heap from the array based on the following pseudo code. $maxHeapify(A, i)$ move the value of $A[i]$ down to leaves to make a sub-tree of node $i$ a max-heap. Here, $H$ is the size of the heap. 1 maxHeapify(A, i) 2 l = left(i) 3 r = right(i) 4 // select the node which has the maximum value 5 if l ≤ H and A[l] > A[i] 6 largest = l 7 else 8 largest = i 9 if r ≤ H and A[r] > A[largest] 10 largest = r 11 12 if largest ≠ i // value of children is larger than that of i 13 swap A[i] and A[largest] 14 maxHeapify(A, largest) // call recursively The following procedure buildMaxHeap(A) makes $A$ a max-heap by performing maxHeapify in a bottom-up manner. 1 buildMaxHeap(A) 2 for i = H/2 downto 1 3 maxHeapify(A, i)
|
# -*- coding: utf_8 -*-
def maxHeapify(A, i):
l = 2 * i
r = 2 * i + 1
largest = i
if l < len(A):
largest = largest if A[largest] > A[l] else l
if r < len(A):
largest = largest if A[largest] > A[r] else r
if largest != i:
A[largest], A[i] = A[i], A[largest]
maxHeapify(A, largest)
if __name__ == "__main__":
H = int(input())
a = [int(x) for x in input().split()]
print(str(int(H/2)))
for i in range(int(H / 2), 1, -1):
maxHeapify(a, i)
print(" ".join(str(x) for x in a)),
|
s347530122
|
Accepted
| 1,000 | 65,720 | 562 |
# -*- coding: utf_8 -*-
def maxHeapify(A, i):
l = 2 * i - 1
r = 2 * i
largest = i - 1
if l < len(A):
largest = largest if A[largest] > A[l] else l
if r < len(A):
largest = largest if A[largest] > A[r] else r
if largest != i - 1:
A[largest], A[i - 1] = A[i - 1], A[largest]
maxHeapify(A, largest + 1)
if __name__ == "__main__":
H = int(input())
a = [int(x) for x in input().split()]
for i in range(int(H / 2), 0, -1):
maxHeapify(a, i)
print(" " + " ".join(str(x) for x in a)),
|
s961655495
|
p03563
|
u119982147
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 199 |
Takahashi is a user of a site that hosts programming contests. When a user competes in a contest, the _rating_ of the user (not necessarily an integer) changes according to the _performance_ of the user, as follows: * Let the current rating of the user be a. * Suppose that the performance of the user in the contest is b. * Then, the new rating of the user will be the avarage of a and b. For example, if a user with rating 1 competes in a contest and gives performance 1000, his/her new rating will be 500.5, the average of 1 and 1000. Takahashi's current rating is R, and he wants his rating to be exactly G after the next contest. Find the performance required to achieve it.
|
N = int(input())
K = int(input())
A = 1 + ( N * K )
B = 1 * 2 +(N-1)*K
C = 1 * 2^2 +(N-2)*K
D = 1 * 2^3 +(N-3)*K
if A<B:
print(A)
elif B<C:
print(B)
elif C<D:
print(C)
else:
print(D)
|
s226378370
|
Accepted
| 17 | 2,940 | 47 |
R = int(input())
G = int(input())
print(2*G-R)
|
s483933245
|
p03369
|
u581603131
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 47 |
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
|
list = list(input())
print(list.count('o')*100)
|
s015467968
|
Accepted
| 18 | 2,940 | 51 |
list = list(input())
print(list.count('o')*100+700)
|
s129490002
|
p03480
|
u497952650
| 2,000 | 262,144 |
Wrong Answer
| 61 | 3,188 | 236 |
You are given a string S consisting of `0` and `1`. Find the maximum integer K not greater than |S| such that we can turn all the characters of S into `0` by repeating the following operation some number of times. * Choose a contiguous segment [l,r] in S whose length is at least K (that is, r-l+1\geq K must be satisfied). For each integer i such that l\leq i\leq r, do the following: if S_i is `0`, replace it with `1`; if S_i is `1`, replace it with `0`.
|
S = input()
ans = 0
tmp = 0
l = len(S)
for i in range(l-1):
if i == l-2:
ans = max(ans,tmp+1)
tmp = 0
elif S[i] == S[i+1]:
tmp += 1
else:
ans = max(ans,tmp+1)
tmp = 0
print(min(2,ans))
|
s978526344
|
Accepted
| 64 | 3,316 | 198 |
S = input()
l = len(S)
ans = l
"""
if len(set(list(S))) == 1 and S[0] == "0":
print(0)
exit()
"""
for i in range(l-1):
if S[i] != S[i+1]:
ans = min(ans,max(l-i-1,i+1))
print(ans)
|
s074371895
|
p02612
|
u089230684
| 2,000 | 1,048,576 |
Wrong Answer
| 26 | 9,160 | 276 |
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 1<=N<=10000:
yen=N%1000
if yen==0:
print(yen)
print("We can pay the exact price.")
else:
yen1=1000-yen
print(yen1)
print("We will use two 1000-yen bills to pay the price and receive %d yen in change."%yen1)
|
s696910486
|
Accepted
| 28 | 9,144 | 102 |
n = int(input())
x = n % 1000
if x == 0:
print(x)
else:
a = ((n // 1000) + 1) * 1000
print(a - n)
|
s231553901
|
p02388
|
u500935997
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,516 | 37 |
Write a program which calculates the cube of a given integer x.
|
x = 2
y = 3
print(x**3)
print(y**3)
|
s053472942
|
Accepted
| 20 | 5,576 | 30 |
x = int(input())
print(x**3)
|
s889072232
|
p03227
|
u038409959
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 124 |
You are given a string S of length 2 or 3 consisting of lowercase English letters. If the length of the string is 2, print it as is; if the length is 3, print the string after reversing it.
|
# coding: utf-8
s = input().strip()
if len(s) == 2:
print(s)
else:
for c in s:
print(c, end="")
print()
|
s217264639
|
Accepted
| 17 | 2,940 | 105 |
# coding: utf-8
s = input().strip()
if len(s) == 2:
print(s)
else:
ans = s[::-1]
print(ans)
|
s817883969
|
p02612
|
u385873134
| 2,000 | 1,048,576 |
Wrong Answer
| 34 | 9,144 | 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)
|
s186399658
|
Accepted
| 28 | 9,140 | 77 |
import math
N = int(input())
tmp = math.ceil(N / 1000)
print(1000*tmp - N)
|
s105764160
|
p03377
|
u663843442
| 2,000 | 262,144 |
Wrong Answer
| 20 | 3,316 | 149 |
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.
|
if __name__ == '__main__':
A, B, X = map(int, input().split())
if A <= X and X <= A + B:
print("Yes")
else:
print("No")
|
s358697990
|
Accepted
| 17 | 2,940 | 149 |
if __name__ == '__main__':
A, B, X = map(int, input().split())
if A <= X and X <= A + B:
print("YES")
else:
print("NO")
|
s127697841
|
p03610
|
u727238330
| 2,000 | 262,144 |
Wrong Answer
| 28 | 9,124 | 29 |
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(str)
print(s[::2])
|
s507288065
|
Accepted
| 24 | 9,016 | 26 |
s = input()
print(s[::2])
|
s877757471
|
p03998
|
u748311048
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 457 |
Alice, Bob and Charlie are playing _Card Game for Three_ , as below: * At first, each of the three players has a deck consisting of some number of cards. Each card has a letter `a`, `b` or `c` written on it. The orders of the cards in the decks cannot be rearranged. * The players take turns. Alice goes first. * If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says `a`, Alice takes the next turn.) * If the current player's deck is empty, the game ends and the current player wins the game. You are given the initial decks of the players. More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way. Determine the winner of the game.
|
sa = str(input())
sb = str(input())
sc = str(input())
turn = 'a'
while True:
if turn == 'a':
if sa == '':
print('A')
exit()
turn = sa[0]
sa = sa[1:]
elif turn == 'b':
if sb == '':
print('B')
exit()
turn = sb[0]
sb = sb[1:]
else:
if sc == '':
print('C')
exit()
turn = sc[0]
sc = sc[1:]
print(turn)
|
s360822171
|
Accepted
| 18 | 3,064 | 441 |
sa = str(input())
sb = str(input())
sc = str(input())
turn = 'a'
while True:
if turn == 'a':
if sa == '':
print('A')
exit()
turn = sa[0]
sa = sa[1:]
elif turn == 'b':
if sb == '':
print('B')
exit()
turn = sb[0]
sb = sb[1:]
else:
if sc == '':
print('C')
exit()
turn = sc[0]
sc = sc[1:]
|
s185612287
|
p03713
|
u806855121
| 2,000 | 262,144 |
Wrong Answer
| 258 | 3,064 | 347 |
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 = map(int, input().split())
Smin = 10**5 * 10**5
for i in range(1, H):
S1 = i * W
S2 = (H-i) * (W//2)
S3 = (H-i) * (W-W//2)
Smin = min(Smin, max(S1, S2, S3)-min(S1, S2, S3))
for i in range(1, W):
S1 = i * H
S2 = (W-i) * (H//2)
S3 = (W-i) * (H-H//2)
Smin = min(Smin, max(S1, S2, S3)-min(S1, S2, S3))
print(Smin)
|
s169084354
|
Accepted
| 500 | 3,064 | 559 |
H, W = map(int, input().split())
Smin = 10**5 * 10**5
for i in range(1, H):
S1 = i * W
S2 = (H-i) * (W//2)
S3 = (H-i) * (W-W//2)
Smin = min(Smin, max(S1, S2, S3)-min(S1, S2, S3))
S2 = (H-i)//2 * W
S3 = ((H-i)-(H-i)//2) * W
Smin = min(Smin, max(S1, S2, S3)-min(S1, S2, S3))
for i in range(1, W):
S1 = i * H
S2 = (W-i) * (H//2)
S3 = (W-i) * (H-H//2)
Smin = min(Smin, max(S1, S2, S3)-min(S1, S2, S3))
S2 = (W-i)//2 * H
S3 = ((W-i)-(W-i)//2) * H
Smin = min(Smin, max(S1, S2, S3)-min(S1, S2, S3))
print(Smin)
|
s439261611
|
p03544
|
u870297120
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 231 |
It is November 18 now in Japan. By the way, 11 and 18 are adjacent Lucas numbers. You are given an integer N. Find the N-th Lucas number. Here, the i-th Lucas number L_i is defined as follows: * L_0=2 * L_1=1 * L_i=L_{i-1}+L_{i-2} (i≥2)
|
n = int(input())
memo = [-1]*(n+1)
def func(i):
if i == 0:
return 2
if i == 1:
return 1
if memo[i] != -1:
return memo[i]
memo[i] = func(i-2) + func(i-1)
return memo[i]
func(n)
|
s162873342
|
Accepted
| 18 | 3,060 | 238 |
n = int(input())
memo = [-1]*(n+1)
def func(i):
if i == 0:
return 2
if i == 1:
return 1
if memo[i] != -1:
return memo[i]
memo[i] = func(i-2) + func(i-1)
return memo[i]
print(func(n))
|
s884608364
|
p03828
|
u652656291
| 2,000 | 262,144 |
Wrong Answer
| 31 | 9,004 | 92 |
You are given an integer N. Find the number of the positive divisors of N!, modulo 10^9+7.
|
n = int(input())
mod = 10**9 +7
ans = 1
for i in range(1,n+1):
ans *= (i % mod)
print(ans)
|
s251155293
|
Accepted
| 31 | 9,116 | 284 |
mod=10**9+7
import math
n=int(input())
ans=1
l=[0]*n
for ii in range(2,n+1):
i=ii
for j in range(2,int(math.sqrt(i))+1):
if i%j==0:
cnt=0
while i%j==0:
cnt+=1;i//=j
l[j-1]+=cnt
if i!=1:
l[i-1]+=1
for i in l:
ans*=(i+1)
ans%=mod
print(ans)
|
s207795745
|
p02850
|
u365364616
| 2,000 | 1,048,576 |
Wrong Answer
| 758 | 46,796 | 868 |
Given is a tree G with N vertices. The vertices are numbered 1 through N, and the i-th edge connects Vertex a_i and Vertex b_i. Consider painting the edges in G with some number of colors. We want to paint them so that, for each vertex, the colors of the edges incident to that vertex are all different. Among the colorings satisfying the condition above, construct one that uses the minimum number of colors.
|
from collections import defaultdict
def bfs(i, V, E, Ec):
V[i] = 1
Q = [(i, 1)]
while(True):
Qi = []
for q, cq in Q:
c = 1
if cq == c:
c += 1
for j, idx in E[q]:
if V[j] == 0:
V[j] = 1
Qi.append((j, c))
Ec[idx] = c
if c + 1 == cq:
c += 2
else:
c += 1
if len(Qi) == 0:
break
else:
Q = Qi
return V, Ec
n = int(input())
E = defaultdict(list)
for i in range(n - 1):
a, b = map(int, input().split())
E[a].append((b, i))
E[b].append((a, i))
Ec = [None for _ in range(n - 1)]
V = [0 for _ in range(n + 1)]
_, Ec = bfs(1, V, E, Ec)
print(max(Ec))
for c in Ec:
print(c)
|
s955797427
|
Accepted
| 729 | 46,444 | 788 |
from collections import defaultdict
def bfs(i, V, E, Ec):
V[i] = 1
Q = [(i, 0)]
while(True):
Qi = []
for q, cq in Q:
c = 1
for j, idx in E[q]:
if V[j] == 0:
if cq == c:
c += 1
V[j] = 1
Qi.append((j, c))
Ec[idx] = c
c += 1
if len(Qi) == 0:
break
else:
Q = Qi
return V, Ec
n = int(input())
E = defaultdict(list)
for i in range(n - 1):
a, b = map(int, input().split())
E[a].append((b, i))
E[b].append((a, i))
Ec = [None for _ in range(n - 1)]
V = [0 for _ in range(n + 1)]
_, Ec = bfs(1, V, E, Ec)
print(max(Ec))
for c in Ec:
print(c)
|
s528570750
|
p03502
|
u062189367
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 178 |
An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10. Given an integer N, determine whether it is a Harshad number.
|
N = int(input())
n = N
print(len(str(N)))
sum = 0
while N>0:
N1 = N%10
print(N1)
sum += N1
N = (N-N1)/10
if n%sum == 0:
print('Yes')
else:
print('No')
|
s425548392
|
Accepted
| 20 | 3,060 | 147 |
N = int(input())
n = N
sum = 0
while N>0:
N1 = N%10
sum += N1
N = (N-N1)/10
if n%sum == 0:
print('Yes')
else:
print('No')
|
s688883086
|
p03485
|
u164229553
| 2,000 | 262,144 |
Wrong Answer
| 148 | 12,468 | 80 |
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.
|
import numpy as np
a, b = [int(j) for j in input().split()]
print((a+b)//2 + 1)
|
s050048015
|
Accepted
| 150 | 12,424 | 78 |
import numpy as np
a, b = [int(j) for j in input().split()]
print((a+b+1)//2)
|
s059446836
|
p02613
|
u341543478
| 2,000 | 1,048,576 |
Wrong Answer
| 149 | 9,144 | 338 |
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 _ in range(n):
s = input()
if s == 'AC':
ac += 1
elif s == 'WA':
wa += 1
elif s == 'TLE':
tle += 1
elif s == 'RE':
re += 1
print('AC x {}'.format(ac))
print('WA x {}'.format(wa))
print('TLE x {}'.format(tle))
print('re x {}'.format(re))
|
s558229951
|
Accepted
| 147 | 9,100 | 338 |
n = int(input())
ac = 0
wa = 0
tle = 0
re = 0
for _ in range(n):
s = input()
if s == 'AC':
ac += 1
elif s == 'WA':
wa += 1
elif s == 'TLE':
tle += 1
elif s == 'RE':
re += 1
print('AC x {}'.format(ac))
print('WA x {}'.format(wa))
print('TLE x {}'.format(tle))
print('RE x {}'.format(re))
|
s023614965
|
p03543
|
u216631280
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 123 |
We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**?
|
x = input()
count = 0
for i in range(3):
if x[i] == x[i + 1]:
count += 1
if count >= 3:
print('Yes')
else:
print('No')
|
s720323763
|
Accepted
| 17 | 2,940 | 93 |
n = input()
if n[0] == n[1] == n[2] or n[1] == n[2] == n[3]:
print('Yes')
else:
print('No')
|
s570126499
|
p03943
|
u410903849
| 2,000 | 262,144 |
Wrong Answer
| 25 | 9,028 | 115 |
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")
|
s173738768
|
Accepted
| 28 | 9,144 | 115 |
a,b,c = map(int,input().split())
if a + b == c or b + c == a or a + c == b:
print("Yes")
else:
print("No")
|
s359944917
|
p03471
|
u829895669
| 2,000 | 262,144 |
Wrong Answer
| 844 | 3,060 | 205 |
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 = (-1,-1,-1)
for i in range(N + 1):
for j in range(N + 1 - i):
k = N - i - j
if 10000*i + 5000*j + 1000*k == Y:
ans = (i,j,k)
print(ans)
|
s691263109
|
Accepted
| 812 | 3,060 | 207 |
N,Y = map(int, input().split())
a,b,c = -1,-1,-1
for i in range(N + 1):
for j in range(N + 1 - i):
k = N - i - j
if 10000*i + 5000*j + 1000*k == Y:
a,b,c = i,j,k
print(a,b,c)
|
s819916106
|
p03416
|
u123745130
| 2,000 | 262,144 |
Wrong Answer
| 68 | 2,940 | 141 |
Find the number of _palindromic numbers_ among the integers between A and B (inclusive). Here, a palindromic number is a positive integer whose string representation in base 10 (without leading zeros) reads the same forward and backward.
|
x,y=map(int,input().split())
print(x,y)
count_num=0
for i in range(x,y+1):
if str(i)==str(i)[::-1]:
count_num+=1
print(count_num)
|
s065087683
|
Accepted
| 62 | 2,940 | 130 |
x,y=map(int,input().split())
count_num=0
for i in range(x,y+1):
if str(i)==str(i)[::-1]:
count_num+=1
print(count_num)
|
s341567095
|
p02618
|
u608755339
| 2,000 | 1,048,576 |
Wrong Answer
| 37 | 9,200 | 171 |
AtCoder currently hosts three types of contests: ABC, ARC, and AGC. As the number of users has grown, in order to meet the needs of more users, AtCoder has decided to increase the number of contests to 26 types, from AAC to AZC. For convenience, we number these 26 types as type 1 through type 26. AtCoder wants to schedule contests for D days so that user satisfaction is as high as possible. For every day, AtCoder will hold exactly one contest, and each contest will end on that day. The satisfaction is calculated as follows. * The satisfaction at the beginning of day 1 is 0. Satisfaction can be negative. * Holding contests increases satisfaction. The amount of increase will vary depending on a variety of factors. Specifically, we know in advance that holding a contest of type i on day d will increase the satisfaction by s_{d,i}. * If a particular type of contest is not held for a while, the satisfaction decreases. Each contest type i has an integer c_i, and at the end of each day d=1,2,...,D, the satisfaction decreases as follows. Let \mathrm{last}(d,i) be the last day before day d (including d) on which a contest of type i was held. If contests of type i have never been held yet, we define \mathrm{last}(d,i)=0. At the end of day d, the satisfaction decreases by \sum _{i=1}^{26}c_i \times (d-\mathrm{last}(d,i)). Please schedule contests on behalf of AtCoder. If the satisfaction at the end of day D is S, you will get a score of \max(10^6 + S, 0). There are 50 test cases, and the score of a submission is the total scores for each test case. You can make submissions multiple times, and the highest score among your submissions will be your score.
|
N = 26
D = int(input())
C = [int(i) for i in input().split()]
S = []
for i in range(D):
s = max([int(i) for i in input().split()])
S.append(s)
for i in S:
print(i)
|
s983373808
|
Accepted
| 36 | 9,208 | 198 |
N = 26
D = int(input())
C = [int(i) for i in input().split()]
S = []
for i in range(D):
s = [int(i) for i in input().split()]
m = max(s)
i = s.index(m)
S.append(i+1)
for i in S:
print(i)
|
s544779132
|
p04029
|
u111525113
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 45 |
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?
|
num = int(input())
print( num * (num+1) / 2 )
|
s222752577
|
Accepted
| 18 | 2,940 | 50 |
num = int(input())
print(int( num * (num+1) / 2 ))
|
s720440362
|
p03478
|
u874723578
| 2,000 | 262,144 |
Wrong Answer
| 36 | 3,060 | 141 |
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
n,a,b = map(int, input().split())
ans = 0
for i in range(n+1):
if a <= sum(list(map(int,list(str(i))))) <= b:
ans += 1
print(ans)
|
s521830107
|
Accepted
| 36 | 3,060 | 144 |
n,a,b = map(int, input().split())
ans = 0
for i in range(1, n+1):
if a <= sum(list(map(int,list(str(i))))) <= b:
ans += i
print(ans)
|
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