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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s148699409 | p02659 | u181719868 | 2,000 | 1,048,576 | Wrong Answer | 19 | 9,172 | 110 | Compute A \times B, truncate its fractional part, and print the result as an integer. | a, b = input().strip().split()
a, b = [int(a), float(b)]
b = b*100 + 0.000000001
print(b)
print(a*int(b)//100) | s226446227 | Accepted | 21 | 9,160 | 102 | a, b = input().strip().split()
a, b = [int(a), float(b)]
b = b*100 + 0.000000001
print(a*int(b)//100) |
s624865059 | p03387 | u062189367 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 147 | 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. | X = list(map(int, input().split()))
M = max(X)
if 3*M %2 == sum(X)%2:
N = (3*M-sum(X))/2
else:
N = (3*(M+1)-sum(X))/2
print(N)
| s343835015 | Accepted | 17 | 2,940 | 152 | X = list(map(int, input().split()))
M = max(X)
if 3*M %2 == sum(X)%2:
N = (3*M-sum(X))/2
else:
N = (3*(M+1)-sum(X))/2
print(int(N))
|
s807141627 | p02601 | u265323413 | 2,000 | 1,048,576 | Wrong Answer | 34 | 9,188 | 369 | 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. | l = list(map(int, input().split()))
K = int(input())
def check(nums):
if nums[0] < nums[1]:
if nums[1] < nums[2]:
print('yes')
exit(0)
check(l)
for i in range(K):
if l[0] > l[1]:
l[1] *= 2
check(l)
continue
elif l[1] > l[2]:
l[2] *= 2
check(l)
continue
print('no')
exit(0)
| s860558482 | Accepted | 28 | 9,192 | 371 | l = list(map(int, input().split()))
K = int(input())
def check(nums):
if nums[0] < nums[1]:
if nums[1] < nums[2]:
print('Yes')
exit(0)
check(l)
for i in range(K):
if l[0] >= l[1]:
l[1] *= 2
check(l)
continue
elif l[1] >= l[2]:
l[2] *= 2
check(l)
continue
print('No')
exit(0)
|
s351369571 | p02927 | u946969297 | 2,000 | 1,048,576 | Wrong Answer | 23 | 2,940 | 197 | Today is August 24, one of the five Product Days in a year. A date m-d (m is the month, d is the date) is called a Product Day when d is a two-digit number, and all of the following conditions are satisfied (here d_{10} is the tens digit of the day and d_1 is the ones digit of the day): * d_1 \geq 2 * d_{10} \geq 2 * d_1 \times d_{10} = m Takahashi wants more Product Days, and he made a new calendar called Takahashi Calendar where a year consists of M month from Month 1 to Month M, and each month consists of D days from Day 1 to Day D. In Takahashi Calendar, how many Product Days does a year have? | M, D = map(int,input().split())
c=0
for m in range(1,M+1):
for d in range(1,D+1):
if d>=10:
t=str(d)
if int(t[0])*int(t[1]) == m:
c+=1
print(c) | s071819116 | Accepted | 26 | 3,060 | 266 | # coding: utf-8
# Your code here!
M, D = map(int,input().split())
c=0
for m in range(1,M+1):
for d in range(1,D+1):
if d>=10:
t=str(d)
if int(t[0])>1 and int(t[1])>1 and int(t[0])*int(t[1]) == m:
c+=1
print(c)
|
s707555980 | p03997 | u333731247 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 64 | 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*0.5) | s176590500 | Accepted | 17 | 2,940 | 69 | a=int(input())
b=int(input())
h=int(input())
print(int((a+b)*h*0.5)) |
s267532440 | p02690 | u034128150 | 2,000 | 1,048,576 | Wrong Answer | 2,205 | 9,856 | 429 | Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X. | X = int(input())
memo = []
A = 0
while True:
A += 1
A5 = A ** 5
memo.append(A5)
if A5 > X:
try:
i = memo.index(A5 - X)
except ValueError:
continue
else:
B = i
break
else:
try:
i = memo.index(X - A5)
except ValueError:
continue
else:
B = -i
break
print(A, B)
| s361273015 | Accepted | 22 | 9,196 | 424 | X = int(input())
memo = []
A = -1
while True:
A += 1
A5 = A ** 5
memo.append(A5)
if A5 > X:
try:
i = memo.index(A5 - X)
except ValueError:
continue
else:
B = i
break
else:
try:
i = memo.index(X - A5)
except ValueError:
continue
else:
B = -i
break
print(A, B) |
s350388996 | p03149 | u966311314 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,060 | 382 | You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974". | a,b,c,d=[int(n) for n in input().split(' ')]
N = [1, 9, 7, 4]
if a not in N:
print("No")
else:
N.remove(a)
if b not in N:
print("No")
else:
N.remove(b)
if c not in N:
print("No")
else:
N.remove(c)
if d not in N:
print("Np")
else:
print("Yes")
| s120597364 | Accepted | 17 | 2,940 | 118 | N =[int(n) for n in input().split(' ')]
M = [1, 9, 7, 4]
if set(N)==set(M):
print("YES")
else:
print("NO") |
s710377733 | p02408 | u180914582 | 1,000 | 131,072 | Wrong Answer | 30 | 6,716 | 224 | 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': [0 for _ in range(13)],
'H': [0 for _ in range(13)],
'C': [0 for _ in range(13)],
'D': [0 for _ in range(13)],
}
n = int(input())
for _ in range(n):
(s, r) = input().split() | s725018557 | Accepted | 20 | 7,708 | 412 | cards = {
'S': [0 for _ in range(13)],
'H': [0 for _ in range(13)],
'C': [0 for _ in range(13)],
'D': [0 for _ in range(13)],
}
n = int(input())
for _ in range(n):
(s, r) = input().split()
r = int(r)
cards[s][r - 1] = r
for s in ('S', 'H', 'C', 'D'):
for r in range(13):
if cards[s][r] == 0:
print( '{0:s} {1:d}'.format(s, r+1)) |
s831250541 | p03605 | u898967808 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 65 | It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N? | n = input()
print('Yes') if (n[0]==9 or n[1]==9) else print('No') | s926911340 | Accepted | 18 | 2,940 | 69 | n = input()
print('Yes') if (n[0]=='9' or n[1]=='9') else print('No') |
s073214044 | p02865 | u185935459 | 2,000 | 1,048,576 | Wrong Answer | 149 | 2,940 | 110 | How many ways are there to choose two distinct positive integers totaling N, disregarding the order? | N = int(input())
count = 0
for i in range(1,int(N/2)+1):
if i + N-i == N:
count+=1
print(count) | s199673776 | Accepted | 111 | 2,940 | 211 | N = int(input())
count = 0
if N % 2 == 0:
for i in range(1,int(N/2)):
if i != N-i:
count+=1
else:
for i in range(1,int(N/2)+1):
if i != N:
count+=1
print(count) |
s452366740 | p03095 | u920204936 | 2,000 | 1,048,576 | Wrong Answer | 33 | 5,664 | 179 | You are given a string S of length N. Among its subsequences, count the ones such that all characters are different, modulo 10^9+7. Two subsequences are considered different if their characters come from different positions in the string, even if they are the same as strings. Here, a subsequence of a string is a concatenation of **one or more** characters from the string without changing the order. | import collections
line = int(input())
string = list(input())
print(string)
data = collections.Counter(string)
sum = 1
for word in data:
sum *= (data[word] + 1)
print(sum - 1) | s522558992 | Accepted | 28 | 4,340 | 212 | # coding: utf-8
import collections
line = int(input())
string = list(input())
data = collections.Counter(string)
sum = 1
for word in data:
sum *= (data[word] + 1)
answer = (sum - 1)%(10**9 + 7)
print(answer) |
s955481282 | p02263 | u637322311 | 1,000 | 131,072 | Wrong Answer | 30 | 5,556 | 457 | An expression is given in a line. Two consequtive symbols (operand or operator) are separated by a space character. You can assume that +, - and * are given as the operator and an operand is a positive integer less than 106 | def is_operator(ele):
if ele == "+" or ele == "-" or ele == "*":
return True
else:
return False
def calc_reverse_polish_notation(A):
stack = []
for i in A:
if is_operator(i):
a = stack.pop()
b = stack.pop()
ans = eval("{0} {1} {2}".format(b, i, a))
stack.append(ans)
else:
stack.append(i)
return stack.pop()
A = list(map(str,input().split()))
| s227151912 | Accepted | 20 | 5,564 | 497 | def is_operator(ele):
if ele == "+" or ele == "-" or ele == "*":
return True
else:
return False
def calc_reverse_polish_notation(A):
stack = []
for i in A:
if is_operator(i):
a = stack.pop()
b = stack.pop()
ans = eval("{0} {1} {2}".format(b, i, a))
stack.append(ans)
else:
stack.append(i)
return stack.pop()
A = list(map(str,input().split()))
print(calc_reverse_polish_notation(A))
|
s544183134 | p03478 | u034369223 | 2,000 | 262,144 | Wrong Answer | 58 | 3,368 | 271 | 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(' '))
total = 0
for num in range(1,n+1):
split_nums = [int(s) for s in list(str(num))]
sum = 0
for s in split_nums:
sum += s
if a <= sum and sum <= b:
print(split_nums, sum)
total += num
print(total) | s626195763 | Accepted | 39 | 3,060 | 240 | n, a, b = map(int,input().split(' '))
total = 0
for num in range(1,n+1):
split_nums = [int(s) for s in list(str(num))]
sum = 0
for s in split_nums:
sum += s
if a <= sum and sum <= b:
total += num
print(total) |
s511793852 | p03434 | u418826171 | 2,000 | 262,144 | Wrong Answer | 29 | 9,168 | 91 | We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score. | N = int(input())
a = sorted(list(map(int,input().split())))
print(sum(a[::2])-sum(a[1::2])) | s367915473 | Accepted | 27 | 9,160 | 186 | N = int(input())
A = sorted(list(map(int,input().split())), reverse = True)
a = 0
b = 0
for i in range(len(A)):
if i%2 == 0:
a += A[i]
else:
b += A[i]
print(a-b) |
s798921065 | p03385 | u540698208 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 113 | You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`. | def main():
s = sorted(input())
print('Yes' if s=='abc' else 'No')
if __name__ == '__main__':
main() | s565845895 | Accepted | 17 | 2,940 | 121 | def main():
s = sorted(input())
print('Yes' if s==['a','b','c'] else 'No')
if __name__ == '__main__':
main() |
s950159898 | p02255 | u939814144 | 1,000 | 131,072 | Wrong Answer | 30 | 7,500 | 657 | Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step. | def insertion_sort(array, element_number):
for i in range(1, element_number):
v = array[i]
j = i - 1
while(j>=0 and array[j] > v):
array[j+1] = array[j]
j = j - 1
array[j+1] = v
for k in range(0, element_number):
if k < element_number:
print(array[k] + ' ', end='')
else:
print(array[k], end='')
if i < element_number-1:
print('\n', end='')
def test():
element_number = int(input())
input_array = input().split()
insertion_sort(input_array, element_number)
if __name__ == '__main__':
test() | s366757655 | Accepted | 20 | 7,708 | 421 | def insertion_sort(array):
for i in range(len(array)):
v = array[i]
j = i - 1
while(j>=0 and array[j] > v):
array[j+1] = array[j]
j = j - 1
array[j+1] = v
print(' '.join(map(str, array)))
return array
if __name__ == '__main__':
element_number = int(input())
input_array = list(map(int, input().split()))
insertion_sort(input_array) |
s614640395 | p03730 | u891847179 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 199 | 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 = list(map(int, input().split()))
flag = False
for i in range(1, B + 1):
remainder = A * i % B
if remainder == C:
flag = True
if flag:
print("Yes")
else:
print("No")
| s614662087 | Accepted | 28 | 9,152 | 96 | import math
A,B,C = map(int,input().split())
print('YES' if C % math.gcd(A,B) == 0 else 'NO')
|
s966228267 | p02678 | u426175055 | 2,000 | 1,048,576 | Wrong Answer | 1,242 | 47,924 | 535 | 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 defaultdict
import queue
N,M = map(int,input().split())
guide = {}
room = defaultdict(list)
for i in range(M):
a,b = map(int,input().split())
room[a].append(b)
room[b].append(a)
print(room)
q = queue.Queue()
for i in room[1]:
q.put(i)
guide[i] = 1
while not(q.empty()):
x = q.get()
for i in room[x]:
if i not in guide:
guide[i]=x
q.put(i)
for i in range(2,N+1):
print(guide[i])
| s253526217 | Accepted | 1,089 | 47,796 | 531 | from collections import defaultdict
import queue
N,M = map(int,input().split())
guide = {}
room = defaultdict(list)
for i in range(M):
a,b = map(int,input().split())
room[a].append(b)
room[b].append(a)
q = queue.Queue()
for i in room[1]:
q.put(i)
guide[i] = 1
while not(q.empty()):
x = q.get()
for i in room[x]:
if i not in guide:
guide[i]=x
q.put(i)
if len(guide) != N:
print("No")
else:
print("Yes")
for i in range(2,N+1):
print(guide[i])
|
s482964089 | p03493 | u414050834 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 78 | 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()
count = 0
for i in range(3):
if s==1:
count= count + 1 | s499618216 | Accepted | 17 | 2,940 | 86 | a = input()
c=0
for i in range(len(a)):
if a[i] == '1':
c = c + 1
print(c) |
s487294783 | p03578 | u233747425 | 2,000 | 262,144 | Wrong Answer | 2,105 | 35,556 | 348 | Rng is preparing a problem set for a qualification round of CODEFESTIVAL. He has N candidates of problems. The difficulty of the i-th candidate is D_i. There must be M problems in the problem set, and the difficulty of the i-th problem must be T_i. Here, one candidate of a problem cannot be used as multiple problems. Determine whether Rng can complete the problem set without creating new candidates of problems. | N = int(input())
D = list(map(int, input().split()))
M = int(input())
T = list(map(int, input().split()))
cou = 0
if N >= M:
for i in range(M):
if T[i] in D:
D.remove(T[i])
cou += 1
else:
x = 'No'
break
if cou == M:
x = 'Yes'
else:
x = 'No'
print(x) | s960153092 | Accepted | 398 | 56,672 | 811 | N = int(input())
D = list(map(int, input().split()))
M = int(input())
T = list(map(int, input().split()))
D_dic = {k: 0 for k in D}
T_dic = {k: 0 for k in T}
for i in range(N):
d = D[i]
if D_dic[d]:
a = D_dic[d] + 1
D_dic[d] = a
else:
D_dic[d] = 1
for j in range(M):
t = T[j]
if T_dic[t]:
b = T_dic[t] + 1
T_dic[t] = b
else:
T_dic[t] = 1
cou = 0
if N >= M:
for k in range(M):
tk = T[k]
try:
if D_dic[tk]:
c = D_dic[tk] - 1
D_dic[tk] = c
cou += 1
else:
x = 'NO'
break
except KeyError:
x = 'NO'
break
if cou == M:
x = 'YES'
else:
x = 'NO'
print(x) |
s091991145 | p02690 | u058496530 | 2,000 | 1,048,576 | Wrong Answer | 23 | 9,196 | 284 | Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X. | x = int(input())
y = []
count = 0
for i in range(0, 70):
y.append(i*i*i*i*i)
for i in y:
count += 1
for j in y[:count]:
y2 = i + j
y3 = i - j
if y2 == x:
print(str(i) + " " + str(j))
break;
if y3 == x:
print(str(i) + " -" + str(j))
break; | s099140926 | Accepted | 25 | 9,104 | 507 | x = int(input())
y = []
count1 = 0
flag = 0
for i in range(0, 130):
y.append(i*i*i*i*i)
for i in y:
if i == x:
print("0 -" + str(count1))
flag = 1
break;
count1 += 1
count1 = 0
for i in y:
if flag == 1:
break;
count2 = 0
for j in y[:count1]:
y2 = i + j
y3 = i - j
if y2 == x:
print(str(count1) + " -" + str(count2))
flag = 1
break;
if y3 == x:
print(str(count1) + " " + str(count2))
flag = 1
break;
count2 +=1
count1 += 1 |
s504577891 | p03711 | u686036872 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 189 | 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. | x, y = map(int, input().split())
A=[1, 3, 5, 7, 8, 10, 12]
B=[4, 6, 9, 11]
C=[2]
if (x in A and y in A) or (x in B and y in B) or (x in C and y in C):
print("YES")
else:
print("NO") | s126920883 | Accepted | 17 | 2,940 | 119 | A={1, 3, 5, 7, 8, 10, 12}
B={4, 6, 9, 11}
x = set(map(int, input().split()))
print("Yes" if x <= A or x <= B else "No") |
s998814547 | p03448 | u088488125 | 2,000 | 262,144 | Wrong Answer | 29 | 9,112 | 188 | 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_dash=x//50
cnt=0
for i in range(1,a+1):
for j in range(1,b+1):
k=x_dash-10*i-2*j
if 0<=k<=c:
cnt+=1
print(cnt) | s344105730 | Accepted | 26 | 9,172 | 188 | a=int(input())
b=int(input())
c=int(input())
x=int(input())
x_dash=x//50
cnt=0
for i in range(0,a+1):
for j in range(0,b+1):
k=x_dash-10*i-2*j
if 0<=k<=c:
cnt+=1
print(cnt) |
s308634618 | p03386 | u350997995 | 2,000 | 262,144 | Wrong Answer | 2,235 | 3,956 | 117 | 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())
S = [i for i in range(A,B+1)]
S = S[:K]+S[len(S)-K:]
S = set(S)
for s in S: print(s) | s073371606 | Accepted | 17 | 3,060 | 175 | A,B,K = map(int,input().split())
if 2*K<B-A+1:
S = [i for i in range(A,A+K)]+[i for i in range(B-K+1,B+1)]
else:
S = [i for i in range(A,B+1)]
for s in S:
print(s) |
s658176463 | p03486 | u221345507 | 2,000 | 262,144 | Wrong Answer | 19 | 3,064 | 834 | 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=list(str(input()))
t=list(str(input()))
alphabet=['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']
for i in range (len(s)):
for j in range (len(alphabet)):
if s[i]==alphabet[j]:
s[i]=j
for i in range (len(t)):
for j in range (len(alphabet)):
if t[i]==alphabet[j]:
t[i]=j
if len(s)>len(t):
lst=[-1]*(len(s)-len(t))
t=t+lst
if len(t)>len(s):
lst=[-1]*(len(t)-len(s))
s=s+lst
import sys
for i in range (len(s)):
if min(s)<max(t):
print ('Yes')
print(s)
print(t)
sys.exit()
elif min(s)>max(t):
print ('No')
print(s)
print(t)
sys.exit()
else:
if min(s)==max(t):
s.remove(min(s))
t.remove(max(t))
print ('No') | s030038267 | Accepted | 19 | 2,940 | 108 | s=list(input())
t=list(input())
s.sort()
t.sort(reverse=True)
if s<t:
print('Yes')
else:
print('No') |
s561468799 | p03997 | u595716769 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 490 | You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid. | import itertools
num = str(input())
olist = [int(i+1) for i in range(len(num)-1)]
comblist = []
for i, _ in enumerate(olist, 1):
for j in itertools.combinations(olist, r=i):
comblist.append(j)
L = []
for i in range(len(comblist)):
t = num
for j in range(len(comblist[i])):
t = t[0:comblist[i][j]+j] + "," + t[comblist[i][j]+j:]
L.append(t)
out = 0
for i in range(len(L)):
t = L[i].split(",")
for j in range(len(t)):
out += int(t[j])
print(out+int(num)) | s004145649 | Accepted | 17 | 2,940 | 69 | a = int(input())
b = int(input())
h = int(input())
print((a+b)*h//2) |
s310297094 | p03944 | u927870520 | 2,000 | 262,144 | Wrong Answer | 150 | 27,252 | 338 | There is a rectangle in the xy-plane, with its lower left corner at (0, 0) and its upper right corner at (W, H). Each of its sides is parallel to the x-axis or y-axis. Initially, the whole region within the rectangle is painted white. Snuke plotted N points into the rectangle. The coordinate of the i-th (1 ≦ i ≦ N) point was (x_i, y_i). Then, he created an integer sequence a of length N, and for each 1 ≦ i ≦ N, he painted some region within the rectangle black, as follows: * If a_i = 1, he painted the region satisfying x < x_i within the rectangle. * If a_i = 2, he painted the region satisfying x > x_i within the rectangle. * If a_i = 3, he painted the region satisfying y < y_i within the rectangle. * If a_i = 4, he painted the region satisfying y > y_i within the rectangle. Find the area of the white region within the rectangle after he finished painting. | import numpy as np
W,H,N=map(int,input().split())
A=np.zeros((W,H))
B=[list(map(int,input().split())) for _ in range(N)]
print(B)
for b in B:
if b[2]==1:
A[:b[0],:]=1
elif b[2]==2:
A[b[0]:,:]=1
elif b[2]==3:
A[:,:b[1]]=1
elif b[2]==4:
A[:,b[1]:]=1
print(A)
print(np.count_nonzero(A==0)) | s673757909 | Accepted | 116 | 27,240 | 316 | import numpy as np
W,H,N=map(int,input().split())
A=np.zeros((W,H))
B=[list(map(int,input().split())) for _ in range(N)]
for b in B:
if b[2]==1:
A[:b[0],:]=1
elif b[2]==2:
A[b[0]:,:]=1
elif b[2]==3:
A[:,:b[1]]=1
elif b[2]==4:
A[:,b[1]:]=1
print(np.count_nonzero(A==0)) |
s246541079 | p00115 | u300645821 | 1,000 | 131,072 | Wrong Answer | 70 | 14,776 | 958 | 恒星歴 2005.11.5。あなたは宇宙船 UAZ アドバンス号の艦長として敵の宇宙船と交戦しようとしています。 幸い敵の宇宙船はまだこちらに気付かずに静止しています。また、すでに敵の宇宙座標は判明しており強力な直線のビームを放つ「フェザー砲」は発射準備を完了しています。あとは、発射命令を出すばかりです。 ところが、宇宙空間には、敵の設置したエネルギーバリアが存在しています。バリアは三角形をしており「フェザー砲」のビームをはね返してしまいます。また、ビームがバリアに当たれば敵に気付かれて逃げられてしまいます。事前に命中すると判定できなければ、発射命令は出せません。 そこで、UAZ アドバンス号、敵、バリアの位置の宇宙座標(3次元座標 x, y, z) を入力して、ビームがバリアをさけて敵に命中する場合は "HIT"、バリアに当たってしまう場合"MISS"と出力するプログラムを作成してください。 ただし、バリアはアドバンス号から 3 角形に見えるものだけが対象であり、線分につぶれて見えるものはないものとします。また、バリアは 3 角形の頂点を含む境界でも有効であり、ビームをはね返すものとします。また、敵がバリア内にいる場合は"MISS"と出力して下さい。 | #!/usr/bin/python
from fractions import Fraction
import sys
if sys.version_info[0]>=3: raw_input=input
def gauss(a):
if not a or len(a)==0: return None
n=len(a)
for i in range(n):
if a[i][i]==0:
for j in range(i+1,n):
if a[j][i]!=0:
for k in range(i,n+1): a[i][k]+=a[j][k]
break
else:
return None
for j in range(n):
if i!=j:
r = Fraction(a[j][i],a[i][i])
for k in range(i,n+1): a[j][k] = a[j][k] - a[i][k]*r
for i in range(n):
x=Fraction(a[i][i],1)
for j in range(len(a[i])):
a[i][j] /= x
return a
uaz=map(int,raw_input().split())
enemy=[0]+[-x+y for x,y in zip(map(int,raw_input().split()),uaz)]
b0=[1]+[x-y for x,y in zip(map(int,raw_input().split()),uaz)]
b1=[1]+[x-y for x,y in zip(map(int,raw_input().split()),uaz)]
b2=[1]+[x-y for x,y in zip(map(int,raw_input().split()),uaz)]
sol=gauss(list(map(list,zip(b0,b1,b2,enemy,[1,0,0,0]))))
print('MISS' if sol and all(0<=e[-1]<=1 for e in sol) else 'HIT') | s210168187 | Accepted | 70 | 14,776 | 964 | #!/usr/bin/python
from fractions import Fraction
import sys
if sys.version_info[0]>=3: raw_input=input
def gauss(a):
if not a or len(a)==0: return None
n=len(a)
for i in range(n):
if a[i][i]==0:
for j in range(i+1,n):
if a[j][i]!=0:
for k in range(i,n+1): a[i][k]+=a[j][k]
break
else:
return None
for j in range(n):
if i!=j:
r = Fraction(a[j][i],a[i][i])
for k in range(i,n+1): a[j][k] = a[j][k] - a[i][k]*r
for i in range(n):
x=Fraction(a[i][i],1)
for j in range(len(a[i])):
a[i][j] /= x
return a
uaz=list(map(int,raw_input().split()))
enemy=[0]+[-x+y for x,y in zip(map(int,raw_input().split()),uaz)]
b0=[1]+[x-y for x,y in zip(map(int,raw_input().split()),uaz)]
b1=[1]+[x-y for x,y in zip(map(int,raw_input().split()),uaz)]
b2=[1]+[x-y for x,y in zip(map(int,raw_input().split()),uaz)]
sol=gauss(list(map(list,zip(b0,b1,b2,enemy,[1,0,0,0]))))
print('MISS' if sol and all(0<=e[-1]<=1 for e in sol) else 'HIT') |
s223291825 | p03471 | u308588791 | 2,000 | 262,144 | Wrong Answer | 2,108 | 3,064 | 574 | 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. | input_list = list(input().split())
bill_num = int(input_list[0])
total_price = int(input_list[1])
yen_10000 = 10000
yen_5000 = 5000
yen_1000 = 1000
bill_num_list = []
for i in range(bill_num):
for j in range(bill_num):
for k in range(bill_num):
if total_price == (i+1) * yen_10000 + (j+1) * yen_5000 + (k+1) * yen_1000:
bill_num_list = [i+1, j+1, k+1]
bill_num_list_str = list(map(str, bill_num_list))
break
if bill_num_list == []:
print('-1 -1 -1')
else:
print(' '.join(bill_num_list_str))
| s667786482 | Accepted | 1,556 | 3,064 | 627 | input_list = list(input().split())
bill_num = int(input_list[0])
total_price = int(input_list[1])
yen_10000 = 10000
yen_5000 = 5000
yen_1000 = 1000
bill_num_list = []
for i in range(bill_num + 1):
for j in range(bill_num - i + 1):
k = 0
k = bill_num - i - j
total = i * yen_10000 + j * yen_5000 + k * yen_1000
if (i + j + k == bill_num) and (k >= 0) and (total == total_price):
bill_num_list = [i, j, k]
bill_num_list_str = list(map(str, bill_num_list))
break
if bill_num_list == []:
print('-1 -1 -1')
else:
print(' '.join(bill_num_list_str))
|
s672715699 | p03385 | u474925961 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 140 | You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`. | import sys
if sys.platform =='ios':
sys.stdin=open('input_file.txt')
S=input()
if sorted(S)=="abc":
print("Yes")
else:
print("No") | s694683304 | Accepted | 18 | 2,940 | 149 | import sys
if sys.platform =='ios':
sys.stdin=open('input_file.txt')
S=input()
if sorted(S)==["a","b","c"]:
print("Yes")
else:
print("No")
|
s582413000 | p03672 | u729119068 | 2,000 | 262,144 | Wrong Answer | 27 | 9,104 | 134 | 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. | A= [i for i in input()]
while True:
A.pop()
A.pop()
B=len(A)//2
if A[:B]==A[B+1:]:
print(len(A))
break | s876998212 | Accepted | 29 | 9,100 | 125 | A= list(input())
while True:
A.pop()
A.pop()
B=len(A)//2
if A[:B]==A[B:]:
print(len(A))
break |
s958941927 | p03564 | u923659712 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 123 | Square1001 has seen an electric bulletin board displaying the integer 1. He can perform the following operations A and B to change this value: * Operation A: The displayed value is doubled. * Operation B: The displayed value increases by K. Square1001 needs to perform these operations N times in total. Find the minimum possible value displayed in the board after N operations. | n=int(input())
k=int(input())
ans=1
j=0
while ans<k or j<n:
ans*=2
j+=1
if j<n:
print(ans+k*(n-j))
else:
print(ans) | s071452609 | Accepted | 17 | 3,060 | 124 | n=int(input())
k=int(input())
ans=1
j=0
while ans<k and j<n:
ans*=2
j+=1
if j<n:
print(ans+k*(n-j))
else:
print(ans) |
s245659539 | p04029 | u127856129 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 50 | There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total? | a=int(input())
b=a
while b==0:
a+=a-1
b-=1
| s238917579 | Accepted | 17 | 2,940 | 63 | a=int(input())
c=0
while a>=0:
c+=a
a=a-1
print(int(c))
|
s742167322 | p04043 | u848647227 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 159 | Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order. | ar = input().split(" ")
f = 0
s = 0
for a in ar:
if a == "5":
f += 1
elif a == "7":
s += 1
if f == 5 and s == 1:
print("YES")
else:
print("NO") | s986110086 | Accepted | 16 | 2,940 | 160 | ar = input().split(" ")
f = 0
s = 0
for a in ar:
if a == "5":
f += 1
elif a == "7":
s += 1
if f == 2 and s == 1:
print("YES")
else:
print("NO")
|
s452262139 | p02607 | u668503853 | 2,000 | 1,048,576 | Wrong Answer | 35 | 9,096 | 121 | We have N squares assigned the numbers 1,2,3,\ldots,N. Each square has an integer written on it, and the integer written on Square i is a_i. How many squares i satisfy both of the following conditions? * The assigned number, i, is odd. * The written integer is odd. | N=int(input())
A=list(map(int,input().split()))
ans=0
for i in range(N):
if i%2==0 and A[i]%2==0:
ans+=1
print(ans) | s153287542 | Accepted | 27 | 9,088 | 122 | N=int(input())
A=list(map(int,input().split()))
ans=0
for i in range(N):
if i%2==0 and A[i]%2==1:
ans+=1
print(ans)
|
s103624836 | p03605 | u756988562 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 93 | It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N? | N = input()
# print(N.find("9"))
if N.find("9") != -1:
print("Yse")
else:
print("No") | s953068310 | Accepted | 17 | 2,940 | 72 | N = input()
if N.find("9") != -1:
print("Yes")
else:
print("No") |
s092050398 | p03080 | u754022296 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 78 | There are N people numbered 1 to N. Each person wears a red hat or a blue hat. You are given a string s representing the colors of the people. Person i wears a red hat if s_i is `R`, and a blue hat if s_i is `B`. Determine if there are more people wearing a red hat than people wearing a blue hat. | s = input()
if s.count("R") > s.count("B"):
print("Yes")
else:
print("No") | s656733610 | Accepted | 17 | 2,940 | 96 | n = int(input())
s = input()
if s.count("R") > s.count("B"):
print("Yes")
else:
print("No")
|
s476881730 | p03303 | u923270446 | 2,000 | 1,048,576 | Wrong Answer | 19 | 3,316 | 145 | You are given a string S consisting of lowercase English letters. We will write down this string, starting a new line after every w letters. Print the string obtained by concatenating the letters at the beginnings of these lines from top to bottom. | s = list(input())
w = int(input())
index = 0
l = []
while index <= len(s) - 1:
print(index)
l.append(s[index])
index += w
print(*l, sep="") | s950096218 | Accepted | 17 | 3,188 | 130 | s = list(input())
w = int(input())
index = 0
l = []
while index <= len(s) - 1:
l.append(s[index])
index += w
print(*l, sep="") |
s106434850 | p03565 | u673361376 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 378 | E869120 found a chest which is likely to contain treasure. However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters. He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`. One more thing he found is a sheet of paper with the following facts written on it: * Condition 1: The string S contains a string T as a contiguous substring. * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1. Print the string S. If such a string does not exist, print `UNRESTORABLE`. | S = input()
T = input()
lenS = len(S)
lenT = len(T)
for i in range(lenS - lenT, -1, -1):
flag = True
for ii in range(lenT):
if S[i+ii] == '?' or S[i+ii] == T[ii]:
continue
else:
flag = False
break
if flag == True:
print((S[:ii] + T + S[ii + lenT:]).replace('?', 'a'))
exit()
print('UNRESTORABLE')
| s105418727 | Accepted | 17 | 3,064 | 558 | def is_possible(start_i):
for i in range(lenT):
if S[start_i + i] != '?' and S[start_i + i] != T[i]:
return False
return True
if __name__ == '__main__':
S = input()
T = input()
lenS = len(S)
lenT = len(T)
tmp_i = None
for i in range(lenS):
if i + lenT > lenS:
break
else:
if is_possible(i):
tmp_i = i
if tmp_i is None:
print('UNRESTORABLE')
else:
print(S[:tmp_i].replace('?', 'a') + T + S[tmp_i + lenT:].replace('?', 'a'))
|
s606295094 | p02741 | u386944085 | 2,000 | 1,048,576 | Wrong Answer | 18 | 3,060 | 247 | Print the K-th element of the following sequence of length 32: 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51 | a = int(input())
if a%2==0 and (a!=25 or a!=27):
print(1)
elif a==32:
print(51)
elif a==24:
print(15)
elif a==16:
print(14)
elif a==28 or a==30:
print(4)
elif (a!=4 and a%4==0) or a==18 or a==27:
print(5)
else:
print(2) | s450369327 | Accepted | 17 | 3,064 | 277 | a = int(input())
if (a%2==1 and (a!=25 and a!=27 and a!=9 and a!=21)) or a==2:
print(1)
elif a==32:
print(51)
elif a==24:
print(15)
elif a==16:
print(14)
elif a==28 or a==30:
print(4)
elif (a!=4 and a%4==0) or a==18 or a==27:
print(5)
else:
print(2) |
s201628865 | p03564 | u957198490 | 2,000 | 262,144 | Wrong Answer | 18 | 3,064 | 98 | Square1001 has seen an electric bulletin board displaying the integer 1. He can perform the following operations A and B to change this value: * Operation A: The displayed value is doubled. * Operation B: The displayed value increases by K. Square1001 needs to perform these operations N times in total. Find the minimum possible value displayed in the board after N operations. | N = int(input())
K = int(input())
ans = 1
for i in range(N):
ans = min(ans+K,2*ans)
print(min) | s586971185 | Accepted | 17 | 2,940 | 98 | N = int(input())
K = int(input())
ans = 1
for i in range(N):
ans = min(ans+K,2*ans)
print(ans) |
s300253588 | p03352 | u384679440 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 145 | 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. | X = int(input())
ans = 0
if X == 1:
ans = 1
else:
for b in range(1, 32):
for p in range(2, 11):
ans = max(ans, pow(b, p))
print(ans) | s375264900 | Accepted | 17 | 3,060 | 168 | X = int(input())
ans = 1
if X > 1:
for b in range(1, 32):
for p in range(2, 11):
if ans <= X and pow(b, p) <= X:
ans = max(ans, pow(b, p))
print(ans) |
s862398009 | p02396 | u019678978 | 1,000 | 131,072 | Wrong Answer | 20 | 7,356 | 117 | In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem. | dataset = input().split("\n")
c = 1
for i in dataset :
print("Case " + str(c) + ":" + " " + str(i))
c = c + 1 | s769361923 | Accepted | 60 | 7,960 | 179 | import sys
c = 1
for i in sys.stdin.readlines() :
if i.strip() == "0" :
break
else :
print("Case " + str(c) + ":" + " " + str(i.strip()))
c = c + 1 |
s981878642 | p02972 | u665038048 | 2,000 | 1,048,576 | Wrong Answer | 217 | 13,568 | 225 | 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 = input()
r = [None] * (n + 1)
for i in range(n, 0, -1):
r[i] = (ord(a[2 * i - 2]) + sum(r[i * 2::i])) % 2
ans = []
for i, x in enumerate(r):
if x:
ans.append(i)
print(r.count(1))
print(ans) | s828291755 | Accepted | 235 | 12,028 | 189 | n = int(input())
a = input()
r = [None] * (n + 1)
for i in range(n, 0, -1):
r[i] = (ord(a[2 * i - 2]) + sum(r[i * 2::i])) % 2
print(r.count(1))
print(*(i for i, x in enumerate(r) if x)) |
s075612101 | p04031 | u298297089 | 2,000 | 262,144 | Wrong Answer | 22 | 3,060 | 173 | Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal **integers** by transforming some of them. He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2). Find the minimum total cost to achieve his objective. | N = int(input())
A = list(map(int, input().split()))
mn = 10**9
for i in range(N):
cost = 0
for j in range(N):
cost += (A[i]-A[j])**2
mn = min(cost, mn)
print(mn) | s836977922 | Accepted | 24 | 2,940 | 215 | n = int(input())
a = list(map(int, input().split()))
ans = float('INF')
for i in range(min(a), max(a)+1):
cost = 0
for aa in a:
cost += (i - aa)**2
if cost < ans :
ans = cost
print(ans)
|
s391730759 | p03474 | u379535139 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 256 | The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom. | A, B = tuple(int(num) for num in input().split())
S = input()
s_list = list(S)
for i, s in enumerate(s_list):
if s_list[A] != '-':
print('No')
break
if isinstance(s, int):
continue
else:
print('No')
break
else:
print('Yes') | s578597921 | Accepted | 18 | 3,060 | 278 | A, B = tuple(int(num) for num in input().split())
S = input()
s_list = list(S)
for i, s in enumerate(s_list):
if i == A:
if s_list[A] != '-':
print('No')
break
else:
try:
s = int(s)
except:
print('No')
break
else:
print('Yes') |
s158833321 | p02383 | u981139449 | 1,000 | 131,072 | Wrong Answer | 20 | 7,452 | 810 | Write a program to simulate rolling a dice, which can be constructed by the following net. As shown in the figures, each face is identified by a different label from 1 to 6. Write a program which reads integers assigned to each face identified by the label and a sequence of commands to roll the dice, and prints the integer on the top face. At the initial state, the dice is located as shown in the above figures. | class Dice:
def __init__(self):
self.top = 1
self.front = 2
self.left = 3
@property
def bottom(self):
return 7 - self.top
@property
def back(self):
return 7 - self.front
@property
def right(self):
return 7 - self.left
def move(self, direction):
if direction == 'N':
bottom = self.bottom
self.top = self.front
self.front = bottom
elif direction == 'W':
right = self.right
self.left = self.top
self.top = right
elif direction == 'E':
bottom = self.bottom
self.top = self.left
self.left = bottom
elif direction == 'S':
back = self.back
self.front = self.top
self.top = back
dice = Dice()
numbers = input().split()
for cmd in input():
dice.move(cmd)
print(numbers[dice.top - 1]) | s410676898 | Accepted | 20 | 7,544 | 857 | class Dice:
def __init__(self):
self.top = 1
self.front = 2
self.left = 4
@property
def bottom(self):
return 7 - self.top
@property
def back(self):
return 7 - self.front
@property
def right(self):
return 7 - self.left
def move(self, direction):
if direction == 'N':
bottom = self.bottom
self.top = self.front
self.front = bottom
elif direction == 'W':
right = self.right
self.left = self.top
self.top = right
elif direction == 'E':
bottom = self.bottom
self.top = self.left
self.left = bottom
elif direction == 'S':
back = self.back
self.front = self.top
self.top = back
def __repr__(self):
print(self.__dict__)
dice = Dice()
numbers = input().split()
for cmd in input():
dice.move(cmd)
print(numbers[dice.top - 1]) |
s370081771 | p03473 | u714587753 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 89 | How many hours do we have until New Year at M o'clock (24-hour notation) on 30th, December? | # Try AtCoder
# author: Leonardone @ NEETSDKASU
m = int(input())
ans = m + 24
print(ans) | s697034813 | Accepted | 18 | 2,940 | 95 | # Try AtCoder
# author: Leonardone @ NEETSDKASU
m = int(input())
ans = 24 - m + 24
print(ans) |
s539533811 | p03863 | u009460507 | 2,000 | 262,144 | Wrong Answer | 171 | 4,772 | 166 | There is a string s of length 3 or greater. No two neighboring characters in s are equal. Takahashi and Aoki will play a game against each other. The two players alternately performs the following operation, Takahashi going first: * Remove one of the characters in s, excluding both ends. However, a character cannot be removed if removal of the character would result in two neighboring equal characters in s. The player who becomes unable to perform the operation, loses the game. Determine which player will win when the two play optimally. | line = list(input())
count = 0
for i, s in enumerate(line[0:-2]):
if line[i] == line[i+2]:
count+=1
if count % 2 == 0:
print("Second")
else:
print("First") | s183371648 | Accepted | 26 | 4,084 | 195 | line = list(input())
if line[0] == line[-1]:
if len(line) % 2 == 0:
print("First")
else:
print("Second")
else:
if len(line) % 2 == 0:
print("Second")
else:
print("First") |
s296743179 | p02385 | u617472286 | 1,000 | 131,072 | Wrong Answer | 20 | 7,744 | 2,005 | Write a program which reads the two dices constructed in the same way as [Dice I](description.jsp?id=ITP1_11_A), and determines whether these two dices are identical. You can roll a dice in the same way as [Dice I](description.jsp?id=ITP1_11_A), and if all integers observed from the six directions are the same as that of another dice, these dices can be considered as identical. | class Dice(object):
def __init__(self, line):
self.top = 1
self.bottom = 6
self.south = 2
self.east = 3
self.west = 4
self.north = 5
self.convert = [int(s) for s in line.split()]
def move(self, direction):
if 'N' == direction:
self.top, self.north, self.bottom, self.south = self.south, self.top, self.north, self.bottom
elif 'S' == direction:
self.top, self.north, self.bottom, self.south = self.north, self.bottom, self.south, self.top
elif 'W' == direction:
self.top, self.east, self.bottom, self.west = self.east, self.bottom, self.west, self.top
elif 'E' == direction:
self.top, self.east, self.bottom, self.west = self.west, self.top, self.east, self.bottom
def rv(self, key):
index_list = {
'top' : self.top,
'bottom' : self.bottom,
'south' : self.south,
'east' : self.east,
'west' : self.west,
'notrh' : self.north
}
return self.convert[index_list[key] - 1]
def search(self, dice):
for direction in 'NNNNWNNNN':
self.move(direction)
if self.rv('south') == dice.rv('south'):
break
if self.rv('south') != dice.rv('south') or self.rv('notrh') != dice.rv('notrh'):
return 'No'
for direction in 'WWWW':
self.move(direction)
if self.rv('top') == dice.rv('top'):
break
if self.rv('top') != dice.rv('top'):
return 'No'
if self.rv('bottom') != dice.rv('bottom'):
return 'No'
if self.rv('east') != dice.rv('east'):
return 'No'
if self.rv('notrh') != dice.rv('notrh'):
return 'No'
return 'Yes'
dice = Dice('1 2 3 4 5 6')
dice2 = Dice('6 5 4 3 2 1')
print(dice.search(dice2)) | s904341754 | Accepted | 50 | 7,900 | 1,993 | class Dice(object):
def __init__(self, line):
self.top = 1
self.bottom = 6
self.south = 2
self.east = 3
self.west = 4
self.north = 5
self.convert = [int(s) for s in line.split()]
def move(self, direction):
if 'N' == direction:
self.top, self.north, self.bottom, self.south = self.south, self.top, self.north, self.bottom
elif 'S' == direction:
self.top, self.north, self.bottom, self.south = self.north, self.bottom, self.south, self.top
elif 'W' == direction:
self.top, self.east, self.bottom, self.west = self.east, self.bottom, self.west, self.top
elif 'E' == direction:
self.top, self.east, self.bottom, self.west = self.west, self.top, self.east, self.bottom
def rv(self, key):
index_list = {
'top' : self.top,
'bottom' : self.bottom,
'south' : self.south,
'east' : self.east,
'west' : self.west,
'notrh' : self.north
}
return self.convert[index_list[key] - 1]
def search(self, dice):
for direction in 'NNNNWNNNN':
self.move(direction)
if self.rv('south') == dice.rv('south'):
break
if self.rv('south') != dice.rv('south') or self.rv('notrh') != dice.rv('notrh'):
return 'No'
for direction in 'WWWW':
self.move(direction)
if self.rv('top') == dice.rv('top'):
break
if self.rv('top') != dice.rv('top'):
return 'No'
if self.rv('bottom') != dice.rv('bottom'):
return 'No'
if self.rv('east') != dice.rv('east'):
return 'No'
if self.rv('notrh') != dice.rv('notrh'):
return 'No'
return 'Yes'
dice = Dice(input())
dice2 = Dice(input())
print(dice.search(dice2)) |
s219043578 | p03474 | u699522269 | 2,000 | 262,144 | Wrong Answer | 31 | 9,132 | 104 | The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom. | a,b = map(int,input().split())
s = input()
print("Yes" if s[:a].isdigit() and s[a:].isdigit() else "No") | s689455340 | Accepted | 28 | 9,148 | 120 | a,b = map(int,input().split())
s = input()
print("Yes" if s[:a].isdigit() and s[a+1:].isdigit() and s[a]=="-" else "No") |
s292286873 | p03470 | u520499660 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 67 | An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have? | s = input().split()
s_set = set(s)
length = len(s)
print(length) | s249922377 | Accepted | 17 | 2,940 | 159 | N = int(input())
s = [input() for i in range(N)]
s_set = set(s)
l = len(s_set)
print(l) |
s818555372 | p02256 | u027874809 | 1,000 | 131,072 | Wrong Answer | 20 | 5,592 | 231 | Write a program which finds the greatest common divisor of two natural numbers _a_ and _b_ | alllist = list(map(int, input().split()))
alllist = sorted(alllist, reverse=True)
def maxsize(a, b):
if a % b == 0:
return a // b
return maxsize(b, a % b)
result = maxsize(alllist[0], alllist[1])
print(result)
| s800806644 | Accepted | 20 | 5,600 | 202 | def gcd(a, b):
if b == 0: return a
return gcd(b, a % b)
alllist = list(map(int, input().split()))
alllist = sorted(alllist, reverse=True)
result = gcd(alllist[0], alllist[1])
print(result)
|
s579857577 | p03494 | u504256702 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 101 | 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(input().split())
print(list(map(lambda x: int(x) % 2 == 0, A)).count(True)) | s602369998 | Accepted | 19 | 3,060 | 159 | N = int(input())
A = list(map(int,input().split()))
for i in range(1, max(A)):
if not all(list(map(lambda x: x % 2**i == 0, A))):
print(i - 1)
break |
s653483552 | p02612 | u349301655 | 2,000 | 1,048,576 | Wrong Answer | 29 | 9,140 | 41 | 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())
A = 1000 - N
print(A)
| s065321056 | Accepted | 24 | 9,064 | 104 | N = int(input())
if N % 1000 == 0:
a = N
else:
a = (N // 1000 + 1) * 1000
A = a - N
print(A)
|
s791970557 | p03399 | u623601489 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 100 | You planned a trip using trains and buses. The train fare will be A yen (the currency of Japan) if you buy ordinary tickets along the way, and B yen if you buy an unlimited ticket. Similarly, the bus fare will be C yen if you buy ordinary tickets along the way, and D yen if you buy an unlimited ticket. Find the minimum total fare when the optimal choices are made for trains and buses. | a1=int(input());a2=int(input());b1=int(input());b2=int(input())
print([a1,a2][a1<a2]+[b1,b2][b1<b2]) | s504693314 | Accepted | 17 | 2,940 | 100 | a1=int(input());a2=int(input());b1=int(input());b2=int(input())
print([a1,a2][a1>a2]+[b1,b2][b1>b2]) |
s643469008 | p03574 | u266874640 | 2,000 | 262,144 | Wrong Answer | 29 | 3,828 | 657 | 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())
dx = [1,1,1,0,-1,-1,-1,0]
dy = [0,1,-1,1,0,1,-1,-1]
ans = [[0] * (w + 2) for _ in range(h + 2)]
s = [input() for _ in range(h)]
for i in range(h):
s[i] = '.' + s[i] + '.'
s = ['.' * (w + 2)] + s + ['.' * (w + 2)]
for i in range(1, h+1):
for j in range(1, w+1):
if s[i][j] == '.':
for A in range(8):
if(s[i + dy[A]][j + dx[A]]=='#'):
ans[i][j] += 1
print(ans[i][j])
for i in range(1, h+1):
for j in range(1, w+1):
if s[i][j] == '.':
print(ans[i][j],end='')
else:
print('#',end='')
print() | s584923169 | Accepted | 27 | 3,444 | 633 | h, w = map(int,input().split())
dx = [1,1,1,0,-1,-1,-1,0]
dy = [0,1,-1,1,0,1,-1,-1]
ans = [[0] * (w + 2) for _ in range(h + 2)]
s = [input() for _ in range(h)]
for i in range(h):
s[i] = '.' + s[i] + '.'
s = ['.' * (w + 2)] + s + ['.' * (w + 2)]
for i in range(1, h+1):
for j in range(1, w+1):
if s[i][j] == '.':
for A in range(8):
if(s[i + dy[A]][j + dx[A]]=='#'):
ans[i][j] += 1
for i in range(1, h+1):
for j in range(1, w+1):
if s[i][j] == '.':
print(ans[i][j],end='')
else:
print('#',end='')
print() |
s917530177 | p03997 | u131666536 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 71 | 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(int((a+b)/h)) | s068566483 | Accepted | 16 | 2,940 | 73 | a = int(input())
b = int(input())
h = int(input())
print(int((a+b)*h/2)) |
s757391007 | p03827 | u072717685 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 158 | You have an integer variable x. Initially, x=0. Some person gave you a string S of length N, and using the string you performed the following operation N times. In the i-th operation, you incremented the value of x by 1 if S_i=`I`, and decremented the value of x by 1 if S_i=`D`. Find the maximum value taken by x during the operations (including before the first operation, and after the last operation). | n = int(input())
s = ""
current_num = 0
max_num = 0
for c in s:
current_num += 1 if c == 'I' else -1
max_num = max(max_num, current_num)
print(max_num)
| s343031759 | Accepted | 17 | 3,060 | 160 | n = int(input())
s = input()
current_num = 0
max_num = 0
for c in s:
current_num += 1 if c == 'I' else -1
max_num = max(max_num, current_num)
print(max_num) |
s338421015 | p04029 | u759412327 | 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) | s517701840 | Accepted | 27 | 9,076 | 34 | N = int(input())
print(N*(N+1)//2) |
s829069076 | p03545 | u311379832 | 2,000 | 262,144 | Wrong Answer | 24 | 3,444 | 885 | Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted. | import sys
from collections import defaultdict
a = input()
alst = [0, 1]
blst = [0, 1]
clst = [0, 1]
ans = ''
for i in alst:
for j in blst:
for k in clst:
tmp = 0
ans = ''
if i == 0:
tmp = int(a[0]) + int(a[1])
ans = a[0] + '+' + a[1]
else:
tmp = int(a[0]) - int(a[1])
ans = a[0] + '-' + a[1]
if j == 0:
tmp = tmp + int(a[2])
ans = ans + '+' + a[2]
else:
tmp = tmp - int(a[2])
ans = ans + '-' + a[2]
if k == 0:
tmp = tmp + int(a[3])
ans = ans + '+' + a[3]
else:
tmp = tmp - int(a[3])
ans = ans + '-' + a[3]
if tmp == 7:
print(ans)
sys.exit() | s609445727 | Accepted | 25 | 3,444 | 892 | import sys
from collections import defaultdict
a = input()
alst = [0, 1]
blst = [0, 1]
clst = [0, 1]
ans = ''
for i in alst:
for j in blst:
for k in clst:
tmp = 0
ans = ''
if i == 0:
tmp = int(a[0]) + int(a[1])
ans = a[0] + '+' + a[1]
else:
tmp = int(a[0]) - int(a[1])
ans = a[0] + '-' + a[1]
if j == 0:
tmp = tmp + int(a[2])
ans = ans + '+' + a[2]
else:
tmp = tmp - int(a[2])
ans = ans + '-' + a[2]
if k == 0:
tmp = tmp + int(a[3])
ans = ans + '+' + a[3]
else:
tmp = tmp - int(a[3])
ans = ans + '-' + a[3]
if tmp == 7:
print(ans + '=7')
sys.exit() |
s827462640 | p03474 | u598684283 | 2,000 | 262,144 | Wrong Answer | 29 | 9,176 | 192 | The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom. | a,b = map(int,input().split())
s = input()
if s[a] == "-":
s = s.replace("-","")
print(s)
if s.isdecimal():
print("Yes")
else:
print("No")
else:
print("No") | s074705187 | Accepted | 31 | 9,100 | 199 | a,b = map(int,input().split())
s = input()
if len(s) == a + b + 1 and s[a] == "-":
s = s.replace("-","")
if s.isdecimal() and len(s) == a + b:
print("Yes")
exit(0)
print("No") |
s508842651 | p02694 | u918817732 | 2,000 | 1,048,576 | Wrong Answer | 27 | 9,044 | 135 | Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time? | import math
X=int(input())
origin=100
count=0
while(origin<X):
origin+=int(origin*0.01)
print(origin)
count+=1
print(count) | s337246194 | Accepted | 22 | 8,976 | 117 | import math
X=int(input())
origin=100
count=0
while(origin<X):
origin+=int(origin*0.01)
count+=1
print(count) |
s574129129 | p02409 | u731896389 | 1,000 | 131,072 | Wrong Answer | 20 | 7,752 | 353 | You manage 4 buildings, each of which has 3 floors, each of which consists of 10 rooms. Write a program which reads a sequence of tenant/leaver notices, and reports the number of tenants for each room. For each notice, you are given four integers b, f, r and v which represent that v persons entered to room r of fth floor at building b. If v is negative, it means that −v persons left. Assume that initially no person lives in the building. | data = [[[0 for r in range(10)] for f in range(3)] for b in range(4)]
n = int(input())
for i in range(n):
b,f,r,v = [int(i) for i in input().split()]
data[b-1][f-1][r-1] += v
for b in range(4):
for f in range(3):
for r in range(10):
print(data[b][f][r],end="")
print()
if b < 3:
print("#"*20) | s796106610 | Accepted | 30 | 7,700 | 348 | data = [[[0 for r in range(10)] for f in range(3)] for b in range(4)]
n = int(input())
for i in range(n):
b,f,r,v = [int(i) for i in input().split()]
data[b-1][f-1][r-1] += v
for b in range(4):
for f in range(3):
for r in range(10):
print("",data[b][f][r],end="")
print()
if b < 3:
print("#"*20) |
s921440469 | p03997 | u071061942 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 86 | 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())
ans = int(a * b * h / 2)
print(ans) | s408827162 | Accepted | 17 | 2,940 | 89 | a = int(input())
b = int(input())
h = int(input())
ans = int((a + b) * h / 2)
print(ans)
|
s733457349 | p03193 | u185802209 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 85 | There are N rectangular plate materials made of special metal called AtCoder Alloy. The dimensions of the i-th material are A_i \times B_i (A_i vertically and B_i horizontally). Takahashi wants a rectangular plate made of AtCoder Alloy whose dimensions are exactly H \times W. He is trying to obtain such a plate by choosing one of the N materials and cutting it if necessary. When cutting a material, the cuts must be parallel to one of the sides of the material. Also, the materials have fixed directions and cannot be rotated. For example, a 5 \times 3 material cannot be used as a 3 \times 5 plate. Out of the N materials, how many can produce an H \times W plate if properly cut? | n = input()
c = 0
for i in range(len(n)):
if n[i] == '2':
c += 1
print(c) | s105019531 | Accepted | 22 | 3,316 | 145 | n, h, w = map(int,input().split())
c = 0
for i in range(n):
a, b = map(int,input().split())
if a >= h and b >= w:
c += 1
print(c) |
s354274737 | p02612 | u546236742 | 2,000 | 1,048,576 | Wrong Answer | 30 | 8,984 | 72 | 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. | a = int(input())
if a % 1000 == 0:
print("0")
else:
print(a % 1000) | s304848367 | Accepted | 27 | 9,016 | 79 | a = int(input())
if a % 1000 == 0:
print("0")
else:
print(1000 - a % 1000) |
s126044874 | p03227 | u778814286 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 170 | 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. | ###template###
import sys
input = sys.stdin.readline
def mi(): return map(int, input().split())
###template###
s = input()
if len(s)==2: print(s)
else: print(s[::-1])
| s529575779 | Accepted | 17 | 2,940 | 179 | ###template###
import sys
input = sys.stdin.readline
def mi(): return map(int, input().split())
###template###
s = input().rstrip()
if len(s)==2: print(s)
else: print(s[::-1])
|
s884952301 | p03574 | u140251125 | 2,000 | 262,144 | Wrong Answer | 24 | 3,572 | 1,065 | 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. | # input
H, W = map(int, input().split())
S = [list(input()) for i in range(H)]
for i in range(H):
S[i] = ['.'] + S[i] + ['.']
S = [['.' for i in range(W + 2)]] + S + [['.' for i in range(W + 2)]]
S_dummy = S
for i in range(1, H + 1):
for j in range(1, W + 1):
if S[i][j] == '.':
tmp = 0
if S[i - 1][j - 1] == '#':
tmp += 1
if S[i - 1][j] == '#':
tmp += 1
if S[i - 1][j + 1] == '#':
tmp += 1
if S[i][j - 1] == '#':
tmp += 1
if S[i][j + 1] == '#':
tmp += 1
tmp += 1
if S[i + 1][j - 1] == '#':
tmp += 1
if S[i + 1][j] == '#':
tmp += 1
if S[i + 1][j + 1] == '#':
tmp += 1
S_dummy[i][j] = tmp
for i in range(1, H + 1):
for j in range(1, W + 1):
print(S_dummy[i][j], end = "")
print('')
| s770262164 | Accepted | 24 | 3,572 | 1,039 | # input
H, W = map(int, input().split())
S = [list(input()) for i in range(H)]
for i in range(H):
S[i] = ['.'] + S[i] + ['.']
S = [['.' for i in range(W + 2)]] + S + [['.' for i in range(W + 2)]]
S_dummy = S
for i in range(1, H + 1):
for j in range(1, W + 1):
if S[i][j] == '.':
tmp = 0
if S[i - 1][j - 1] == '#':
tmp += 1
if S[i - 1][j] == '#':
tmp += 1
if S[i - 1][j + 1] == '#':
tmp += 1
if S[i][j - 1] == '#':
tmp += 1
if S[i][j + 1] == '#':
tmp += 1
if S[i + 1][j - 1] == '#':
tmp += 1
if S[i + 1][j] == '#':
tmp += 1
if S[i + 1][j + 1] == '#':
tmp += 1
S_dummy[i][j] = tmp
for i in range(1, H + 1):
for j in range(1, W + 1):
print(S_dummy[i][j], end = "")
print('') |
s490491800 | p03671 | u744898490 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 471 | Snuke is buying a bicycle. The bicycle of his choice does not come with a bell, so he has to buy one separately. He has very high awareness of safety, and decides to buy two bells, one for each hand. The store sells three kinds of bells for the price of a, b and c yen (the currency of Japan), respectively. Find the minimum total price of two different bells. | abc = input().split(' ')
abc =[ int(x) for x in abc]
s= 0
min_ = min(abc)
if abc[0] == abc[1] == abc[2] :
print(abc[0] + abc[1])
elif (abc[0] == min_ and abc[1] == min_ or abc[0] == min_ and abc[2] == min_ or
abc[2] == min_ and abc[1] == min_):
for i in abc:
if i != min_:
s += i
print(s + min_)
else:
for i in abc:
if i != min(abc):
s += i
print(s)
| s383875278 | Accepted | 17 | 3,060 | 306 | abc = input().split(' ')
max_ = max(abc)
abc =[ int(x) for x in abc]
s= 0
min_ = min(abc)
if abc[0] == abc[1] == abc[2] :
print(abc[0] + abc[1])
elif abc.count(max(abc)) == 2:
print(max(abc) + min(abc))
else:
for i in abc:
if i != max(abc):
s += i
print(s)
|
s117999702 | p03760 | u580362735 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 121 | Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions. You are given two strings O and E. O contains the characters at the odd- numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password. | O = list(input())
E = list(input())
ans = ['0']*(len(O)+len(E))
print(ans)
ans[::2] = O
ans[1::2] = E
print(''.join(ans)) | s123506731 | Accepted | 17 | 3,060 | 110 | O = list(input())
E = list(input())
ans = ['0']*(len(O)+len(E))
ans[::2] = O
ans[1::2] = E
print(''.join(ans)) |
s203442992 | p03543 | u003928116 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 46 | 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**? | print("Yes" if int(input())%1111==0 else "No") | s409561440 | Accepted | 17 | 2,940 | 145 | n=input()
list=[]
list.append(n[0:3])
list.append(n[1:4])
if len(set(list[0]))==1 or len(set(list[1]))==1:
print("Yes")
else:
print("No") |
s070111792 | p03448 | u368931345 | 2,000 | 262,144 | Wrong Answer | 53 | 3,064 | 492 | 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("500円玉の枚数:"))
B = int(input("100円玉の枚数:"))
C = int(input("50円玉の枚数:"))
X = int(input("金額(50の倍数):"))
if ((A+B+C<1) or (A<0 or A>50) or (B<0 or B>50) or (C<0 or C>50) or (X<50 or X > 20000 or X%50!=0)):
print("条件を満たしてください")
count = 0
for i in range(A+1):
for j in range(B+1):
for k in range(C+1):
Y = i*500 + j*100 + k*50
if (X==Y):
count = count+1
print (count) | s705668734 | Accepted | 53 | 3,064 | 408 | A = int(input())
B = int(input())
C = int(input())
X = int(input())
if ((A+B+C<1) or (A<0 or A>50) or (B<0 or B>50) or (C<0 or C>50) or (X<50 or X > 20000 or X%50!=0)):
print("条件を満たしてください")
count = 0
for i in range(A+1):
for j in range(B+1):
for k in range(C+1):
Y = i*500 + j*100 + k*50
if (X==Y):
count = count+1
print (count) |
s507814861 | p02606 | u084411645 | 2,000 | 1,048,576 | Wrong Answer | 28 | 9,144 | 53 | How many multiples of d are there among the integers between L and R (inclusive)? | print(sum([int(i)%2 for i in input().split()[::2]]))
| s620569202 | Accepted | 30 | 9,164 | 72 | l, r, d = [int(i) for i in input().split(" ")]
l -= 1
print(r//d - l//d) |
s926041676 | p03387 | u820560680 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 194 | 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. | X = sorted(list(map(int, input().split())))
if X[0] % 2 == X[1] % 2 == X[2] % 2:
print((X[2] - X[0]) // 2 + (X[2] - X[1]) // 2)
else:
print((X[2] - X[0]) // 2 + (X[2] - X[1]) // 2 + 2)
| s614648322 | Accepted | 17 | 3,064 | 276 | X = sorted(list(map(int, input().split())))
if X[0] % 2 == X[1] % 2 == X[2] % 2:
print((X[2] - X[0]) // 2 + (X[2] - X[1]) // 2)
elif X[0] % 2 == X[1] % 2:
print((X[2] - X[0]) // 2 + (X[2] - X[1]) // 2 + 1)
else:
print((X[2] - X[0]) // 2 + (X[2] - X[1]) // 2 + 2)
|
s692943060 | p03141 | u153094838 | 2,000 | 1,048,576 | Wrong Answer | 555 | 21,604 | 535 | There are N dishes of cuisine placed in front of Takahashi and Aoki. For convenience, we call these dishes Dish 1, Dish 2, ..., Dish N. When Takahashi eats Dish i, he earns A_i points of _happiness_ ; when Aoki eats Dish i, she earns B_i points of happiness. Starting from Takahashi, they alternately choose one dish and eat it, until there is no more dish to eat. Here, both of them choose dishes so that the following value is maximized: "the sum of the happiness he/she will earn in the end" minus "the sum of the happiness the other person will earn in the end". Find the value: "the sum of the happiness Takahashi earns in the end" minus "the sum of the happiness Aoki earns in the end". | if __name__ == '__main__':
n = int(input())
c = []
d = []
e = []
for i in range(0, n):
a, b = map(int, input().split())
c.append(a)
d.append(b)
e.append(abs(a - b))
li1, li2 = sorted(e), sorted(range(len(e)), key=lambda k: e[k])
turn = 0
happy = 0
for i in range(n - 1, -1, -1):
if (turn == 0):
happy += c[li2[i]]
else:
happy -= d[li2[i]]
print(happy)
turn = (turn + 1) % 2
print(happy)
| s287668855 | Accepted | 623 | 21,528 | 966 | n = int(input())
c = []
d = []
e = []
for i in range(0, n):
a, b = map(int, input().split())
c.append(a)
d.append(b)
e.append(a + b)
li1, li2 = sorted(e), sorted(range(len(e)), key=lambda k: e[k])
turn = 0
happy = 0
for i in range(n - 1, -1, -1):
now = li1[i]
if (turn == 0):
max = 0
maxKey = 0
for j in range(i, -1, -1):
if (now == li1[j] and max < c[li2[j]]):
max = c[li2[j]]
maxKey = j
else:
li2[i], li2[maxKey] = li2[maxKey], li2[i]
break
happy += c[li2[i]]
else:
max = 0
maxKey = 0
for j in range(i, -1, -1):
if (now == li1[j] and max < d[li2[j]]):
max = d[li2[j]]
maxKey = j
else:
li2[i], li2[maxKey] = li2[maxKey], li2[i]
break
happy -= d[li2[i]]
turn = (turn + 1) % 2
print(happy) |
s217841835 | p03229 | u845333844 | 2,000 | 1,048,576 | Wrong Answer | 247 | 7,516 | 615 | You are given N integers; the i-th of them is A_i. Find the maximum possible sum of the absolute differences between the adjacent elements after arranging these integers in a row in any order you like. | n=int(input())
l=[]
for i in range(n):
l.append(int(input()))
l.sort()
if n%2==0:
m=n//2
sum1=0
sum2=0
for j in range(m):
sum1 += l[j]
sum2 += l[n-1-j]
max=2*sum2-2*sum1-l[m]+l[m-1]
print(max)
else:
if n==3:
max1=l[2]+l[1]-2*l[0]
max2=2*l[2]-l[1]-l[0]
if (max1>max2):
print(max1)
else:
print(max2)
else:
m=(n-1)//2
sum1=0
sum2=0
for j in range(m+1):
sum1 += l[j]
sum2 += l[n-1-j]
max1=2*sum2-l[m]-l[m+1]-2*sum1+2*l[m]
max2=2*sum2-2*l[m]-2*sum1+l[m-1]+l[m-2]
if (max1>max2):
print(max1)
else:
print(max2) | s582328272 | Accepted | 240 | 7,512 | 613 | n=int(input())
l=[]
for i in range(n):
l.append(int(input()))
l.sort()
if n%2==0:
m=n//2
sum1=0
sum2=0
for j in range(m):
sum1 += l[j]
sum2 += l[n-1-j]
max=2*sum2-2*sum1-l[m]+l[m-1]
print(max)
else:
if n==3:
max1=l[2]+l[1]-2*l[0]
max2=2*l[2]-l[1]-l[0]
if (max1>max2):
print(max1)
else:
print(max2)
else:
m=(n-1)//2
sum1=0
sum2=0
for j in range(m+1):
sum1 += l[j]
sum2 += l[n-1-j]
max1=2*sum2-l[m]-l[m+1]-2*sum1+2*l[m]
max2=2*sum2-2*l[m]-2*sum1+l[m]+l[m-1]
if (max1>max2):
print(max1)
else:
print(max2) |
s293193774 | p03813 | u596536048 | 2,000 | 262,144 | Wrong Answer | 27 | 8,976 | 161 | 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. | print('レートを入力してください')
rate = int(input())
print('参加するコンテストは')
if rate < 1200:
print('ABC')
else:
print('ARC') | s520749761 | Accepted | 28 | 9,060 | 75 | rate = int(input())
if rate < 1200:
print('ABC')
else:
print('ARC') |
s137852854 | p03196 | u299172013 | 2,000 | 1,048,576 | Wrong Answer | 95 | 3,572 | 660 | There are N integers a_1, a_2, ..., a_N not less than 1. The values of a_1, a_2, ..., a_N are not known, but it is known that a_1 \times a_2 \times ... \times a_N = P. Find the maximum possible greatest common divisor of a_1, a_2, ..., a_N. | import functools
import itertools
import math
import sys
import collections
import enum
def primefac(x):
fac = []
for i in range(2, int(math.sqrt(x)) + 1 + 1):
if x % i != 0:
continue
ex = 0
while x % i == 0:
ex += 1
x = x // i
fac.append((i, ex))
if x != 1:
fac.append((x, 1))
return fac
if __name__ == "__main__":
N, P = [int(x) for x in input().split(" ")]
fac = primefac(P)
def f(x, y):
return x[0] * y[0]
try:
print(functools.reduce(f, [(val, ex // N) for val, ex in fac if ex >= N]))
except TypeError:
print(0)
| s323009319 | Accepted | 95 | 3,684 | 636 | import functools
import itertools
import math
import sys
import collections
import enum
def primefac(x):
fac = []
for i in range(2, int(math.sqrt(x)) + 1 + 1):
if x % i != 0:
continue
ex = 0
while x % i == 0:
ex += 1
x = x // i
fac.append((i, ex))
if x != 1:
fac.append((x, 1))
return fac
if __name__ == "__main__":
N, P = [int(x) for x in input().split(" ")]
fac = primefac(P)
def f(x, y):
return x[0] * y[0]
ans = 1
for val, ex in fac:
if ex >= N:
ans *= val ** (ex // N)
print(ans)
|
s104301344 | p03386 | u119982147 | 2,000 | 262,144 | Wrong Answer | 2,104 | 3,060 | 106 | 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())
for i in range(a, b+1):
if i <= a+k or b-k < i:
print(i)
| s154860690 | Accepted | 17 | 3,060 | 134 | a, b, k = map(int, input().split())
for i in range(a, min(b+1, a+k)):
print(i)
for i in range(max(b-k+1, a+k), b+1):
print(i) |
s317534148 | p03469 | u771532493 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 43 | On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it. | S=input()
S.replace('2017','2018')
print(S) | s923110640 | Accepted | 17 | 2,940 | 34 | S=input()
s=S[4:]
print('2018'+s)
|
s042623209 | p03471 | u044638697 | 2,000 | 262,144 | Time Limit Exceeded | 2,104 | 3,060 | 372 | 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 = (int(i) for i in input().split())
n_10000,n_5000 = 0,0
n_1000 = N-n_10000-n_5000
x = True
while x:
while x:
if n_1000 < 0:
print(-1,-1,-1)
x = False
elif n_10000 +n_5000 +n_1000 == N and 10000*n_10000+5000*n_5000+1000*n_1000 == Y:
print(n_10000,n_5000,n_1000)
n_5000 += 1
n_10000 += 1
| s512747921 | Accepted | 798 | 3,060 | 268 | N,Y = (int(i) for i in input().split())
a,b,c = -1,-1,-1
for n_10000 in range(N+1):
for n_5000 in range(N-n_10000+1):
n_1000 = N-n_10000-n_5000
if 10000*n_10000+5000*n_5000 +1000*n_1000 == Y:
a,b,c = n_10000,n_5000,n_1000
print(a,b,c)
|
s370324292 | p03828 | u254871849 | 2,000 | 262,144 | Wrong Answer | 34 | 3,064 | 867 | You are given an integer N. Find the number of the positive divisors of N!, modulo 10^9+7. | import math
mod = int(1e9+7)
N = int(input())
n = math.factorial(N)
count_of_prime_factors = {}
while True:
if n > 1:
count = 0
while n % 2 == 0:
n //= 2
count += 1
if count != 0:
count_of_prime_factors.update({2: count})
if n == 1:
break
for i in range(3, N+1, 2):
count = 0
while n % i == 0:
n //= i
count += 1
if count != 0:
count_of_prime_factors.update({i: count})
if n == 1:
break
else:
count_of_prime_factors.update({n: 1})
break
else:
print(1)
exit()
ans = 1
for prime_factor, count in count_of_prime_factors.items():
ans *= count + 1
print(ans % mod)
print(count_of_prime_factors)
| s588928019 | Accepted | 438 | 95,816 | 897 | import sys
import numpy as np
def sieve_of_eratosthenes(n=10 ** 7):
sieve = np.ones(n + 1); sieve[:2] = 0
for i in range(2, int(np.sqrt(n)) + 1):
if sieve[i]: sieve[i*2::i] = 0
return sieve, np.flatnonzero(sieve)
is_prime, prime_numbers = sieve_of_eratosthenes()
def prime_factorize(n):
res = dict()
if n < 2: return res
border = int(n ** 0.5)
for p in prime_numbers:
if p > border: break
while n % p == 0: res[p] = res.get(p, 0) + 1; n //= p
if n == 1: return res
res[n] = 1
return res
def prime_factorize_factorial(n):
res = dict()
for i in range(2, n + 1):
for p, c in prime_factorize(i).items(): res[p] = res.get(p, 0) + c
return res
MOD = 10 ** 9 + 7
n = int(sys.stdin.readline().rstrip())
def main():
res = 1
for c in prime_factorize_factorial(n).values():
res *= c + 1; res %= MOD
print(res)
if __name__ == '__main__':
main() |
s277798673 | p03370 | u780269042 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 109 | Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts. Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition: * For each of the N kinds of doughnuts, make at least one doughnut of that kind. At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition. | a,b = map(int, input().split())
price = [int(input()) for i in range(a)]
print(sum(price)+ min(price)*(a-b)) | s990861932 | Accepted | 17 | 2,940 | 135 | a,b = map(int, input().split())
price = [int(input()) for i in range(a)]
rema = b - sum(price)
print(len(price)+ int(rema/min(price))) |
s411320398 | p03474 | u790812284 | 2,000 | 262,144 | Wrong Answer | 19 | 3,188 | 190 | 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. | import re
a,b=map(int,input().split())
s=input()
if re.search(r"^([0-9])*$",s[:a]) and s[a]=="-" and re.search(r"^([0-9])$",s[a+1:]) and len(s)==a+b+1:
print("Yes")
else:
print("No") | s246625522 | Accepted | 20 | 3,188 | 191 | import re
a,b=map(int,input().split())
s=input()
if re.search(r"^([0-9])*$",s[:a]) and s[a]=="-" and re.search(r"^([0-9])*$",s[a+1:]) and len(s)==a+b+1:
print("Yes")
else:
print("No") |
s164395746 | p03992 | u284563808 | 2,000 | 262,144 | Wrong Answer | 23 | 3,064 | 32 | This contest is `CODE FESTIVAL`. However, Mr. Takahashi always writes it `CODEFESTIVAL`, omitting the single space between `CODE` and `FESTIVAL`. So he has decided to make a program that puts the single space he omitted. You are given a string s with 12 letters. Output the string putting a single space between the first 4 letters and last 8 letters in the string s. | s=input()
print(s[:4]+'_'+s[4:]) | s743541937 | Accepted | 22 | 3,064 | 32 | s=input()
print(s[:4]+' '+s[4:]) |
s664005555 | p02646 | u095094246 | 2,000 | 1,048,576 | Wrong Answer | 20 | 9,188 | 218 | Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally. | a,v=map(int,input().split())
b,w=map(int,input().split())
t=int(input())
if a > b:
if a - v*t <= b - w*t:
print('Yes')
else:
print('No')
else:
if a + v*t >= b+w*t:
print('Yes')
else:
print('No') | s960063076 | Accepted | 25 | 9,180 | 218 | a,v=map(int,input().split())
b,w=map(int,input().split())
t=int(input())
if a > b:
if a - v*t <= b - w*t:
print('YES')
else:
print('NO')
else:
if a + v*t >= b+w*t:
print('YES')
else:
print('NO') |
s388256206 | p03997 | u417014669 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 62 | You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid. | a=int(input())
b=int(input())
h=int(input())
print((a+b)*h/2) | s313057474 | Accepted | 17 | 2,940 | 68 | a=int(input())
b=int(input())
h=int(input())
print(int((a+b)*h/2)) |
s154788525 | p02614 | u886297662 | 1,000 | 1,048,576 | Wrong Answer | 64 | 9,516 | 390 | We have a grid of H rows and W columns of squares. The color of the square at the i-th row from the top and the j-th column from the left (1 \leq i \leq H, 1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white if c_{i,j} is `.`, and black if c_{i,j} is `#`. Consider doing the following operation: * Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns. You are given a positive integer K. How many choices of rows and columns result in exactly K black squares remaining after the operation? Here, we consider two choices different when there is a row or column chosen in only one of those choices. | import itertools
import copy
H, W, K = map(int, input().split())
C = [list(input()) for i in range(H)]
clist = list(itertools.product([0, 1], repeat=H+W))
ans = 0
for q in clist:
black = 0
h = q[:H]
w = q[H:]
for i in range(H):
for j in range(W):
if h[i] == w[j] == 0 and C[i][j] == '#':
black += 1
if black == K:
ans += 1
print(q)
print(ans)
| s629532714 | Accepted | 61 | 9,436 | 391 | import itertools
import copy
H, W, K = map(int, input().split())
C = [list(input()) for i in range(H)]
clist = list(itertools.product([0, 1], repeat=H+W))
ans = 0
for q in clist:
black = 0
h = q[:H]
w = q[H:]
for i in range(H):
for j in range(W):
if h[i] == w[j] == 0 and C[i][j] == '#':
black += 1
if black == K:
ans += 1
#print(q)
print(ans)
|
s100830734 | p02612 | u131453093 | 2,000 | 1,048,576 | Wrong Answer | 31 | 9,136 | 45 | 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())
ans = N % 1000
print(ans)
| s987012828 | Accepted | 31 | 9,160 | 142 | N = int(input())
for i in range(1,11):
money = i * 1000
if money < N:
continue
else:
print(money-N)
break |
s739056628 | p03214 | u223646582 | 2,525 | 1,048,576 | Wrong Answer | 17 | 3,064 | 217 | Niwango-kun is an employee of Dwango Co., Ltd. One day, he is asked to generate a thumbnail from a video a user submitted. To generate a thumbnail, he needs to select a frame of the video according to the following procedure: * Get an integer N and N integers a_0, a_1, ..., a_{N-1} as inputs. N denotes the number of the frames of the video, and each a_i denotes the representation of the i-th frame of the video. * Select t-th frame whose representation a_t is nearest to the average of all frame representations. * If there are multiple such frames, select the frame with the smallest index. Find the index t of the frame he should select to generate a thumbnail. | N=int(input())
a=[int(i) for i in input().split()]
ave=(sum(a)/N)
ave_a=[abs(i-ave) for i in a]
print(ave_a)
ans=101
thum=101
for i in range(N):
if ave_a[i]<thum:
thum=ave_a[i]
ans=i
print(ans) | s776883576 | Accepted | 17 | 3,064 | 204 | N=int(input())
a=[int(i) for i in input().split()]
ave=(sum(a)/N)
ave_a=[abs(i-ave) for i in a]
ans=101
thum=101
for i in range(N):
if ave_a[i]<thum:
thum=ave_a[i]
ans=i
print(ans) |
s291100256 | p02418 | u083560765 | 1,000 | 131,072 | Wrong Answer | 20 | 5,544 | 47 | Write a program which finds a pattern $p$ in a ring shaped text $s$. | r=input()
a=input()
if a in r+r:
print('yes')
| s156498370 | Accepted | 20 | 5,548 | 66 | r=input()
a=input()
if a in r+r:
print('Yes')
else:
print('No')
|
s292796852 | p03698 | u771532493 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 76 | You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different. | s=input()
if list(s)==list(set(list(s))):
print('yes')
else:
print('no') | s208295104 | Accepted | 18 | 2,940 | 74 | s=input()
if len(s)==len(set(list(s))):
print('yes')
else:
print('no') |
s851361019 | p03486 | u519968172 | 2,000 | 262,144 | Wrong Answer | 28 | 8,928 | 243 | 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(list(input()))
t=sorted(list(input()),reverse=True)
for i in range(min(len(s),len(t))):
if ord(s[i])<ord(t[i]):
print("YES")
break
elif ord(s[i])>ord(t[i]):
print("NO")
break
else:
print("Yes")
| s362593085 | Accepted | 26 | 9,056 | 295 | s=sorted(list(input()))
t=sorted(list(input()),reverse=True)
for i in range(min(len(s),len(t))):
if ord(s[i])<ord(t[i]):
print("Yes")
break
elif ord(s[i])>ord(t[i]):
print("No")
break
else:
if len(s)>=len(t):
print("No")
else:
print("Yes")
|
s914565321 | p02665 | u447679353 | 2,000 | 1,048,576 | Wrong Answer | 730 | 669,276 | 969 | Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1. | # coding: utf-8
N = int(input())
A = list(map(int,input().split()))
leaf= [-1 for i in range(N+1)]
notleaf = [-1 for i in range(N+1)]
flag=0
if N == 0 and A[0] == 1:
print(1)
elif N != 0 and A[0]>=1:
print(-1)
else:
leaf[0]=0
notleaf[0]=1
for d in range(1, N+1):
leaf[d] = A[d]
notleaf[d] = 2*notleaf[d-1] - leaf[d]
if notleaf[d]<0:
flag=1
print(-1)
break
#if flag==0 and notleaf[N] != 0:
# print(-1)
else:
sleaf = 0
for i in range(N+1):
sleaf += leaf[i]
d1 = 0
for i in range(N+1):
d1 += (i+1) * leaf[i]
sleaf -= 1
for i in range(1,N+1):
if sleaf>0:
d1 -= i*( min(sleaf, leaf[i]+notleaf[i] - notleaf[i-1]))
sleaf -= (leaf[i]+notleaf[i] - notleaf[i-1])
else:
print(d1)
break
| s970050042 | Accepted | 756 | 669,532 | 968 | # coding: utf-8
N = int(input())
A = list(map(int,input().split()))
leaf= [-1 for i in range(N+1)]
notleaf = [-1 for i in range(N+1)]
flag=0
if N == 0 and A[0] == 1:
print(1)
elif N==0 and A[0]!=1:
print(-1)
elif N != 0 and A[0]>=1:
print(-1)
else:
leaf[0]=0
notleaf[0]=1
for d in range(1, N+1):
leaf[d] = A[d]
notleaf[d] = 2*notleaf[d-1] - leaf[d]
if notleaf[d]<0:
flag=1
print(-1)
break
if flag == 1:
pass
else:
sleaf = 0
for i in range(N+1):
sleaf += leaf[i]
d1 = 0
for i in range(N+1):
d1 += (i+1) * leaf[i]
sleaf -= 1
for i in range(1,N+1):
if sleaf>0:
d1 -= i*( min(sleaf, leaf[i]+notleaf[i] - notleaf[i-1]))
sleaf -= (leaf[i]+notleaf[i] - notleaf[i-1])
else:
break
print(d1)
|
s634242000 | p02612 | u843135954 | 2,000 | 1,048,576 | Wrong Answer | 33 | 9,136 | 242 | We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required. | import sys
stdin = sys.stdin
sys.setrecursionlimit(10**6)
ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
nn = lambda: list(stdin.readline().split())
ns = lambda: stdin.readline().rstrip()
n = ni()
print(n%1000) | s800128180 | Accepted | 31 | 9,188 | 269 | import sys
stdin = sys.stdin
sys.setrecursionlimit(10**6)
ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
nn = lambda: list(stdin.readline().split())
ns = lambda: stdin.readline().rstrip()
n = ni()
print(0 if n%1000 == 0 else 1000-n%1000) |
s358697139 | p03962 | u834301346 | 2,000 | 262,144 | Wrong Answer | 111 | 27,176 | 87 | AtCoDeer the deer recently bought three paint cans. The color of the one he bought two days ago is a, the color of the one he bought yesterday is b, and the color of the one he bought today is c. Here, the color of each paint can is represented by an integer between 1 and 100, inclusive. Since he is forgetful, he might have bought more than one paint can in the same color. Count the number of different kinds of colors of these paint cans and tell him. | import numpy as np
array = list(map(int, input().split(' ')))
print(np.unique(array))
| s375079367 | Accepted | 118 | 26,760 | 90 | import numpy as np
array = list(map(int, input().split(' ')))
print(len(np.unique(array))) |
s514523648 | p03377 | u802963389 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 85 | There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals. | A, B, X = map(int, input().split())
if A + B <= X:
print("YES")
else:
print("NO") | s348493141 | Accepted | 17 | 2,940 | 97 | A, B, X = map(int, input().split())
if A + B >= X and X >= A:
print("YES")
else:
print("NO")
|
s084076747 | p02614 | u292746386 | 1,000 | 1,048,576 | Wrong Answer | 202 | 27,220 | 610 | We have a grid of H rows and W columns of squares. The color of the square at the i-th row from the top and the j-th column from the left (1 \leq i \leq H, 1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white if c_{i,j} is `.`, and black if c_{i,j} is `#`. Consider doing the following operation: * Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns. You are given a positive integer K. How many choices of rows and columns result in exactly K black squares remaining after the operation? Here, we consider two choices different when there is a row or column chosen in only one of those choices. | import numpy as np
H, W, K = [int(i) for i in input().split()]
A = np.zeros([H, W], dtype=bool)
for i in range(H):
A[i] = np.array([False if i=="." else True for i in input().split()])
T = A.sum()
T_row = A.sum(1)
T_col = A.sum(0)
def calc(I, J):
return T - T_row[I].sum() - T_col[J].sum() + A[I][:,J].sum()
def repr(i, W):
s = np.binary_repr(i, width=W)[::-1]
return [i for i, j in enumerate(s) if j == "1"]
count = 0
for i in range(2**H-1):
for j in range(2**W-1):
row = repr(i, H)
col = repr(j, W)
if calc(row, col) == K:
count += 1
print(count) | s506025372 | Accepted | 191 | 27,312 | 602 | import numpy as np
H, W, K = [int(i) for i in input().split()]
A = np.zeros([H, W], dtype=bool)
for i in range(H):
A[i] = np.array([False if i=="." else True for i in input()])
T = A.sum()
T_row = A.sum(1)
T_col = A.sum(0)
def calc(I, J):
return T - T_row[I].sum() - T_col[J].sum() + A[I][:,J].sum()
def repr(i, W):
s = np.binary_repr(i, width=W)[::-1]
return [i for i, j in enumerate(s) if j == "1"]
count = 0
for i in range(2**H-1):
for j in range(2**W-1):
row = repr(i, H)
col = repr(j, W)
if calc(row, col) == K:
count += 1
print(count) |
s334933765 | p03720 | u312078744 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 221 | 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())
mat = [[0] * n] * n
for i in range(m):
a, b = map(int, input().split())
mat[a - 1][b - 1] = 1
mat[b - 1][a - 1] = 1
for i in range(n):
ans = sum(mat[i])
print(ans)
| s225321322 | Accepted | 153 | 12,384 | 282 | import numpy as np
n, m = map(int, input().split())
mat = [[0] * n] * n
mat = np.array(mat)
for i in range(m):
a, b = map(int, input().split())
mat[a - 1, b - 1] += 1
mat[b - 1, a - 1] += 1
# print(mat)
for i in range(n):
ans = np.sum(mat[i, :])
print(ans)
|
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