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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s109806385
|
p03493
|
u313014073
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 130
|
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 = int(input())
ans = 0
if s%10==1:
ans += 1
s = s//10
if s%10==1:
ans += 1
s = s//10
if s==1:
ans += 1
print(ans+1)
|
s649126219
|
Accepted
| 17
| 3,060
| 128
|
s = int(input())
ans = 0
if s%10==1:
ans += 1
s = s//10
if s%10==1:
ans += 1
s = s//10
if s==1:
ans += 1
print(ans)
|
s505091998
|
p02664
|
u357751375
| 2,000
| 1,048,576
|
Wrong Answer
| 76
| 10,600
| 278
|
For a string S consisting of the uppercase English letters `P` and `D`, let the _doctoral and postdoctoral quotient_ of S be the total number of occurrences of `D` and `PD` in S as contiguous substrings. For example, if S = `PPDDP`, it contains two occurrences of `D` and one occurrence of `PD` as contiguous substrings, so the doctoral and postdoctoral quotient of S is 3. We have a string T consisting of `P`, `D`, and `?`. Among the strings that can be obtained by replacing each `?` in T with `P` or `D`, find one with the maximum possible doctoral and postdoctoral quotient.
|
t = list(input())
a = 0
b = 0
for i in range(len(t)):
if i == 0 and t[i] == '?':
t[i] == 'D'
if i > 0 and t[i] == '?':
if t[i-1] == 'P':
t[i] == 'D'
else:
t[i] == 'P'
a = t.count('PD')
b = t.count('D')
print(a + b)
|
s336987182
|
Accepted
| 130
| 10,824
| 367
|
t = list(input())
a = 0
b = 0
for i in range(len(t)):
if i == 0 and t[i] == '?':
t[i] = 'D'
if 0 < i < len(t) - 1 and t[i] == '?':
if t[i-1] == 'P':
t[i] = 'D'
elif t[i+1] == 'P':
t[i] = 'D'
else:
t[i] = 'P'
if i == len(t) -1 and t[i] == '?':
t[i] = 'D'
print(''.join(t))
|
s109406949
|
p03543
|
u887207211
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 97
|
We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**?
|
N = input()
if(N[0] == N[1] == N[2] and N[1] == N[2] == N[3]):
print("Yes")
else:
print("No")
|
s873112876
|
Accepted
| 18
| 2,940
| 97
|
N = input()
ans = "No"
if(N[0] == N[1] == N[2] or N[1] == N[2] == N[3]):
ans = "Yes"
print(ans)
|
s099013181
|
p03408
|
u403355272
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 176
|
Takahashi has N blue cards and M red cards. A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i. Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen. Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces `atcoder`, he will not earn money even if there are blue cards with `atcoderr`, `atcode`, `btcoder`, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.) At most how much can he earn on balance? Note that the same string may be written on multiple cards.
|
N = int(input())
s = [input() for i in range(N)]
M = int(input())
t = [input() for i in range(M)]
le = set(s)
for i in le:
count = max(0,s.count(i)-t.count(i))
print(count)
|
s243422848
|
Accepted
| 19
| 3,064
| 204
|
N = int(input())
s = [input() for i in range(N)]
M = int(input())
t = [input() for i in range(M)]
le = list(set(s))
count = max(s.count(le[i]) - t.count(le[i]) for i in range(len(le)))
print(max(0,count))
|
s460855247
|
p03359
|
u617659131
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 71
|
In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year. For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year. In this country, a date is called _Takahashi_ when the month and the day are equal as numbers. For example, 5-5 is Takahashi. How many days from 2018-1-1 through 2018-a-b are Takahashi?
|
a,b = map(int, input().split())
if a < b:
print(a)
else:
print(a-1)
|
s481378971
|
Accepted
| 17
| 2,940
| 72
|
a,b = map(int, input().split())
if a <= b:
print(a)
else:
print(a-1)
|
s173799372
|
p02396
|
u655518263
| 1,000
| 131,072
|
Wrong Answer
| 120
| 7,516
| 136
|
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.
|
for i in range(10000):
x = input()
if x=="0":
break
else:
print ("Case " + str(i) + ": " + str(x))
|
s405489008
|
Accepted
| 130
| 7,408
| 138
|
for i in range(20000):
x = input()
if x=="0":
break
else:
print ("Case " + str(i+1) + ": " + str(x))
|
s559009799
|
p02833
|
u524870111
| 2,000
| 1,048,576
|
Wrong Answer
| 22
| 3,444
| 414
|
For an integer n not less than 0, let us define f(n) as follows: * f(n) = 1 (if n < 2) * f(n) = n f(n-2) (if n \geq 2) Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N).
|
import sys
stdin = sys.stdin
import copy
ns = lambda: stdin.readline().rstrip()
ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
n = ni()
if n % 2 != 0:
print(0)
else:
n2 = copy.copy(n)
n5 = copy.copy(n)
c2, c5 = 0, 0
while n2 > 1:
n2 //= 2
c2 += n2
while n5 > 1:
n5 //= 5
c5 += n5//2
print(c2, c5)
print(min(c2, c5))
|
s332664996
|
Accepted
| 22
| 3,444
| 396
|
import sys
stdin = sys.stdin
import copy
ns = lambda: stdin.readline().rstrip()
ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
n = ni()
if n % 2 != 0:
print(0)
else:
n2 = copy.copy(n)
n5 = copy.copy(n)
c2, c5 = 0, 0
while n2 > 1:
n2 //= 2
c2 += n2
while n5 > 1:
n5 //= 5
c5 += n5//2
print(min(c2, c5))
|
s546983249
|
p02697
|
u420059236
| 2,000
| 1,048,576
|
Wrong Answer
| 263
| 35,044
| 339
|
You are going to hold a competition of one-to-one game called AtCoder Janken. _(Janken is the Japanese name for Rock-paper-scissors.)_ N players will participate in this competition, and they are given distinct integers from 1 through N. The arena has M playing fields for two players. You need to assign each playing field two distinct integers between 1 and N (inclusive). You cannot assign the same integer to multiple playing fields. The competition consists of N rounds, each of which proceeds as follows: * For each player, if there is a playing field that is assigned the player's integer, the player goes to that field and fight the other player who comes there. * Then, each player adds 1 to its integer. If it becomes N+1, change it to 1. You want to ensure that no player fights the same opponent more than once during the N rounds. Print an assignment of integers to the playing fields satisfying this condition. It can be proved that such an assignment always exists under the constraints given.
|
#E
import numpy as np
n, m = list(map(int, input().split()))
lists = list(np.arange(1, n+1))
list1 = lists[:n//2]
list2 = lists[n//2:]
num = m
for i in range(m//2 +1):
if num == 0:
break
print(list2[i],list2[len(list2)-1-i])
num -= 1
if num == 0:
break
print(list1[i],list1[len(list1)-1-i])
num -= 1
|
s778166086
|
Accepted
| 254
| 34,968
| 317
|
#E
import numpy as np
n, m = list(map(int, input().split()))
lists = list(np.arange(1, n+1))
list1 = lists[:m]
list2 = lists[m:]
num = m
for i in range(m//2 +1):
if num == 0:
break
print(list2[i],list2[m+1-i-1])
num -= 1
if num == 0:
break
print(list1[i],list1[m-i-1])
num -= 1
|
s593428709
|
p03565
|
u624696727
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 509
|
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`.
|
if __name__ == '__main__':
S = input()
T = input()
T_length = len(T)
head = -1
for i in range(len(S)-T_length+1):
counter = 0
for j in range(T_length):
if(S[i+j]==T[j] or S[i+j]=='?'):
counter += 1
if(counter == T_length):
head = i
print(head)
ans = ''
i=0
if(head == -1):
print("UNRESTORABLE")
else :
while i < len(S):
if(i!=head and S[i]=='?'):
ans += 'a'
i += 1
elif(i==head):
ans += T
i += T_length
else :
ans += S[i]
i+=1
print(ans)
|
s822191175
|
Accepted
| 17
| 3,064
| 496
|
if __name__ == '__main__':
S = input()
T = input()
T_length = len(T)
head = -1
for i in range(len(S)-T_length+1):
counter = 0
for j in range(T_length):
if(S[i+j]==T[j] or S[i+j]=='?'):
counter += 1
if(counter == T_length):
head = i
ans = ''
i=0
if(head == -1):
print("UNRESTORABLE")
else :
while i < len(S):
if(i!=head and S[i]=='?'):
ans += 'a'
i += 1
elif(i==head):
ans += T
i += T_length
else :
ans += S[i]
i+=1
print(ans)
|
s775407025
|
p03048
|
u854093727
| 2,000
| 1,048,576
|
Wrong Answer
| 1,587
| 2,940
| 153
|
Snuke has come to a store that sells boxes containing balls. The store sells the following three kinds of boxes: * Red boxes, each containing R red balls * Green boxes, each containing G green balls * Blue boxes, each containing B blue balls Snuke wants to get a total of exactly N balls by buying r red boxes, g green boxes and b blue boxes. How many triples of non-negative integers (r,g,b) achieve this?
|
R,G,B,N = map(int,input().split())
ans = 0
for i in range(N):
for e in range(N-R*i):
if (N-i*i-e*G)%B == 0:
ans += 1
print(ans)
|
s538161571
|
Accepted
| 1,531
| 3,316
| 158
|
R,G,B,N = map(int,input().split())
ans=0
for i in range(N//R+1):
for e in range((N-R*i)//G+1):
if (N-i*R-e*G)%B==0:
ans+=1
print(ans)
|
s009341856
|
p03050
|
u565387531
| 2,000
| 1,048,576
|
Wrong Answer
| 19
| 2,940
| 132
|
Snuke received a positive integer N from Takahashi. A positive integer m is called a _favorite number_ when the following condition is satisfied: * The quotient and remainder of N divided by m are equal, that is, \lfloor \frac{N}{m} \rfloor = N \bmod m holds. Find all favorite numbers and print the sum of those.
|
N = int(input())
ans = 0
while N > 2:
if N % 2 == 1:
ans += N-1
break
ans += N - 1
N = N / 2
print(ans)
|
s483211388
|
Accepted
| 115
| 3,268
| 372
|
N = int(input())
def make_divisors(n):
divisors = []
for i in range(1, int(n**0.5)+1):
if n % i == 0:
divisors.append(i)
if i != n // i:
divisors.append(n//i)
divisors.sort()
return divisors
ans = 0
for i in make_divisors(N):
if i > 1 and N // (i - 1) == N % (i - 1):
ans += i - 1
print(ans)
|
s901412886
|
p02409
|
u391228754
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,660
| 321
|
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.
|
n = int(input())
a =[[[0 for i in range(10)]for j in range(3)]for k in range(4)]
for i in range(n):
b, f, r, v = map(int, input().split())
a[b-1][f-1][r-1] += v
for i in range(4):
for j in range(3):
for k in range(10):
print(" ", a[i][j][k], end='')
print()
print("#" * 20)
|
s098810223
|
Accepted
| 30
| 7,716
| 339
|
n = int(input())
a =[[[0 for i in range(10)]for j in range(3)]for k in range(4)]
for i in range(n):
b, f, r, v = map(int, input().split())
a[b-1][f-1][r-1] += v
for i in range(4):
for j in range(3):
for k in range(10):
print("", a[i][j][k], end='')
print()
if i != 3:
print("#" * 20)
|
s862731170
|
p03644
|
u223904637
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 151
|
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
n=int(input())
def ans(n):
r=0
while n%2==0:
n=n//2
r+=1
return r
a=0
for i in range(1,n+1):
a=max(a,ans(i))
print(a)
|
s760629846
|
Accepted
| 17
| 2,940
| 154
|
n=int(input())
def ans(n):
r=0
while n%2==0:
n=n//2
r+=1
return r
a=0
for i in range(1,n+1):
a=max(a,ans(i))
print(2**a)
|
s586041216
|
p03591
|
u785220618
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 72
|
Ringo is giving a present to Snuke. Ringo has found out that Snuke loves _yakiniku_ (a Japanese term meaning grilled meat. _yaki_ : grilled, _niku_ : meat). He supposes that Snuke likes grilled things starting with `YAKI` in Japanese, and does not like other things. You are given a string S representing the Japanese name of Ringo's present to Snuke. Determine whether S starts with `YAKI`.
|
s = input()
if s[:4] == 'YAKi':
print('Yes')
else:
print('No')
|
s211891099
|
Accepted
| 17
| 2,940
| 72
|
s = input()
if s[:4] == 'YAKI':
print('Yes')
else:
print('No')
|
s431884264
|
p02408
|
u098047375
| 1,000
| 131,072
|
Wrong Answer
| 30
| 5,728
| 870
|
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.
|
n = int(input())
a = []
for i in range(n):
x, y = input().split()
a.append([x, y])
trump = []
for i in range(1,14):
trump.append(['S', str(i)])
trump.append(['H', str(i)])
trump.append(['C', str(i)])
trump.append(['D', str(i)])
miss = []
for i in range(len(trump)):
for j in range(len(a)):
if trump[i] != a[j]:
miss.append(trump[i])
miss.sort()
for i in range(len(miss)):
x = miss[i]
if x[0] == "S":
print(x[0] + ' '+ str(x[1]))
for i in range(len(miss)):
x = miss[i]
if x[0] == "H":
print(x[0] + ' '+ str(x[1]))
for i in range(len(miss)):
x = miss[i]
if x[0] == "C":
print(x[0] + ' '+ str(x[1]))
for i in range(len(miss)):
x = miss[i]
if x[0] == "D":
print(x[0] + ' '+ str(x[1]))
|
s219355041
|
Accepted
| 30
| 5,612
| 981
|
n = int(input())
a = []
for i in range(n):
x, y = input().split()
a.append([x, y])
trump = []
for i in range(1,14):
trump.append(['S', str(i)])
trump.append(['H', str(i)])
trump.append(['C', str(i)])
trump.append(['D', str(i)])
dif = []
for i in range(1,14):
dif.append(['S', str(i)])
dif.append(['H', str(i)])
dif.append(['C', str(i)])
dif.append(['D', str(i)])
for i in range(len(trump)):
for j in range(len(a)):
if trump[i] == a[j]:
dif.remove(trump[i])
for i in range(len(dif)):
x = dif[i]
if x[0] == "S":
print(x[0] +' '+ x[1])
for i in range(len(dif)):
x = dif[i]
if x[0] == "H":
print(x[0] + ' '+ x[1])
for i in range(len(dif)):
x = dif[i]
if x[0] == "C":
print(x[0] + ' '+ x[1])
for i in range(len(dif)):
x = dif[i]
if x[0] == "D":
print(x[0] + ' '+ x[1])
|
s994069358
|
p03485
|
u209594105
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 87
|
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
str = input()
a = int(str.split()[0])
b = int(str.split()[1])
print(int((a+b)/2 + 1))
|
s598930198
|
Accepted
| 17
| 2,940
| 89
|
str = input()
a = int(str.split()[0])
b = int(str.split()[1])
print(int((a+b)/2 + 0.5))
|
s583723899
|
p03863
|
u667024514
| 2,000
| 262,144
|
Wrong Answer
| 20
| 3,316
| 92
|
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.
|
s = str(input())
if ((len(s)%2) ^ (s[0] == s[-1])):
print("Second")
else:
print("First")
|
s533556623
|
Accepted
| 20
| 3,316
| 87
|
s = str(input())
if ((len(s)%2) ^ (s[0] == s[-1])):print("First")
else:print("Second")
|
s702494304
|
p03447
|
u780812005
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 98
|
You went shopping to buy cakes and donuts with X yen (the currency of Japan). First, you bought one cake for A yen at a cake shop. Then, you bought as many donuts as possible for B yen each, at a donut shop. How much do you have left after shopping?
|
def resolve():
X = int(input())
A = int(input())
B = int(input())
print((X-A)%B)
|
s252322656
|
Accepted
| 17
| 3,064
| 143
|
def resolve():
X = int(input())
A = int(input())
B = int(input())
print((X-A)%B)
if __name__ == "__main__":
resolve()
|
s246777939
|
p03730
|
u289288647
| 2,000
| 262,144
|
Wrong Answer
| 25
| 9,132
| 162
|
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`.
|
import sys
A, B, C = map(int, input().split())
i = 1
while (A*i)%B != C:
i += 1
if i > 100:
print('NO')
sys.exit()
print(i)
print('YES')
|
s458312441
|
Accepted
| 24
| 9,152
| 153
|
import sys
A, B, C = map(int, input().split())
i = 1
while (A*i)%B != C:
i += 1
if i > 100:
print('NO')
sys.exit()
print('YES')
|
s801461782
|
p02389
|
u546130344
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,592
| 113
|
Write a program which calculates the area and perimeter of a given rectangle.
|
a, b = input().strip().split(' ')
a, b = [int(a), int(b)]
men = a * b
syu = 2 * (a + b)
print(men)
print(syu)
|
s792587245
|
Accepted
| 20
| 5,588
| 124
|
a, b = input().strip().split(' ')
a, b = [int(a), int(b)]
men = a * b
syu = 2 * (a + b)
print(str(men) + ' ' + str(syu))
|
s331072957
|
p02613
|
u581511366
| 2,000
| 1,048,576
|
Wrong Answer
| 139
| 9,128
| 227
|
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
def main():
N = int(input())
di = {"AC": 0, "WA": 0, "TLE": 0, "RE": 0}
for _ in range(N):
di[input()] += 1
for k, v in di.items():
print(f"{k} × {v}")
if __name__ == "__main__":
main()
|
s566055044
|
Accepted
| 65
| 9,192
| 290
|
import sys
def input(): return sys.stdin.readline().strip()
def main():
N = int(input())
di = {"AC": 0, "WA": 0, "TLE": 0, "RE": 0}
for _ in range(N):
di[input()] += 1
for k, v in di.items():
print(f"{k} x {v}")
if __name__ == "__main__":
main()
|
s237064084
|
p02678
|
u164471280
| 2,000
| 1,048,576
|
Wrong Answer
| 2,211
| 38,924
| 618
|
There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.
|
n, m = map(int, input().split())
ANS = [-1]*(n+1)
P = []
L = [[] for i in range(n+1)]
for i in range(m):
a, b = map(int, input().split())
if a == 1:
ANS[b] = 1
P.append(b)
elif b == 1:
ANS[a] = 1
P.append(a)
else:
L[a].append(b)
L[b].append(a)
print(L, ANS, P)
i = 0
while len(P) < n-1:
for j in range(1, n+1):
if P[i] in L[j] and ANS[j] == -1:
P.append(j)
L[j].remove(P[i])
ANS[j] = P[i]
# print(i, P[i], j, L, ANS, P)
i += 1
print('Yes')
for i in range(2, n+1):
print(ANS[i])
|
s881104583
|
Accepted
| 677
| 33,932
| 454
|
from collections import deque
n, m = map(int, input().split())
L = [[] for i in range(n)]
for i in range(m):
a, b = map(int, input().split())
L[a-1].append(b-1)
L[b-1].append(a-1)
d = deque([0])
D = [-1] * n
D[0] = 0
while len(d) > 0 :
t = d.popleft()
for i in L[t]:
if D[i] == -1:
D[i] = t
d.append(i)
# print(d, D)
print('Yes')
for i in range(1, n):
print(D[i]+1)
|
s491493729
|
p03555
|
u014646120
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 169
|
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
C11C12C13=input()
C21C22C23=input()
if (C11C12C13[0]==C21C22C23[2] and C11C12C13[2]==C21C22C23[0] and C11C12C13[1]==C21C22C23[1]):
print("Yes")
else:
print("No")
|
s886548662
|
Accepted
| 17
| 2,940
| 170
|
C11C12C13=input()
C21C22C23=input()
if (C11C12C13[0]==C21C22C23[2] and C11C12C13[2]==C21C22C23[0] and C11C12C13[1]==C21C22C23[1]):
print("YES")
else:
print("NO")
|
s947509104
|
p03377
|
u217526092
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 126
|
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.
|
# coding: utf-8
A, B, X = [int(i) for i in input().split()]
if X <= A and X - A <= B:
print('Yes')
else:
print('No')
|
s396284460
|
Accepted
| 17
| 2,940
| 118
|
# coding: utf-8
A, B, X = map(int, input().split())
if X >= A and X - A <= B:
print('YES')
else:
print('NO')
|
s885983711
|
p03698
|
u785066634
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 362
|
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
s=list(input())
for i in range(len(s)):
for j in range(i+1,len(s)):
if s[i]!=s[j]:
pass
else:
print('s[i]',s[i],'s[j]',s[j])
print('No')
exit()
print('Yes')
|
s158482473
|
Accepted
| 17
| 2,940
| 363
|
s=list(input())
for i in range(len(s)):
for j in range(i+1,len(s)):
if s[i]!=s[j]:
pass
else:
#print('s[i]',s[i],'s[j]',s[j])
print('no')
exit()
print('yes')
|
s871666740
|
p03478
|
u635273885
| 2,000
| 262,144
|
Wrong Answer
| 35
| 3,060
| 208
|
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())
sum = 0
cnt = 0
for i in range(1, N+1):
words_i = str(i)
for word_i in words_i:
sum += int(word_i)
if A <= sum <= B:
cnt += 1
print(cnt)
|
s317170526
|
Accepted
| 34
| 3,060
| 216
|
N, A, B = map(int, input().split())
sums = 0
cnt = 0
for i in range(1, N+1):
words_i = str(i)
for word_i in words_i:
sums += int(word_i)
if A <= sums <= B:
cnt += i
sums = 0
print(cnt)
|
s768035195
|
p03150
|
u077075933
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,064
| 393
|
A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once. Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string.
|
arr = input()
KEY = "keyence"
def solve():
index = 0
state = 0
for letter in arr:
print(arr, state)
if letter == KEY[index]:
index+=1
if index == 7:
return "YES"
if state == 1:
state = 2
else:
if state == 0:
state = 1
elif state == 1:
pass
elif state == 2:
return "NO"
return "NO"
print(solve())
|
s575481287
|
Accepted
| 17
| 2,940
| 205
|
def solve():
arr = input()
KEY = "keyence"
N = len(arr)
L = N-7
for s in range(N-L):
extracted = arr[:s]+arr[s+L:]
if(extracted == KEY):
return "YES"
return "NO"
print(solve())
|
s622944542
|
p03712
|
u444238096
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 186
|
You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}. Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1.
|
H, W = map(int, input().split())
mylist=[]
for i in range(H):
var = input()
mylist.append(var)
print("*"*(W+2))
for i in range(H):
print ("*"+mylist[i]+"*")
print("*"*(W+2))
|
s861065883
|
Accepted
| 18
| 3,060
| 186
|
H, W = map(int, input().split())
mylist=[]
for i in range(H):
var = input()
mylist.append(var)
print("#"*(W+2))
for i in range(H):
print ("#"+mylist[i]+"#")
print("#"*(W+2))
|
s195518383
|
p03860
|
u566321790
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 46
|
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
name = str(input())
print('A' + name[0] + 'C')
|
s907578629
|
Accepted
| 17
| 2,940
| 63
|
a, b, c = map(str, input().split())
print(a[0] + b[0] + c[0])
|
s166928088
|
p04029
|
u244203620
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 31
|
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
a=int(input())
print(a*(a+1)/2)
|
s308328497
|
Accepted
| 17
| 2,940
| 36
|
a=int(input())
print(int(a*(a+1)/2))
|
s398098923
|
p04043
|
u909851424
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 220
|
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.
|
args = input().split()
five = 0
seven = 0
for arg in args:
print(arg)
if arg == "5":
five = five + 1
else:
seven = seven + 1
if five == 2 and seven == 1:
print("YES")
else:
print("NO")
|
s881434021
|
Accepted
| 17
| 2,940
| 205
|
args = input().split()
five = 0
seven = 0
for arg in args:
if arg == "5":
five = five + 1
else:
seven = seven + 1
if five == 2 and seven == 1:
print("YES")
else:
print("NO")
|
s016189797
|
p03129
|
u325227960
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 80
|
Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.
|
a,b=map(int,input().split())
if a<=b+1 :
print("No")
else :
print("Yes")
|
s179685167
|
Accepted
| 18
| 2,940
| 121
|
a,b=map(int,input().split())
if (a%2==1 and a>=(b-1)*2) or (a%2==0 and a>=b*2) :
print("YES")
else :
print("NO")
|
s786448602
|
p03433
|
u670481241
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 69
|
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
n = int(input())
a = int(input())
r = "YES" if n % 500 <= a else "NO"
|
s193103906
|
Accepted
| 17
| 3,064
| 78
|
n = int(input())
a = int(input())
r = "Yes" if n % 500 <= a else "No"
print(r)
|
s044497336
|
p03854
|
u844789719
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,188
| 137
|
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
if input().replace('dreamer', '').replace('eraser', '').replace('dream', '').replace('erase', ''):
print('No')
else:
print('Yes')
|
s715515245
|
Accepted
| 67
| 3,188
| 246
|
Cr = input()
while Cr:
if Cr[-7:] == 'dreamer':
Cr = Cr[0:-7]
elif Cr[-6:] == 'eraser':
Cr = Cr[0:-6]
elif Cr[-5:] in ['dream', 'erase']:
Cr = Cr[0:-5]
else:
print('NO')
exit()
print('YES')
|
s056220284
|
p03448
|
u320098990
| 2,000
| 262,144
|
Wrong Answer
| 26
| 9,188
| 286
|
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.
|
in_a = input()
in_b = input()
in_c = input()
in_x = input()
A = int(in_a)
B = int(in_b)
C = int(in_c)
X = int(in_x)
count = 0
for i in range(A+1):
for j in range(B+1):
k = ( X - 500*i - 100*j ) / 50
if k.is_integer() and k <= C:
count +=1
print(count)
|
s212332322
|
Accepted
| 27
| 9,084
| 392
|
in_a = input()
in_b = input()
in_c = input()
in_x = input()
A = int(in_a)
B = int(in_b)
C = int(in_c)
X = int(in_x)
count = 0
for i in range(A+1):
for j in range(B+1):
k = ( X - 500*i - 100*j ) / 50
if k < 0:
break;
elif k.is_integer() and k <= C:
count +=1
# print('clear', k, count)
print(count)
|
s059936424
|
p03378
|
u125568248
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 209
|
There are N + 1 squares arranged in a row, numbered 0, 1, ..., N from left to right. Initially, you are in Square X. You can freely travel between adjacent squares. Your goal is to reach Square 0 or Square N. However, for each i = 1, 2, ..., M, there is a toll gate in Square A_i, and traveling to Square A_i incurs a cost of 1. It is guaranteed that there is no toll gate in Square 0, Square X and Square N. Find the minimum cost incurred before reaching the goal.
|
n, m, x = map(int, input().split())
a = list(map(int, input().split()))
c = 0
for b in a:
c += 1
if x < b:
if c < (m/2):
print(c)
else:
print(m-c)
break
|
s089895199
|
Accepted
| 17
| 3,060
| 224
|
n, m, x = map(int, input().split())
a = list(map(int, input().split()))
c = 0
for b in a:
c += 1
if x < b:
c -= 1
if c < (m/2):
print(c)
else:
print(m-c)
break
|
s511206849
|
p03456
|
u521241621
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 149
|
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.
|
import math
d = input()
n1, n2 = d.split(" ")
num = n1 + n2
square = math.sqrt(int(num))
if square.is_integer():
print("Yex")
else:
print("No")
|
s431885803
|
Accepted
| 17
| 2,940
| 150
|
import math
d = input()
n1, n2 = d.split(" ")
num = n1 + n2
square = math.sqrt(int(num))
if square.is_integer():
print("Yes")
else:
print("No")
|
s910387894
|
p03605
|
u785205215
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 69
|
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 = str(input())
if "9" in n:
print("yes")
else:
print("no")
|
s400029980
|
Accepted
| 18
| 2,940
| 68
|
n = str(input())
if "9" in n:
print("Yes")
else:
print("No")
|
s173766216
|
p03160
|
u790053529
| 2,000
| 1,048,576
|
Wrong Answer
| 119
| 14,724
| 349
|
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
def cost(n, nums):
dp = [0 for _ in range(n)]
dp[0] = 0
dp[1] = abs(nums[1] - nums[0])
for i in range(2, n):
dp[i] = min(
dp[i-1] + abs(nums[i]-nums[i-1]),
dp[i-2] + abs(nums[i]-nums[i-2]))
print(dp)
return dp[-1]
n = int(input())
nums = [int(i) for i in input().split()]
print(cost(n, nums))
|
s106191914
|
Accepted
| 109
| 13,924
| 336
|
def cost(n, nums):
dp = [0 for _ in range(n)]
dp[0] = 0
dp[1] = abs(nums[1] - nums[0])
for i in range(2, n):
dp[i] = min(
dp[i-1] + abs(nums[i]-nums[i-1]),
dp[i-2] + abs(nums[i]-nums[i-2]))
return dp[-1]
n = int(input())
nums = [int(i) for i in input().split()]
print(cost(n, nums))
|
s718813099
|
p03485
|
u871841829
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 73
|
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
import math
a,b = map(int, input().split())
print(int(math.ceil(a+b/2)))
|
s001366706
|
Accepted
| 17
| 2,940
| 75
|
import math
a,b = map(int, input().split())
print(int(math.ceil((a+b)/2)))
|
s217672762
|
p03129
|
u514222361
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 159
|
Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.
|
# -*- coding: utf-8 -*-
import sys
i_str = input().split(' ')
a = int(i_str[0])
n = int(i_str[0])
if (a+1) / 2 >= n:
print("YES")
else:
print("NO")
|
s714794118
|
Accepted
| 17
| 2,940
| 158
|
# -*- coding: utf-8 -*-
import sys
i_str = input().split(' ')
a = int(i_str[0])
n = int(i_str[1])
if (a+1) / 2 >= n:
print("YES")
else:
print("NO")
|
s350989426
|
p02603
|
u094534261
| 2,000
| 1,048,576
|
Wrong Answer
| 35
| 9,172
| 170
|
To become a millionaire, M-kun has decided to make money by trading in the next N days. Currently, he has 1000 yen and no stocks - only one kind of stock is issued in the country where he lives. He is famous across the country for his ability to foresee the future. He already knows that the price of one stock in the next N days will be as follows: * A_1 yen on the 1-st day, A_2 yen on the 2-nd day, ..., A_N yen on the N-th day. In the i-th day, M-kun can make the following trade **any number of times** (possibly zero), **within the amount of money and stocks that he has at the time**. * Buy stock: Pay A_i yen and receive one stock. * Sell stock: Sell one stock for A_i yen. What is the maximum possible amount of money that M-kun can have in the end by trading optimally?
|
n=int(input())
a=list(map(int,input().split()))
ans=1000
for i in range(1,n):
if a[i]>a[i-1]:
ans += (a[i]-a[i-1])*(ans//a[i-1])
print(ans)
print(ans)
|
s705836349
|
Accepted
| 28
| 9,176
| 151
|
n=int(input())
a=list(map(int,input().split()))
ans=1000
for i in range(1,n):
if a[i]>a[i-1]:
ans += (a[i]-a[i-1])*(ans//a[i-1])
print(ans)
|
s979325252
|
p04029
|
u381585104
| 2,000
| 262,144
|
Wrong Answer
| 27
| 9,012
| 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)
|
s430656580
|
Accepted
| 26
| 9,060
| 35
|
n = int(input())
print(n*(n+1)//2)
|
s391187463
|
p03494
|
u412481017
| 2,000
| 262,144
|
Wrong Answer
| 20
| 2,940
| 244
|
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.
|
# -*- coding: utf-8 -*-
a = int(input())
b = input().split(" ")
i=1
while 1:
flag=0
for item in b:
if int(item)%2**i!=0:
flag=1
if flag==1:
break
i=i+1
print(i)
|
s036415896
|
Accepted
| 23
| 3,060
| 265
|
# -*- coding: utf-8 -*-
a = int(input())
b = input().split(" ")
i=1
while 1:
flag=0
for item in b:
if int(item)%2**i!=0:
#print(item)
flag=1
if flag==1:
break
i=i+1
print(i-1)
|
s063705457
|
p04029
|
u608355135
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 45
|
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
a = int(input())
b = a * (a + 1 ) /2
print(b)
|
s071993112
|
Accepted
| 17
| 2,940
| 51
|
a = int(input())
b = a * (a + 1) / 2
print(int(b))
|
s427311382
|
p03251
|
u730043482
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 274
|
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.
|
M, N, X, Y = map(int, input().split())
xs = list(map(int,input().split()))
ys = list(map(int,input().split()))
result = "War"
if X > Y:
result = "No War"
xMax = max(xs)
yMin = min(ys)
if xMax > yMin:
if result == "War":
result = "No War"
print(result)
|
s104885324
|
Accepted
| 18
| 3,064
| 622
|
def main():
M, N, X, Y = map(int, input().split())
xs = list(map(int,input().split()))
ys = list(map(int,input().split()))
zs = []
for n in range(X+1,Y+1):
zs.append(n)
xMax = max(xs)
yMin = min(ys)
tmp = []
for z in zs:
if z > xMax:
tmp.append(z)
if z <= yMin:
tmp.append(z)
flag = False
for z in zs:
if tmp.count(z) > 1:
flag = True
if len(zs) == 0:
print("War")
return 0
if flag:
print("No War")
else:
print("War")
if __name__ == "__main__":
main()
|
s171405590
|
p02645
|
u030278108
| 2,000
| 1,048,576
|
Wrong Answer
| 24
| 9,560
| 94
|
When you asked some guy in your class his name, he called himself S, where S is a string of length between 3 and 20 (inclusive) consisting of lowercase English letters. You have decided to choose some three consecutive characters from S and make it his nickname. Print a string that is a valid nickname for him.
|
import random
S = str(input())
A = list(S)
N = random.sample(A, k=3)
print(N[0]+N[1]+N[2])
|
s859550189
|
Accepted
| 25
| 9,536
| 59
|
import random
S = str(input())
A = list(S)
print(S[0:3])
|
s749438849
|
p02833
|
u557565572
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 179
|
For an integer n not less than 0, let us define f(n) as follows: * f(n) = 1 (if n < 2) * f(n) = n f(n-2) (if n \geq 2) Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N).
|
n = int(input())
ans = 0
if n%2==0:
n /= 2
fives = 5
while 1:
if fives > n:
break
ans += n // fives
fives *= 5
print(ans)
|
s330181816
|
Accepted
| 17
| 3,060
| 182
|
n = int(input())
ans = 0
if n % 2 == 0:
n = n // 2
fives = 5
while 1:
if fives > n:
break
ans += n // fives
fives *= 5
print(ans)
|
s491704532
|
p02694
|
u949618949
| 2,000
| 1,048,576
|
Wrong Answer
| 21
| 9,024
| 87
|
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?
|
X = int(input())
S = 100
C = 0
while X >= S:
S = int(S * 1.01)
C += 1
print(C)
|
s320381814
|
Accepted
| 23
| 9,156
| 86
|
X = int(input())
S = 100
C = 0
while X > S:
S = int(S * 1.01)
C += 1
print(C)
|
s315689426
|
p04043
|
u771538568
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 124
|
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
a,b,c=map(int,input().split())
d=min(a,b,c)
e=max(a,b,c)
if d==5 and e==7 and a+b+c==17:
print("Yes")
else:
print("No")
|
s492901655
|
Accepted
| 17
| 3,060
| 123
|
a,b,c=map(int,input().split())
d=min(a,b,c)
e=max(a,b,c)
if d==5 and e==7 and a+b+c==17:
print("YES")
else:
print("NO")
|
s719363869
|
p03371
|
u358254559
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 362
|
"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively. Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.
|
A,B,C,X,Y = map(int, input().split())
res=0
if X > 0:
if A > 2 * C:
res += C * 2*X
Y -= X
else:
res += X * A
if Y > 0:
if B > 2 * C:
res += C*2*Y
else:
res+= Y * B
print(res)
|
s020746378
|
Accepted
| 17
| 3,064
| 573
|
A,B,C,X,Y = map(int, input().split())
res=0
if X>0 and Y>0:
if A+B > 2*C:
minxy = min(X,Y)
res += 2 * C * minxy
X -= minxy
Y -= minxy
# here, X == 0 or Y == 0
if X > 0:
if A > 2 * C:
res += C * 2*X
Y -= X
else:
res += X * A
if Y > 0:
if B > 2 * C:
res += C*2*Y
else:
res+= Y * B
print(res)
|
s867493645
|
p03555
|
u983918956
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 122
|
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
s,t = [input() for i in range(2)]
if s[0] == t[2] and s[1] == t[1] and s[2] == t[0]:
print("Yes")
else:
print("No")
|
s460643616
|
Accepted
| 17
| 3,060
| 119
|
s,t = [input() for i in range(2)]
if s[0] == t[2] and s[1] == t[1] and s[2] == t[0]:
print("YES")
else:
print("NO")
|
s420631101
|
p03455
|
u177775074
| 2,000
| 262,144
|
Wrong Answer
| 24
| 8,928
| 85
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a,b = map(int,input().split())
if a*b // 2 == 0:
print("Even")
else:
print("Odd")
|
s696990860
|
Accepted
| 24
| 9,100
| 85
|
a,b = map(int,input().split())
if a*b % 2 == 0:
print("Even")
else:
print("Odd")
|
s807936665
|
p03610
|
u153147777
| 2,000
| 262,144
|
Wrong Answer
| 29
| 3,572
| 70
|
You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.
|
s = input()
print(''.join([c for i, c in enumerate(s) if i % 2 == 1]))
|
s159722426
|
Accepted
| 29
| 3,572
| 70
|
s = input()
print(''.join([c for i, c in enumerate(s) if i % 2 == 0]))
|
s094706954
|
p03456
|
u014333473
| 2,000
| 262,144
|
Wrong Answer
| 27
| 8,916
| 59
|
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.
|
a,b=input().split();print('EOvdedn'[eval(a+'*'+b)%2!=0::2])
|
s523308169
|
Accepted
| 25
| 9,332
| 69
|
a,b=input().split();print('YNeos'[int(a+b)!=int(int(a+b)**.5)**2::2])
|
s993646128
|
p03814
|
u513081876
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,516
| 94
|
Snuke has decided to construct a string that starts with `A` and ends with `Z`, by taking out a substring of a string s (that is, a consecutive part of s). Find the greatest length of the string Snuke can construct. Here, the test set guarantees that there always exists a substring of s that starts with `A` and ends with `Z`.
|
S = input()
SS = S[::-1]
ind_A = S.index('A')
ind_Z = SS.index('Z')
print(len(S[ind_A:ind_Z]))
|
s979347026
|
Accepted
| 44
| 3,500
| 179
|
s = input()
for ind, i in enumerate(s):
if i == 'A':
num1 = ind
break
for ind, i in enumerate(s):
if i == 'Z':
num2 = ind
print(num2 - num1 + 1)
|
s621267076
|
p03455
|
u829393411
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 84
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a,b=map(int,input().split())
if a%2==0 or b%2==0:
print("Odd")
else:
print("Even")
|
s657888436
|
Accepted
| 17
| 2,940
| 87
|
a , b =map(int,input().split())
if a%2==0 or b%2==0:
print("Even")
else:
print("Odd")
|
s839372469
|
p02394
|
u042885182
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,480
| 671
|
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
|
# coding: utf-8
# Here your code !
import math
def func():
try:
line=input().rstrip()
numbers=line.split(" ")
width=float(numbers[0])
height=float(numbers[1])
x=float(numbers[2])
y=float(numbers[3])
r=abs(float(numbers[4]))
except:
print("input error")
return -1
wsign=math.copysign(1,width)
hsign=math.copysign(1,height)
result=(abs(x)+r <= wsign*width)
result=result and (abs(x)-r >= 0)
result=result and (abs(y)+r <= hsign*height)
result=result and (abs(y)-r >= 0)
if result:
print("Yes")
else:
print("No")
|
s925843859
|
Accepted
| 20
| 7,660
| 666
|
# coding: utf-8
# Here your code !
import math
def func():
try:
line=input().rstrip()
numbers=line.split(" ")
width=float(numbers[0])
height=float(numbers[1])
x=float(numbers[2])
y=float(numbers[3])
r=abs(float(numbers[4]))
except:
print("input error")
return -1
wsign=math.copysign(1,width)
hsign=math.copysign(1,height)
result="No"
if(wsign*x+r <= wsign*width):
if(wsign*x-r >= 0):
if(hsign*y+r <= hsign*height):
if(hsign*y-r >= 0):
result="Yes"
print(result)
func()
|
s690956531
|
p03048
|
u227085629
| 2,000
| 1,048,576
|
Time Limit Exceeded
| 2,104
| 2,940
| 167
|
Snuke has come to a store that sells boxes containing balls. The store sells the following three kinds of boxes: * Red boxes, each containing R red balls * Green boxes, each containing G green balls * Blue boxes, each containing B blue balls Snuke wants to get a total of exactly N balls by buying r red boxes, g green boxes and b blue boxes. How many triples of non-negative integers (r,g,b) achieve this?
|
r,g,b,n = map(int,input().split())
c = 0
for i in range(3001):
for j in range(3001):
for k in range(3001):
if i*r + j*g + k*b == n:
c += 1
print(c)
|
s519749978
|
Accepted
| 1,936
| 2,940
| 166
|
r,g,b,n = map(int,input().split())
c = 0
for i in range(n//r+1):
for j in range((n-i*r)//g+1):
k = (n-i*r-j*g)
if k%b == 0 and k >= 0:
c += 1
print(c)
|
s811323533
|
p03680
|
u762008592
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 176
|
Takahashi wants to gain muscle, and decides to work out at AtCoder Gym. The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up. These buttons are numbered 1 through N. When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i. When Button i is not lighten up, nothing will happen by pressing it. Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up. Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons.
|
N = int(input())
a_n=list(map(int,input().split()))
if 2 not in a_n:
print(-1)
else:
for i in range(N):
if a_n[i]==2:
print(i+1)
break
|
s488074818
|
Accepted
| 193
| 7,084
| 231
|
N = int(input())
a_n=[int(input()) for i in range(N)]
import sys
i=1
count=0
for k in range(N):
if a_n[i-1]==2:
count+=1
print(count)
sys.exit()
else:
i=a_n[i-1]
count+=1
print(-1)
|
s351464665
|
p02612
|
u888933875
| 2,000
| 1,048,576
|
Wrong Answer
| 25
| 8,924
| 28
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n=int(input())
print(n%1000)
|
s476217151
|
Accepted
| 26
| 9,084
| 72
|
n=int(input())
if n%1000 ==0:
print(n%1000)
else:
print(1000-n%1000)
|
s137406738
|
p03845
|
u325206354
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 220
|
Joisino is about to compete in the final round of a certain programming competition. In this contest, there are N problems, numbered 1 through N. Joisino knows that it takes her T_i seconds to solve problem i(1≦i≦N). Also, there are M kinds of drinks offered to the contestants, numbered 1 through M. If Joisino takes drink i(1≦i≦M), her brain will be stimulated and the time it takes for her to solve problem P_i will become X_i seconds. It does not affect the time to solve the other problems. A contestant is allowed to take exactly one of the drinks before the start of the contest. For each drink, Joisino wants to know how many seconds it takes her to solve all the problems if she takes that drink. Here, assume that the time it takes her to solve all the problems is equal to the sum of the time it takes for her to solve individual problems. Your task is to write a program to calculate it instead of her.
|
x=int(input())
a=list(map(int,input().split()))
d=int(input())
n=0
A=a
j=[list(map(int,input().split())) for i in range(d)]
for T in range(d):
s=j[n][0]
A[s-1]=j[n][1]
print(sum(A))
n+=1
A[s-1]=a[s-1]
|
s348028837
|
Accepted
| 18
| 3,064
| 315
|
x=int(input())
first=list(map(int,input().split()))
drink_count=int(input())
count=0
Total=sum(first)
drink=[list(map(int,input().split())) for i in range(drink_count)]
for T in range(drink_count):
change_num = drink[count][0]
score = first[change_num-1]-drink[count][1]
print(Total-score)
count+=1
|
s124398564
|
p04043
|
u645127738
| 2,000
| 262,144
|
Wrong Answer
| 23
| 8,952
| 34
|
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
a,b,c=input().split()
print(a,b,c)
|
s222205043
|
Accepted
| 24
| 9,028
| 186
|
a,b,c=input().split()
if a=="5" and b=="5" and c=="7":
print("YES")
elif a=="5" and b=="7" and c=="5":
print("YES")
elif a=="7" and b=="5" and c=="5":
print("YES")
else:
print("NO")
|
s432033443
|
p03636
|
u690037900
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 58
|
The word `internationalization` is sometimes abbreviated to `i18n`. This comes from the fact that there are 18 letters between the first `i` and the last `n`. You are given a string s of length at least 3 consisting of lowercase English letters. Abbreviate s in the same way.
|
a = list(input())
num = len(a)
print(a[0]+str(num)+a[-1])
|
s359295347
|
Accepted
| 17
| 2,940
| 49
|
a = list(input())
print(a[0]+str(len(a)-2)+a[-1])
|
s164495797
|
p02742
|
u487767879
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 103
|
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:
|
H, W = list(map(int, input().split()))
res = int((H+1)/2) * W
if W%2==1:
res -= int(W/2)
print(res)
|
s929512122
|
Accepted
| 17
| 3,064
| 144
|
H, W = list(map(int, input().split()))
if H==1 or W==1:
print(1)
exit()
res = int((H+1)/2) * W
if H%2==1:
res -= int(W/2)
print(res)
|
s658069974
|
p03795
|
u901598613
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 45
|
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
n=int(input())
x=500*n
y=n//15*200
print(x-y)
|
s289064772
|
Accepted
| 17
| 2,940
| 45
|
n=int(input())
x=800*n
y=n//15*200
print(x-y)
|
s600245217
|
p03860
|
u302292660
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 72
|
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
con = input().split()
cap = con[2][0]
name = "A" + cap + "C"
print(name)
|
s642963723
|
Accepted
| 19
| 3,060
| 72
|
con = input().split()
cap = con[1][0]
name = "A" + cap + "C"
print(name)
|
s869260217
|
p03485
|
u629276590
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 100
|
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
a,b=map(int,input().split())
x=(a+b)/2
if x*10%10==5:
print(x*10//10+1)
else:
print(x)
|
s956269430
|
Accepted
| 17
| 2,940
| 109
|
a,b=map(int,input().split())
x=(a+b)/2
if x*10%10==5:
print(int(x*10//10+1))
else:
print(int(x))
|
s209397682
|
p03610
|
u934052933
| 2,000
| 262,144
|
Wrong Answer
| 40
| 3,700
| 128
|
You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.
|
s = input()
counter = 0
res = []
for i in s:
counter += 1
if counter % 2 != 0:
res.append(i)
print("res:","".join(res))
|
s160233783
|
Accepted
| 39
| 3,572
| 121
|
s = input()
counter = 0
res = []
for i in s:
counter += 1
if counter % 2 != 0:
res.append(i)
print("".join(res))
|
s658881056
|
p01101
|
u398978447
| 8,000
| 262,144
|
Wrong Answer
| 40
| 5,648
| 389
|
Mammy decided to give Taro his first shopping experience. Mammy tells him to choose any two items he wants from those listed in the shopping catalogue, but Taro cannot decide which two, as all the items look attractive. Thus he plans to buy the pair of two items with the highest price sum, not exceeding the amount Mammy allows. As getting two of the same item is boring, he wants two different items. You are asked to help Taro select the two items. The price list for all of the items is given. Among pairs of two items in the list, find the pair with the highest price sum not exceeding the allowed amount, and report the sum. Taro is buying two items, not one, nor three, nor more. Note that, two or more items in the list may be priced equally.
|
n=5
m=5
sum=0
A=[]
while n!=0:
n,m=[int(i) for i in input().split()]
if n==0 and m==0:
break
val=[int(i) for i in input().split()]
m1=m2=val[0]
for i in range(1,n):
if val[i]>=m1:
m1=val[i]
elif val[i]>m2:
m2=val[i];
sum=m1+m2
if sum>m:
sum=0
A.append(sum)
for i in range(len(A)):
print(A[i])
|
s113845160
|
Accepted
| 1,990
| 5,716
| 481
|
n=5
m=5
z=0
sum=0
A=[]
while n!=0:
n,m=[int(i) for i in input().split()]
if n==0 and m==0:
break
val=[int(i) for i in input().split()]
ans1=0
mm=0
for i in range(0,n):
for j in range(i+1,n):
if i==j:
break
ans1=val[i]+val[j]
if ans1>mm and ans1<=m:
mm=ans1
ans1=0
A.append(mm)
z=z+1
for i in range(z):
if A[i]==0:
A[i]="NONE"
print(A[i])
|
s337911984
|
p03438
|
u214617707
| 2,000
| 262,144
|
Wrong Answer
| 24
| 4,600
| 238
|
You are given two integer sequences of length N: a_1,a_2,..,a_N and b_1,b_2,..,b_N. Determine if we can repeat the following operation zero or more times so that the sequences a and b become equal. Operation: Choose two integers i and j (possibly the same) between 1 and N (inclusive), then perform the following two actions **simultaneously** : * Add 2 to a_i. * Add 1 to b_j.
|
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
k = 0
for i in a:
if i % 2 == 0:
k += 1
for j in b:
if j % 2 == 1:
k -= 1
if k == 0:
print("YES")
else:
print("NO")
|
s603291144
|
Accepted
| 27
| 4,596
| 271
|
N = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
p, q = 0, 0
for i in range(N):
if a[i] > b[i]:
p += a[i] - b[i]
elif a[i] < b[i]:
q += (b[i] - a[i]) // 2
if p <= q:
print('Yes')
else:
print('No')
|
s905779868
|
p02394
|
u303842929
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,604
| 349
|
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
|
value = input().split()
W = int(value[0])
H = int(value[1])
coordinate = [[0 for _ in range(W)] for _ in range(H)]
x = int(value[2])
y = int(value[3])
r =int(value[4])
try:
tmp = coordinate[y][x + r]
tmp = coordinate[y][x - r]
tmp = coordinate[y + r][x]
tmp = coordinate[y - r][x]
print("YES")
except Exception:
print("NO")
|
s213636211
|
Accepted
| 20
| 5,644
| 455
|
value = input().split()
W = int(value[0])
H = int(value[1])
coordinate = [[0 for _ in range(W + 1)] for _ in range(H + 1)]
x = int(value[2])
y = int(value[3])
r =int(value[4])
try:
if x < 0 or y < 0:
print("No")
else:
tmp = coordinate[y][x + r]
tmp = coordinate[y][x - r]
tmp = coordinate[y + r][x]
tmp = coordinate[y - r][x]
print("Yes")
except Exception:
print("No")
|
s625713369
|
p02272
|
u003309334
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,768
| 783
|
Write a program of a Merge Sort algorithm implemented by the following pseudocode. You should also report the number of comparisons in the Merge function. Merge(A, left, mid, right) n1 = mid - left; n2 = right - mid; create array L[0...n1], R[0...n2] for i = 0 to n1-1 do L[i] = A[left + i] for i = 0 to n2-1 do R[i] = A[mid + i] L[n1] = SENTINEL R[n2] = SENTINEL i = 0; j = 0; for k = left to right-1 if L[i] <= R[j] then A[k] = L[i] i = i + 1 else A[k] = R[j] j = j + 1 Merge-Sort(A, left, right){ if left+1 < right then mid = (left + right)/2; call Merge-Sort(A, left, mid) call Merge-Sort(A, mid, right) call Merge(A, left, mid, right)
|
def merge(A, left, mid, right):
cnt = 0
n1 = mid - left
n2 = right - mid
L, R = [0]*(n1+1), [0]*(n2+1)
for i in range(n1):
L[i] = A[left + i]
for i in range(n2):
R[i] = A[mid + i]
L[n1] = float("inf")
R[n2] = float("inf")
i = 0
j = 0
for k in range(left, right):
cnt = cnt + 1
if L[i] <= R[j]:
A[k] = L[i]
i = i + 1
else:
A[k] = R[j]
j = j + 1
return cnt
def mergeSort(A, left, right, cnt):
if left+1 < right:
mid = int((left + right)/2)
mergeSort(A, left, mid, cnt)
mergeSort(A, mid, right, cnt)
cnt[0] = cnt[0] + merge(A, left, mid, right)
if __name__ == "__main__":
n = int(input())
S = list(map(int, input().split()))
cnt = [0]
mergeSort(S, 0, len(S), cnt)
print(S)
print(cnt[0])
|
s963555256
|
Accepted
| 4,750
| 68,732
| 673
|
count = 0
def merge(A, left, mid, right):
global count
n1 = mid - left
n2 = right - mid
L = A[left:left+n1]
R = A[mid:mid+n2]
L.append(float("inf"))
R.append(float("inf"))
i = 0
j = 0
for k in range(left, right):
count += 1
if L[i] <= R[j]:
A[k] = L[i]
i += 1
else:
A[k] = R[j]
j += 1
def mergeSort(A, left, right):
if left + 1 < right:
mid = int((left + right)/2)
mergeSort(A, left, mid)
mergeSort(A, mid, right)
merge(A, left, mid, right)
if __name__ == "__main__":
n = int(input())
S = list(map(int, input().split()))
mergeSort(S, 0, len(S))
print(" ".join(map(str, S)))
print(count)
|
s055126744
|
p02388
|
u176580590
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,508
| 59
|
Write a program which calculates the cube of a given integer x.
|
def f(x):
if 1 <= x <= 100:
print("error")
return x**3
|
s977677854
|
Accepted
| 20
| 5,576
| 29
|
x = int(input())
print(x**3)
|
s922887280
|
p00002
|
u553148578
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,652
| 89
|
Write a program which computes the digit number of sum of two integers a and b.
|
import math
a, b = map(int, input().split())
ans = a + b
print(int(math.log10(ans)) + 1)
|
s271268975
|
Accepted
| 20
| 5,584
| 93
|
while True:
try:
a, b = map(int, input().split())
print(len(str(a+b)))
except:
break
|
s323460057
|
p03455
|
u840497549
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 114
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = input().split(" ")
a= int(a)
b= int(b)
c = a*b
if c%2 ==0:
print("even")
else:
print("odd")
|
s572794260
|
Accepted
| 17
| 2,940
| 110
|
a, b= input().split()
a = int(a)
b = int(b)
if a*b % 2 == 0:
print("Even")
else:
print("Odd")
|
s805571858
|
p03759
|
u846218099
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 64
|
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful.
|
a,b,c=map(int,input().split())
print("YES" if a==b==c else "NO")
|
s005080198
|
Accepted
| 17
| 2,940
| 65
|
a,b,c=map(int,input().split())
print("YES" if b-a==c-b else "NO")
|
s403126582
|
p02409
|
u923668099
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,668
| 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.
|
# coding: utf-8
# Here your code !
num_j = [[[0 for _ in range(10)] for _ in range(3)] for _ in range(4)]
n = int(input())
for _ in range(n):
b,r,f,v = map(int,input().split())
num_j[b-1][r-1][f-1] += v
for i in range(4):
for j in range(3):
print(" ",end="")
print(" ".join(map(str,num_j[i][j])))
print("#"*20)
|
s898076251
|
Accepted
| 30
| 7,712
| 355
|
# coding: utf-8
# Here your code !
num_j = [[[0 for _ in range(10)] for _ in range(3)] for _ in range(4)]
n = int(input())
for _ in range(n):
b,r,f,v = [int(x) for x in input().split()]
num_j[b-1][r-1][f-1] += v
for i in range(4):
for f in num_j[i]:
print(""," ".join([str(x) for x in f]))
if i != 3:
print("#"*20)
|
s095193276
|
p04043
|
u539675863
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 112
|
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
a,b,c = sorted(map(int,input().split()))
if a == 5 and b == 7 and c == 5:
print("YES")
else:
print("NO")
|
s180536902
|
Accepted
| 17
| 2,940
| 116
|
go = list(map(int,input().split()))
if go.count(5) == 2 and go.count(7) == 1:
print("YES")
else:
print("NO")
|
s376567918
|
p03044
|
u950708010
| 2,000
| 1,048,576
|
Wrong Answer
| 1,782
| 25,028
| 1,013
|
We have a tree with N vertices numbered 1 to N. The i-th edge in the tree connects Vertex u_i and Vertex v_i, and its length is w_i. Your objective is to paint each vertex in the tree white or black (it is fine to paint all vertices the same color) so that the following condition is satisfied: * For any two vertices painted in the same color, the distance between them is an even number. Find a coloring of the vertices that satisfies the condition and print it. It can be proved that at least one such coloring exists under the constraints of this problem.
|
class UnionFind(object):
def __init__(self, n=1):
self.par = [i for i in range(n)]
self.rank = [0 for _ in range(n)]
def find(self, x):
if self.par[x] == x:
return x
else:
self.par[x] = self.find(self.par[x])
return self.par[x]
def union(self, x, y):
x = self.find(x)
y = self.find(y)
if x != y:
if self.rank[x] < self.rank[y]:
x, y = y, x
if self.rank[x] == self.rank[y]:
self.rank[x] += 1
self.par[y] = x
def is_same(self, x, y):
return self.find(x) == self.find(y)
def solve():
n = int(input())
uf = UnionFind(n)
ans = [0]*n
tmp = []
for i in range(n-1):
x,y,z = (int(i) for i in input().split())
if z%2 == 1:
tmp.append((x,y,z))
else:
uf.union(x-1,y-1)
ans[x-1] = 1
ans[y-1] = 1
while tmp:
x,y,z = tmp.pop(0)
for i in ans:
print(i)
solve()
|
s944346894
|
Accepted
| 485
| 36,016
| 719
|
import sys
input = sys.stdin.readline
from collections import deque
def solve():
n = int(input())
treenode = [[] for _ in range(n)]
treecolor = [-2]*n
for i in range(n-1):
u,v,w = (int(i) for i in input().split())
u -= 1
v -= 1
treenode[u].append((v,w%2))
treenode[v].append((u,w%2))
query = deque([(0,0,0)])
while query:
a,b,c = query.popleft()
if b%2 == 0:
treecolor[a] = c
nextc = c
else:
treecolor[a] = 1^c
nextc = 1^c
for nextnode,weight in treenode[a]:
if treecolor[nextnode] != -2:
continue
else:
query.append((nextnode,weight,nextc))
treecolor[nextnode] = -1
for i in treecolor:
print(i)
solve()
|
s682349106
|
p02841
|
u370721525
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 124
|
In this problem, a date is written as Y-M-D. For example, 2019-11-30 means November 30, 2019. Integers M_1, D_1, M_2, and D_2 will be given as input. It is known that the date 2019-M_2-D_2 follows 2019-M_1-D_1. Determine whether the date 2019-M_1-D_1 is the last day of a month.
|
M_1, D_1 = map(int, input().split())
M_2, D_2 = map(int, input().split())
if M_1 != M_2:
print('No')
else:
print('Yes')
|
s678272766
|
Accepted
| 17
| 2,940
| 118
|
M_1, D_1 = map(int, input().split())
M_2, D_2 = map(int, input().split())
if M_1 != M_2:
print(1)
else:
print(0)
|
s118557197
|
p02972
|
u151785909
| 2,000
| 1,048,576
|
Wrong Answer
| 280
| 22,452
| 284
|
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.
|
n = int(input())
a = list(map(int,input().split()))
b = [0]*n
for i in range(n):
b1=b[n-1-i::n-1-i+1]
if a[n-1-i]==0 and sum(b1)%2==0:
b[n-1-i]=0
else:
b[n-1-i]=1
ans =[]
for i in range(n):
if b[i]==1:
ans.append(str(i+1))
print(' '.join(ans))
|
s716800943
|
Accepted
| 313
| 19,508
| 339
|
n = int(input())
a = list(map(int,input().split()))
b = [0]*n
for i in range(n):
b1=b[n-1-i::n-1-i+1]
su=sum(b1)
if (a[n-1-i]==0 and su%2==0) or (a[n-1-i]==1 and su%2==1):
b[n-1-i]=0
else:
b[n-1-i]=1
ans =[]
print(sum(b))
for i in range(n):
if b[i]==1:
ans.append(str(i+1))
print(' '.join(ans))
|
s807375963
|
p03543
|
u496821919
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 267
|
We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**?
|
X = list(input())
for i in range(2**3):
W = ["+"]*3
for j in range(3):
if (i >> j) & 1:
W[j] = "-"
formula = ""
for k,l in zip(X,W+[""]):
formula += k+l
if eval(formula) == 7:
print(formula+"=7")
break
|
s559470700
|
Accepted
| 17
| 2,940
| 214
|
# -*- coding: utf-8 -*-
"""
Created on Wed May 13 11:36:51 2020
@author: shinba
"""
n = list(input())
if (n[0] == n[1] and n[1] == n[2]) or (n[1] == n[2] and n[2] == n[3]):
print("Yes")
else:
print("No")
|
s632455040
|
p03549
|
u331128419
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 90
|
Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`. When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases. Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds. Now, he goes through the following process: * Submit the code. * Wait until the code finishes execution on all the cases. * If the code fails to correctly solve some of the M cases, submit it again. * Repeat until the code correctly solve all the cases in one submission. Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).
|
N, M = map(int, input().split())
time = (M * 1900 + (N-M) * 100) / (0.5 ** M)
print(time)
|
s388875486
|
Accepted
| 17
| 2,940
| 95
|
N, M = map(int, input().split())
time = (M * 1900 + (N-M) * 100) / (0.5 ** M)
print(int(time))
|
s109882458
|
p03162
|
u212786022
| 2,000
| 1,048,576
|
Wrong Answer
| 2,106
| 93,140
| 405
|
Taro's summer vacation starts tomorrow, and he has decided to make plans for it now. The vacation consists of N days. For each i (1 \leq i \leq N), Taro will choose one of the following activities and do it on the i-th day: * A: Swim in the sea. Gain a_i points of happiness. * B: Catch bugs in the mountains. Gain b_i points of happiness. * C: Do homework at home. Gain c_i points of happiness. As Taro gets bored easily, he cannot do the same activities for two or more consecutive days. Find the maximum possible total points of happiness that Taro gains.
|
N = int(input())
T = [list(map(int, input().split())) for i in range(N)]
print(T)
dp = [[0 for i in range(3)] for j in range(N+1)]
print(dp)
dp[1] = T[0]
print(dp)
for i in range(2,N+1):
dp[i][0] = max(dp[i-1][1]+T[i-1][0],dp[i-1][2]+T[i-1][0])
dp[i][1] = max(dp[i-1][0]+T[i-1][1],dp[i-1][2]+T[i-1][1])
dp[i][2] = max(dp[i-1][0]+T[i-1][2],dp[i-1][1]+T[i-1][2])
print(dp)
print(max(dp[-1]))
|
s056289217
|
Accepted
| 633
| 47,348
| 362
|
N = int(input())
T = [list(map(int, input().split())) for i in range(N)]
dp = [[0 for i in range(3)] for j in range(N+1)]
dp[1] = T[0]
for i in range(2,N+1):
dp[i][0] = max(dp[i-1][1]+T[i-1][0],dp[i-1][2]+T[i-1][0])
dp[i][1] = max(dp[i-1][0]+T[i-1][1],dp[i-1][2]+T[i-1][1])
dp[i][2] = max(dp[i-1][0]+T[i-1][2],dp[i-1][1]+T[i-1][2])
print(max(dp[-1]))
|
s212929694
|
p03679
|
u600402037
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 260
|
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though. Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious. Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it. Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
|
import sys
stdin = sys.stdin
ri = lambda: int(rs())
rl = lambda: list(map(int, stdin.readline().split()))
rs = lambda: stdin.readline().rstrip() # ignore trailing spaces
X, A, B = rl()
print('delicious' if A >= B else 'safe' if X >= A + B else 'dangerous')
|
s742560618
|
Accepted
| 17
| 3,060
| 260
|
import sys
stdin = sys.stdin
ri = lambda: int(rs())
rl = lambda: list(map(int, stdin.readline().split()))
rs = lambda: stdin.readline().rstrip() # ignore trailing spaces
X, A, B = rl()
print('delicious' if A >= B else 'safe' if X >= B - A else 'dangerous')
|
s757210784
|
p03151
|
u871841829
| 2,000
| 1,048,576
|
Wrong Answer
| 40
| 11,192
| 193
|
A university student, Takahashi, has to take N examinations and pass all of them. Currently, his _readiness_ for the i-th examination is A_{i}, and according to his investigation, it is known that he needs readiness of at least B_{i} in order to pass the i-th examination. Takahashi thinks that he may not be able to pass all the examinations, and he has decided to ask a magician, Aoki, to change the readiness for as few examinations as possible so that he can pass all of them, while not changing the total readiness. For Takahashi, find the minimum possible number of indices i such that A_i and C_i are different, for a sequence C_1, C_2, ..., C_{N} that satisfies the following conditions: * The sum of the sequence A_1, A_2, ..., A_{N} and the sum of the sequence C_1, C_2, ..., C_{N} are equal. * For every i, B_i \leq C_i holds. If such a sequence C_1, C_2, ..., C_{N} cannot be constructed, print -1.
|
N = int(input())
A = map(int, input().split())
ans = 1
import sys
if 0 in A:
print(0)
sys.exit()
for a in A:
ans *= a
if ans > pow(10, 18):
print(-1)
sys.exit()
print(ans)
|
s425619458
|
Accepted
| 109
| 24,212
| 471
|
N = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
diff = [0] * N
neg = []
pos = []
for i in range(N):
diff[i] = A[i] - B[i]
if diff[i] < 0:
neg.append(diff[i])
else:
pos.append(diff[i])
if sum(diff) < 0:
print(-1)
exit(0)
tobe_supplied = sum(neg)
pos.sort(reverse=True)
cnt = len(neg)
for p in pos:
if tobe_supplied >= 0:
break
cnt += 1
tobe_supplied += p
print(cnt)
|
s627076093
|
p03658
|
u241159583
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 117
|
Snuke has N sticks. The length of the i-th stick is l_i. Snuke is making a snake toy by joining K of the sticks together. The length of the toy is represented by the sum of the individual sticks that compose it. Find the maximum possible length of the toy.
|
N, K = map(int, input().split())
l = list(map(int, input().split()))
l.sort(reverse=True)
print(l)
print(sum(l[:K]))
|
s978412604
|
Accepted
| 17
| 2,940
| 96
|
N, K = map(int, input().split())
l = list(map(int, input().split()))
l.sort()
print(sum(l[-K:]))
|
s336281308
|
p03417
|
u130900604
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 123
|
There is a grid with infinitely many rows and columns. In this grid, there is a rectangular region with consecutive N rows and M columns, and a card is placed in each square in this region. The front and back sides of these cards can be distinguished, and initially every card faces up. We will perform the following operation once for each square contains a card: * For each of the following nine squares, flip the card in it if it exists: the target square itself and the eight squares that shares a corner or a side with the target square. It can be proved that, whether each card faces up or down after all the operations does not depend on the order the operations are performed. Find the number of cards that face down after all the operations.
|
n,m=map(int,input().split())
if (n,m)==(1,1):
print(1)
elif min(n,m)==1:
print(max(n,m)-2)
else:
print((n-2)*(m-2)+4)
|
s039386992
|
Accepted
| 17
| 2,940
| 122
|
n,m=map(int,input().split())
if (n,m)==(1,1):
print(1)
elif min(n,m)==1:
print(max(n,m)-2)
else:
print((n-2)*(m-2))
|
s171021576
|
p02927
|
u375616706
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 631
|
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?
|
# python template for atcoder1
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline
M, D = map(int, input().split())
cand = set()
for m in range(1, M+1):
i = 2
while i*i <= D:
if m % i == 0:
ma = m//i
mi = i
if mi <= 1 or ma <= 1:
pass
elif int(str(mi)+str(ma)) <= D:
cand.add(int(str(m)+str(mi)+str(ma)))
print("mi,ma->", mi, ma)
elif int(str(ma)+str(mi)) <= D:
cand.add(int(str(m)+str(ma)+str(mi)))
print("ma,mi->", ma, mi)
i += 1
print(len(cand))
|
s830721938
|
Accepted
| 18
| 3,064
| 549
|
# python template for atcoder1
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline
M, D = map(int, input().split())
cand = set()
for m in range(1, M+1):
i = 2
while i*i <= D:
if m % i == 0:
ma = m//i
mi = i
if mi <= 1 or ma <= 1:
pass
elif int(str(mi)+str(ma)) <= D:
cand.add(int(str(m)+str(mi)+str(ma)))
elif int(str(ma)+str(mi)) <= D:
cand.add(int(str(m)+str(ma)+str(mi)))
i += 1
print(len(cand))
|
s620046091
|
p03068
|
u297756089
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,316
| 142
|
You are given a string S of length N consisting of lowercase English letters, and an integer K. Print the string obtained by replacing every character in S that differs from the K-th character of S, with `*`.
|
n=int(input())
s=list(input())
k=int(input())
num=s[k-1]
num=str(num)
for i in range (0,n):
if s[i]==num:
s[i]="*"
s=''.join(s)
print(s)
|
s074686960
|
Accepted
| 17
| 3,060
| 142
|
n=int(input())
s=list(input())
k=int(input())
num=s[k-1]
num=str(num)
for i in range (0,n):
if s[i]!=num:
s[i]="*"
s=''.join(s)
print(s)
|
s287026531
|
p03644
|
u017271745
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 191
|
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
N = int(input())
check = 0
ans = 1
for i in range(1, N+1):
cnt = 0
while i%2 == 0:
cnt += 1
i //= 2
if cnt > check:
ans = i
check = cnt
print(ans)
|
s702343433
|
Accepted
| 17
| 3,060
| 201
|
N = int(input())
check = 0
ans = 1
for i in range(1, N+1):
cnt = 0
j = i
while j%2 == 0:
cnt += 1
j //= 2
if cnt > check:
ans = i
check = cnt
print(ans)
|
s864186040
|
p04029
|
u739959951
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 41
|
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
n=int(input())
ans=n*(n+1)/2
print(ans)
|
s964018924
|
Accepted
| 17
| 2,940
| 42
|
n=int(input())
ans=n*(n+1)//2
print(ans)
|
s351915971
|
p02238
|
u855199458
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,752
| 1,284
|
Depth-first search (DFS) follows the strategy to search ”deeper” in the graph whenever possible. In DFS, edges are recursively explored out of the most recently discovered vertex $v$ that still has unexplored edges leaving it. When all of $v$'s edges have been explored, the search ”backtracks” to explore edges leaving the vertex from which $v$ was discovered. This process continues until all the vertices that are reachable from the original source vertex have been discovered. If any undiscovered vertices remain, then one of them is selected as a new source and the search is repeated from that source. DFS timestamps each vertex as follows: * $d[v]$ records when $v$ is first discovered. * $f[v]$ records when the search finishes examining $v$’s adjacency list. Write a program which reads a directed graph $G = (V, E)$ and demonstrates DFS on the graph based on the following rules: * $G$ is given in an adjacency-list. Vertices are identified by IDs $1, 2,... n$ respectively. * IDs in the adjacency list are arranged in ascending order. * The program should report the discover time and the finish time for each vertex. * When there are several candidates to visit during DFS, the algorithm should select the vertex with the smallest ID. * The timestamp starts with 1.
|
# -*- coding: utf-8 -*-
N = int(input())
adj = [[0]*N for _ in range(N)]
searched = {n: 0 for n in range(N)}
edge = {}
def find_all(list_, x, j=-1):
return [i for i, n in enumerate(list_) if n==1 and n != j]
for n in range(N):
inp = list(map(int, input().split()))
edge[n] = inp[1]
for i in inp[2:]:
adj[inp[0]-1][i-1] = 1
time = 1
stack = []
unsearch = set(range(N))
uncomplete = set(range(N))
result = {n:{"found": 0, "finish": 0} for n in range(N)}
while len(unsearch) > 0 or len(stack) > 0:
if len(stack) == 0:
i = min(unsearch)
result[i]["found"] = time
unsearch.remove(i)
time += 1
while True:
j = i
for n in sorted(unsearch):
if adj[i][n] == 1:
j = n
result[n]["found"] = time
unsearch.remove(n)
stack.append(i)
time += 1
break
if i == j:
result[i]["finish"] = time
if len(stack) != 0:
i = stack.pop()
time += 1
else:
break
else:
i = j
for n in range(N):
print("{} {} {}".format(n, result[n]["found"], result[n]["finish"]))
|
s581546860
|
Accepted
| 20
| 7,832
| 951
|
# -*- coding: utf-8 -*-
import sys
sys.setrecursionlimit(100000)
N = int(input())
adj = [[0]*N for _ in range(N)]
searched = {n: 0 for n in range(N)}
edge = {}
for n in range(N):
inp = list(map(int, input().split()))
edge[n] = inp[1]
for i in inp[2:]:
adj[inp[0]-1][i-1] = 1
time = 1
stack = []
unsearch = set(range(N))
uncomplete = set(range(N))
result = {n:{"found": 0, "finish": 0} for n in range(N)}
def DFS(i):
global time
global unsearch
itr = sorted(unsearch)
for j in itr:
if adj[i][j] == 1 and j in unsearch:
result[j]["found"] = time
time += 1
unsearch.remove(j)
DFS(j)
result[i]["finish"] = time
time += 1
while len(unsearch):
i = min(unsearch)
unsearch.remove(i)
result[i]["found"] = time
time += 1
DFS(i)
for n in range(N):
print("{} {} {}".format(n+1, result[n]["found"], result[n]["finish"]))
|
s257070697
|
p03605
|
u227085629
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 71
|
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?
|
a=input()
if a[0] == 9 or a[1] == 9:
print('Yes')
else:
print('No')
|
s769215574
|
Accepted
| 17
| 2,940
| 78
|
a=int(input())
if a//10 == 9 or a%10 == 9:
print('Yes')
else:
print('No')
|
s778618072
|
p02271
|
u125481461
| 5,000
| 131,072
|
Wrong Answer
| 20
| 5,596
| 632
|
Write a program which reads a sequence _A_ of _n_ elements and an integer _M_ , and outputs "yes" if you can make _M_ by adding elements in _A_ , otherwise "no". You can use an element only once. You are given the sequence _A_ and _q_ questions where each question contains _M i_.
|
def can_construct_q(q, array, i, sum_sofar):
if sum_sofar == q or sum_sofar + array[i] == q:
return True
elif sum_sofar > q or i >= len(array) - 1:
return False
if can_construct_q(q, array, i + 1, sum_sofar + array[i]):
return True
if can_construct_q(q, array, i + 1, sum_sofar):
return True
def main():
input()
array_a = list(map(int, input().split()))
input()
array_q = map(int, input().split())
for q in array_q:
print(q)
if can_construct_q(q, array_a, 0, 0):
print('yes')
else:
print('no')
return
main()
|
s302068738
|
Accepted
| 360
| 5,624
| 731
|
def main():
input()
array_a = list(map(int, input().split()))
input()
array_q = map(int, input().split())
def can_construct_q (q, i, sum_sofar):
if sum_sofar == q or sum_sofar + array_a[i] == q:
return True
elif sum_sofar > q or i >= len (array_a) - 1:
return False
if can_construct_q(q, i + 1, sum_sofar + array_a[i]):
return True
if can_construct_q(q, i + 1, sum_sofar):
return True
sum_array_a = sum(array_a)
for q in array_q:
#print(q)
if sum_array_a < q:
print('no')
elif can_construct_q(q, 0, 0):
print('yes')
else:
print('no')
return
main()
|
s744106404
|
p03544
|
u011062360
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 99
|
It is November 18 now in Japan. By the way, 11 and 18 are adjacent Lucas numbers. You are given an integer N. Find the N-th Lucas number. Here, the i-th Lucas number L_i is defined as follows: * L_0=2 * L_1=1 * L_i=L_{i-1}+L_{i-2} (i≥2)
|
n = int(input())
l = [2, 1]
for _ in range(n):
k = l[-1] + l[-2]
l.append(k)
print(l[n-1])
|
s212252926
|
Accepted
| 17
| 2,940
| 98
|
n = int(input())
l = [2, 1]
for _ in range(n):
k = l[-1] + l[-2]
l.append(k)
print(l[-2])
|
s610403565
|
p03095
|
u905582793
| 2,000
| 1,048,576
|
Wrong Answer
| 28
| 4,340
| 134
|
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.
|
from collections import Counter
n = int(input())
s = list(input())
c = Counter(s)
ans = 1
for x in c.values():
ans *= x
print(ans-1)
|
s550374413
|
Accepted
| 26
| 4,260
| 151
|
from collections import Counter
n = int(input())
s = list(input())
c = Counter(s)
ans = 1
for x in c.values():
ans = ans*(x+1)%(10**9+7)
print(ans-1)
|
s465284411
|
p02972
|
u864213970
| 2,000
| 1,048,576
|
Wrong Answer
| 714
| 38,316
| 703
|
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())
arr = [int(x) for x in input().split()]
balls = {}
result = []
reslen = 0
for i in range(n, n // 2, -1):
if arr[i - 1] == 1:
balls[i - 1] = 1
result.append(str(i))
reslen += 1
else:
balls[i - 1] = 0
for i in range(n // 2, 0, -1):
s = sum([balls[i * m - 1] for m in range(2, n // i + 1)])
if arr[i - 1] == 1:
if (s % 2) == 1:
balls[i - 1] = 0
else:
balls[i - 1] = 1
result.append(str(i))
reslen += 1
if arr[i - 1] == 0:
if (s % 2) == 0:
balls[i - 1] = 0
else:
balls[i - 1] = 1
result.append(str(i))
reslen += 1
print(balls)
print('---')
print(reslen)
if reslen != 0:
print(' '.join(result))
|
s090300482
|
Accepted
| 672
| 38,440
| 677
|
n = int(input())
arr = [int(x) for x in input().split()]
balls = {}
result = []
reslen = 0
for i in range(n, n // 2, -1):
if arr[i - 1] == 1:
balls[i - 1] = 1
result.append(str(i))
reslen += 1
else:
balls[i - 1] = 0
for i in range(n // 2, 0, -1):
s = sum([balls[i * m - 1] for m in range(2, n // i + 1)])
if arr[i - 1] == 1:
if (s % 2) == 1:
balls[i - 1] = 0
else:
balls[i - 1] = 1
result.append(str(i))
reslen += 1
if arr[i - 1] == 0:
if (s % 2) == 0:
balls[i - 1] = 0
else:
balls[i - 1] = 1
result.append(str(i))
reslen += 1
print(reslen)
if reslen != 0:
print(' '.join(result))
|
s980547050
|
p03494
|
u867132034
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 125
|
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.
|
A=list(map(int,input().split()))
A
count=0
while all([a%2==0 for a in A]):
A = [a/2 for a in A]
count+=1
print(count)
|
s489635690
|
Accepted
| 18
| 3,060
| 133
|
N=input()
A=list(map(int,input().split()))
count=0
while all([a%2==0 for a in A]):
A = [a/2 for a in A]
count+=1
print(count)
|
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