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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s715257289
|
p03720
|
u722189950
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 306 |
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?
|
#ABC061B
N,M = map(int, input().split())
ab = [list(map(int, input().split())) for _ in range(M)]
ans =[]
for i in range(1,M+1):
count=0
for j in range(M):
if ab[j][0] ==i:
count +=1
if ab[j][1] ==i:
count +=1
ans.append(count)
for a in ans:
print(a)
|
s348504684
|
Accepted
| 18 | 3,064 | 306 |
#ABC061B
N,M = map(int, input().split())
ab = [list(map(int, input().split())) for _ in range(M)]
ans =[]
for i in range(1,N+1):
count=0
for j in range(M):
if ab[j][0] ==i:
count +=1
if ab[j][1] ==i:
count +=1
ans.append(count)
for a in ans:
print(a)
|
s981539652
|
p02401
|
u299798926
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,644 | 255 |
Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part.
|
while 1:
x,y,z=(i for i in input().split())
if y=='?':
break
elif y=='+':
print(int(x)+int(z))
elif y =='-':
print(int(x)-int(z))
elif y=='*':
print(int(x)*int(z))
else :
print(int(x)/int(z))
|
s613605444
|
Accepted
| 20 | 7,728 | 256 |
while 1:
x,y,z=(i for i in input().split())
if y=='?':
break
elif y=='+':
print(int(x)+int(z))
elif y =='-':
print(int(x)-int(z))
elif y=='*':
print(int(x)*int(z))
else :
print(int(x)//int(z))
|
s761455329
|
p02612
|
u188305619
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 9,148 | 62 |
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.
|
def solver(N):
return N % 1000
print(solver(int(input())))
|
s543326485
|
Accepted
| 30 | 9,108 | 167 |
def solver(N):
for i in range(1,11):
charge = 1000 * i - N
if charge >= 0 and charge < 1000:
return charge
print(solver(int(input())))
|
s972167454
|
p02260
|
u733159526
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,528 | 513 |
Write a program of the Selection Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: SelectionSort(A) 1 for i = 0 to A.length-1 2 mini = i 3 for j = i to A.length-1 4 if A[j] < A[mini] 5 mini = j 6 swap A[i] and A[mini] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 6 of the pseudocode in the case where i ≠ mini.
|
dummy = input()
s_list = list(map(int,input().split()))
print(s_list)
count = 0
for i in range(len(s_list)-1):
min_i = i
for j in range(i,len(s_list)):
if s_list[j] < s_list[min_i]:
min_i = j
if s_list[i] != s_list[min_i]:
count += 1
min_num = s_list[min_i]
s_list[min_i] = s_list[i]
s_list[i] = min_num
for i,row in enumerate(s_list):
if i == 0:
print(row,end='')
else:
print('',row,end='')
print()
print(count)
|
s088543365
|
Accepted
| 20 | 7,696 | 787 |
dummy = input()
s_list = list(map(int,input().split()))
#print(s_list)
count = 0
if len(s_list) == 1:
for i,row in enumerate(s_list):
if i == 0:
print(row,end='')
else:
print('',row,end='')
print()
print(count)
else:
for i in range(len(s_list)-1):
min_i = i
for j in range(i,len(s_list)):
if s_list[j] < s_list[min_i]:
min_i = j
if s_list[i] != s_list[min_i]:
count += 1
min_num = s_list[min_i]
s_list[min_i] = s_list[i]
s_list[i] = min_num
for i,row in enumerate(s_list):
if i == 0:
print(row,end='')
else:
print('',row,end='')
print()
print(count)
|
s992672627
|
p03681
|
u135116520
| 2,000 | 262,144 |
Wrong Answer
| 2,111 | 3,564 | 319 |
Snuke has N dogs and M monkeys. He wants them to line up in a row. As a Japanese saying goes, these dogs and monkeys are on bad terms. _("ken'en no naka", literally "the relationship of dogs and monkeys", means a relationship of mutual hatred.)_ Snuke is trying to reconsile them, by arranging the animals so that there are neither two adjacent dogs nor two adjacent monkeys. How many such arrangements there are? Find the count modulo 10^9+7 (since animals cannot understand numbers larger than that). Here, dogs and monkeys are both distinguishable. Also, two arrangements that result from reversing each other are distinguished.
|
N,M=map(int,input().split())
MOD=10**9+7
if N>=M:
s=1
for i in range(2,N+1):
s=s*i
s=s%MOD
t=N-M+2
for j in range(N-M+3,N+1):
t=t*j
t=t%MOD
print((s*t)%MOD)
else:
s=1
for i in range(2,M+1):
s=s*i
s=s%MOD
t=M-N+2
for j in range(N-M+3,N+1):
t=t*j
t=t%MOD
print((s*t)%MOD)
|
s453481225
|
Accepted
| 56 | 3,064 | 219 |
N,M=map(int,input().split())
MOD=10**9+7
s=1
for i in range(2,N+1):
s=(s*i)%MOD
t=1
for i in range(2,M+1):
t=(t*i)%MOD
if abs(N-M)>1:
print(0)
exit()
if abs(N-M)==1:
print((s*t)%MOD)
else:
print((2*s*t)%MOD)
|
s003587895
|
p03679
|
u432805419
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 143 |
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.
|
a = list(map(int,input().split()))
if a[1] <= a[2]:
print("delicious")
elif (a[2] - a[1]) <= a[0]:
print("safe")
else:
print("dangerous")
|
s872505864
|
Accepted
| 17 | 2,940 | 143 |
a = list(map(int,input().split()))
if a[1] >= a[2]:
print("delicious")
elif (a[2] - a[1]) <= a[0]:
print("safe")
else:
print("dangerous")
|
s743298146
|
p03997
|
u952130512
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 61 |
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)
|
s178966861
|
Accepted
| 17 | 2,940 | 66 |
a=int(input())
b=int(input())
h=int(input())
print(int((a+b)*h/2))
|
s737522188
|
p02410
|
u572790226
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,740 | 289 |
Write a program which reads a $ n \times m$ matrix $A$ and a $m \times 1$ vector $b$, and prints their product $Ab$. A column vector with m elements is represented by the following equation. \\[ b = \left( \begin{array}{c} b_1 \\\ b_2 \\\ : \\\ b_m \\\ \end{array} \right) \\] A $n \times m$ matrix with $m$ column vectors, each of which consists of $n$ elements, is represented by the following equation. \\[ A = \left( \begin{array}{cccc} a_{11} & a_{12} & ... & a_{1m} \\\ a_{21} & a_{22} & ... & a_{2m} \\\ : & : & : & : \\\ a_{n1} & a_{n2} & ... & a_{nm} \\\ \end{array} \right) \\] $i$-th element of a $m \times 1$ column vector $b$ is represented by $b_i$ ($i = 1, 2, ..., m$), and the element in $i$-th row and $j$-th column of a matrix $A$ is represented by $a_{ij}$ ($i = 1, 2, ..., n,$ $j = 1, 2, ..., m$). The product of a $n \times m$ matrix $A$ and a $m \times 1$ column vector $b$ is a $n \times 1$ column vector $c$, and $c_i$ is obtained by the following formula: \\[ c_i = \sum_{j=1}^m a_{ij}b_j = a_{i1}b_1 + a_{i2}b_2 + ... + a_{im}b_m \\]
|
n, m = map(int, input().split())
A = []
B = []
for line in range(n):
A.append(list(map(int, input().split())))
for line in range(m):
B.append(int(input()))
print(A,B)
ret = []
for i in range(n):
ret.append(sum(A[i][j] * B[j] for j in range(m)))
print('\n'.join(map(str, ret)))
|
s020490375
|
Accepted
| 40 | 8,024 | 290 |
n, m = map(int, input().split())
A = []
B = []
for line in range(n):
A.append(list(map(int, input().split())))
for line in range(m):
B.append(int(input()))
ret = []
for i in range(n):
ret.append(sum(A[i][j] * B[j] for j in range(m)))
print('\n'.join(map(str, ret)))
|
s341311826
|
p03455
|
u947123009
| 2,000 | 262,144 |
Wrong Answer
| 23 | 9,004 | 92 |
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 == 1:
print("odd")
else:
print("even")
|
s330727924
|
Accepted
| 22 | 9,016 | 92 |
a, b = map(int, input().split())
if a * b % 2 == 1:
print("Odd")
else:
print("Even")
|
s511328791
|
p03477
|
u089142196
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 117 |
A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R. Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan. Print `Left` if the balance scale tips to the left; print `Balanced` if it balances; print `Right` if it tips to the right.
|
A,B,C,D=map(int,input().split())
if A+B>C+D:
print("Left")
if A+B==C+D:
print("Balanced")
else:
print("Right")
|
s397197032
|
Accepted
| 17 | 3,060 | 119 |
A,B,C,D=map(int,input().split())
if A+B>C+D:
print("Left")
elif A+B==C+D:
print("Balanced")
else:
print("Right")
|
s711200638
|
p02613
|
u175217658
| 2,000 | 1,048,576 |
Wrong Answer
| 173 | 16,140 | 316 |
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
N = int(input())
S = [0]*N
AC = int(0)
WA = int(0)
TLE = int(0)
RE = int(0)
for i in range(0,N):
S[i] = input()
if(S[i]=='AC'):
AC += 1
if(S[i]=='WA'):
WA += 1
if(S[i]=='TLE'):
TLE += 1
if(S[i]=='RE'):
RE += 1
print('AC *', AC)
print('WA *', WA)
print('TLE *', TLE)
print('RE *', RE)
|
s585720793
|
Accepted
| 172 | 16,180 | 316 |
N = int(input())
S = [0]*N
AC = int(0)
WA = int(0)
TLE = int(0)
RE = int(0)
for i in range(0,N):
S[i] = input()
if(S[i]=='AC'):
AC += 1
if(S[i]=='WA'):
WA += 1
if(S[i]=='TLE'):
TLE += 1
if(S[i]=='RE'):
RE += 1
print('AC x', AC)
print('WA x', WA)
print('TLE x', TLE)
print('RE x', RE)
|
s451523257
|
p03658
|
u477320129
| 2,000 | 262,144 |
Wrong Answer
| 28 | 8,936 | 91 |
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())
print(sorted(map(int, input().split()), reverse=True)[:K])
|
s514053991
|
Accepted
| 27 | 9,156 | 97 |
N, K = map(int, input().split())
print(sum(sorted(map(int, input().split()), reverse=True)[:K]))
|
s704481811
|
p03471
|
u734876600
| 2,000 | 262,144 |
Wrong Answer
| 1,454 | 3,060 | 299 |
The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.
|
n,y = map(int,input().split())
ans = 0
c = 1
for i in range(y // 10000 + 1):
for j in range(y // 5000 + 1):
if n - i - j >= 0 and (y - 10000 * i - 5000 * j) % 1000 == 0:
if c != 0:
print(i, j, (y- 10000 * i - 5000 * j)//1000)
c = 0
|
s387581795
|
Accepted
| 1,586 | 3,064 | 341 |
n,y = map(int,input().split())
ans = 0
c = 1
for i in range(y // 10000 + 1,-1,-1):
for j in range(y // 5000 + 1,-1,-1):
if n - i - j >= 0 and y - 10000 * i - 5000 * j == 1000 * (n-i-j):
if c != 0:
print(i, j, (y-10000*i-5000*j)//1000)
c = 0
if c == 1:
print('-1 -1 -1')
|
s794220545
|
p03455
|
u770490999
| 2,000 | 262,144 |
Wrong Answer
| 28 | 9,080 | 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("Odd")
else:
print("Even")
|
s307060010
|
Accepted
| 28 | 9,096 | 82 |
a, b = map(int, input().split())
if (a*b) %2:
print("Odd")
else:
print("Even")
|
s207090405
|
p03997
|
u633450100
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 67 |
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)
|
s843983146
|
Accepted
| 17 | 2,940 | 68 |
a = int(input())
b = int(input())
h = int(input())
print((a+b)*h//2)
|
s438535593
|
p03852
|
u259738923
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 59 |
Given a lowercase English letter c, determine whether it is a vowel. Here, there are five vowels in the English alphabet: `a`, `e`, `i`, `o` and `u`.
|
['vowel','consonant'][not input() in ['a','e','i','o','u']]
|
s747245664
|
Accepted
| 16 | 2,940 | 157 |
let = input()
if let == 'a' or let == "i" or let == "u" or let == "e" or let == "o" :
result = "vowel"
else:
result = "consonant"
print(result)
|
s491140513
|
p03129
|
u743272507
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 89 |
Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.
|
n,k = map(int,input().split())
l = n//2
if k <= l:
print("YES")
else:
print("NO")
|
s097373459
|
Accepted
| 19 | 3,060 | 110 |
n,k = map(int,input().split())
l = (n+1)//2
if n == 1: l = 1
if k <= l:
print("YES")
else:
print("NO")
|
s414474377
|
p03488
|
u545368057
| 2,000 | 524,288 |
Wrong Answer
| 2,216 | 435,504 | 804 |
A robot is put at the origin in a two-dimensional plane. Initially, the robot is facing in the positive x-axis direction. This robot will be given an instruction sequence s. s consists of the following two kinds of letters, and will be executed in order from front to back. * `F` : Move in the current direction by distance 1. * `T` : Turn 90 degrees, either clockwise or counterclockwise. The objective of the robot is to be at coordinates (x, y) after all the instructions are executed. Determine whether this objective is achievable.
|
Ss = input()
x,y = map(int, input().split())
direc = 0
Ss = Ss + "T"
prev = Ss[0]
if prev == "F":
cnt = 1
direc = 0
else:
cnt = 0
direc = 1
mvs = []
for S in Ss[1:]:
if prev == "F" and S == "F":
cnt += 1
elif prev == "F" and S == "T":
mvs.append((direc, cnt))
direc ^= 1
cnt = 0
elif prev == "T" and S == "F":
cnt += 1
else:
direc ^= 1
prev = S
print(mvs)
import copy
cs = set()
cs.add(0)
for direc, dist in mvs:
cs_tmp = copy.copy(cs)
for c in cs_tmp:
if direc == 0:
cs.add(c+dist)
cs.add(c-dist)
else:
cs.add(c+8000*dist)
cs.add(c-8000*dist)
print(cs)
print(y*8000+x)
if y*8000+x in cs:
print("Yes")
else:
print("No")
|
s855722534
|
Accepted
| 1,037 | 10,396 | 594 |
from copy import copy
Ss = input() + "T"
X, Y = map(int, input().split())
d = [len(s) for s in Ss.split("T")]
x = d[0]
d_x = d[2::2]
d_y = d[1::2]
xs = set([x])
ys = set([0])
blx = [X in xs]
for d in d_x:
xs_ = set()
for x in xs:
xs_.add(x+d)
xs_.add(x-d)
blx.append(X in xs_)
xs = copy(xs_)
bly = []
for d in d_y:
ys_ = set()
for y in ys:
ys_.add(y+d)
ys_.add(y-d)
bly.append(Y in ys_)
ys = copy(ys_)
if blx[-1] and bly[-1]: print("Yes")
else: print("No")
|
s855467815
|
p03130
|
u058861899
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,060 | 211 |
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads. Determine if we can visit all the towns by traversing each of the roads exactly once.
|
a=[0 for i in range(3)]
b=[0 for i in range(3)]
for i in range(3):
a[i],b[i] = map(int,input().split())
a=(set(a))
b=(set(b))
if len(a)==3 and len(b)==3:
print("YES")
else:
print("NO")
|
s154393106
|
Accepted
| 17 | 2,940 | 186 |
a=""
for i in range(3):
a+=(str(input()))
error=0
for i in range(4):
if a.count(str(i+1))>2:
error=1
if error==0:
print("YES")
else:
print("NO")
|
s561310205
|
p02389
|
u281808376
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,572 | 51 |
Write a program which calculates the area and perimeter of a given rectangle.
|
a_b=input().split()
print(int(a_b[0])*int(a_b[1]))
|
s458951755
|
Accepted
| 20 | 5,592 | 81 |
a_b=input().split()
a=int(a_b[0])
b=int(a_b[1])
print(str(a*b)+" "+str((a+b)*2))
|
s552203100
|
p03644
|
u445404615
| 2,000 | 262,144 |
Wrong Answer
| 20 | 3,060 | 51 |
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.
|
s = input()
print(s[0] + str(len(s[1:-1])) + s[-1])
|
s603145486
|
Accepted
| 17 | 2,940 | 148 |
n = int(input())
for i in range(1,10):
if 2**i == n:
print(2**i)
exit()
if 2**i > n:
print(2**(i-1))
exit()
|
s823410118
|
p03828
|
u619819312
| 2,000 | 262,144 |
Wrong Answer
| 32 | 3,064 | 397 |
You are given an integer N. Find the number of the positive divisors of N!, modulo 10^9+7.
|
n=int(input())
k=[2]
mod=10**9+7
for i in range(2,n+1):
for j in range(len(k)):
if i%k[j]==0:
break
else:
k.append(i)
h=[1 for i in range(len(k))]
print(k)
for i in range(2,n+1):
c=0
while i!=1:
if i%k[c]==0:
i=i/k[c]
h[c]+=1
else:
c+=1
p=1
print(h)
for i in range(len(k)):
p=(p*h[i])%mod
print(p)
|
s507279229
|
Accepted
| 30 | 3,064 | 379 |
n=int(input())
k=[2]
mod=10**9+7
for i in range(2,n+1):
for j in range(len(k)):
if i%k[j]==0:
break
else:
k.append(i)
h=[1 for i in range(len(k))]
for i in range(2,n+1):
c=0
while i!=1:
if i%k[c]==0:
i=i/k[c]
h[c]+=1
else:
c+=1
p=1
for i in range(len(k)):
p=(p*h[i])%mod
print(p)
|
s457101245
|
p03591
|
u223904637
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 93 |
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=list(input())
if len(s)>=4 and s[0:5]==list('YAKI'):
print('Yes')
else:
print('No')
|
s310155536
|
Accepted
| 17 | 2,940 | 93 |
s=list(input())
if len(s)>=4 and s[0:4]==list('YAKI'):
print('Yes')
else:
print('No')
|
s883138851
|
p03958
|
u697690147
| 1,000 | 262,144 |
Wrong Answer
| 29 | 9,096 | 175 |
There are K pieces of cakes. Mr. Takahashi would like to eat one cake per day, taking K days to eat them all. There are T types of cake, and the number of the cakes of type i (1 ≤ i ≤ T) is a_i. Eating the same type of cake two days in a row would be no fun, so Mr. Takahashi would like to decide the order for eating cakes that minimizes the number of days on which he has to eat the same type of cake as the day before. Compute the minimum number of days on which the same type of cake as the previous day will be eaten.
|
a = list(map(int, input().split()))
largest = a[0]
total = 0
for ai in a:
total += ai
if ai > largest:
largest = ai
print(max(largest-(total-largest+1), 0))
|
s792227119
|
Accepted
| 26 | 9,168 | 129 |
k, t = list(map(int, input().split()))
a = list(map(int, input().split()))
largest = max(a)
print(max(largest-(k-largest+1), 0))
|
s586419080
|
p03471
|
u681110193
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 208 |
The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.
|
N,Y = map(int,input().split())
Y /= 1000
if Y // 10 + (Y % 10) // 5 + (Y % 10) % 5 > N:
print(-1,-1,-1)
elif Y // 10 + (Y % 10) // 5 + (Y % 10) % 5 == N:
print(Y // 10 , (Y % 10) // 5 , (Y % 10) % 5)
|
s640926151
|
Accepted
| 859 | 3,064 | 396 |
N,Y = map(int,input().split())
Y /= 1000
x=int(Y//10)
y= int((Y % 10) // 5)
z=int((Y % 10) % 5)
flag = False
if x+y+z > N:
print(-1,-1,-1)
elif x + y + z == N:
print(x , y , z)
else:
for i in range (N):
for j in range(N-i):
if 10*i+5*j+(N-i-j)==Y:
print(i,j,N-i-j)
flag=True
break
if flag:
break
if flag==False:
print(-1,-1,-1)
|
s968620438
|
p03796
|
u532514769
| 2,000 | 262,144 |
Wrong Answer
| 30 | 2,940 | 72 |
Snuke loves working out. He is now exercising N times. Before he starts exercising, his _power_ is 1. After he exercises for the i-th time, his power gets multiplied by i. Find Snuke's power after he exercises N times. Since the answer can be extremely large, print the answer modulo 10^{9}+7.
|
n=int(input())
x=0
for i in range(n):
x=x*(n+1)%1000000007
print(x)
|
s693016857
|
Accepted
| 38 | 2,940 | 72 |
n=int(input())
x=1
for i in range(n):
x=x*(i+1)%1000000007
print(x)
|
s314127090
|
p03474
|
u627417051
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 201 |
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 = list(map(int, input().split()))
S = input()
a = S[0:A]
x = S[A]
b = S[A + 1:]
print(a, x, b)
if list(a).count("-") == 0 and x == "-" and list(b).count("-") == 0:
print("Yes")
else:
print("No")
|
s445066762
|
Accepted
| 17 | 3,060 | 186 |
A, B = list(map(int, input().split()))
S = input()
a = S[0:A]
x = S[A]
b = S[A + 1:]
if list(a).count("-") == 0 and x == "-" and list(b).count("-") == 0:
print("Yes")
else:
print("No")
|
s581175283
|
p02393
|
u226541377
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,516 | 81 |
Write a program which reads three integers, and prints them in ascending order.
|
a,b,c = map(int,input().split())
numbers = [a,b,c]
numbers.sort()
print(numbers)
|
s000777655
|
Accepted
| 30 | 7,640 | 137 |
a,b,c = map(int,input().split())
numbers = [a,b,c]
numbers.sort()
print(str(numbers[0]) + " " + str(numbers[1]) + " " + str(numbers[2]))
|
s376816782
|
p03227
|
u223904637
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 128 |
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.
|
s=list(input())
ans=''
if len(s)==2:
print(','.join(s))
else:
tmp=s[0]
s[0]=s[2]
s[2]=tmp
print(','.join(s))
|
s382167111
|
Accepted
| 17 | 2,940 | 126 |
s=list(input())
ans=''
if len(s)==2:
print(''.join(s))
else:
tmp=s[0]
s[0]=s[2]
s[2]=tmp
print(''.join(s))
|
s419261119
|
p03606
|
u898967808
| 2,000 | 262,144 |
Wrong Answer
| 20 | 2,940 | 107 |
Joisino is working as a receptionist at a theater. The theater has 100000 seats, numbered from 1 to 100000. According to her memo, N groups of audiences have come so far, and the i-th group occupies the consecutive seats from Seat l_i to Seat r_i (inclusive). How many people are sitting at the theater now?
|
n = int(input())
ans = 0
for i in range(n):
a,b = map(int,input().split())
ans += b-a-1
print(ans)
|
s733084589
|
Accepted
| 20 | 3,060 | 107 |
n = int(input())
ans = 0
for i in range(n):
a,b = map(int,input().split())
ans += b-a+1
print(ans)
|
s474325615
|
p02865
|
u844697453
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 37 |
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
a = int(input())
print(a//2+0**(a%2))
|
s923102034
|
Accepted
| 17 | 2,940 | 37 |
a = int(input())
print(a//2-0**(a%2))
|
s325906853
|
p03657
|
u642905089
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 226 |
Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.
|
def sunuke(A, B):
re = 0
if A % 3 == 0:
re = 1
if B % 3 == 0:
re = 1
if (A+B) % 3 == 0:
re = 1
if re ==1:
return "Possible"
else:
return "Impossible"
|
s807234108
|
Accepted
| 17 | 3,060 | 228 |
Imp = list(map(int, input().split()))
A = Imp[0]
B = Imp[1]
re = 0
if A % 3 == 0:
re = 1
if B % 3 == 0:
re = 1
if (A+B) % 3 == 0:
re = 1
if re ==1:
print("Possible")
else:
print("Impossible")
|
s593287197
|
p03680
|
u183896397
| 2,000 | 262,144 |
Wrong Answer
| 208 | 7,084 | 298 |
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 = []
for i in range(N):
a = int(input())
A.append(a)
check = 0
count = 0
botton = 1
print(A[1 -1])
for i in range(N):
count += 1
if A[botton - 1] == 2:
check = 1
break
botton = A[botton - 1]
if check == 0:
print(-1)
else:
print(count)
|
s839357284
|
Accepted
| 217 | 7,084 | 283 |
N = int(input())
A = []
for i in range(N):
a = int(input())
A.append(a)
check = 0
count = 0
botton = 1
for i in range(N):
count += 1
if A[botton - 1] == 2:
check = 1
break
botton = A[botton - 1]
if check == 0:
print(-1)
else:
print(count)
|
s579680241
|
p03860
|
u881378225
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 31 |
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.
|
s = input()
print("A"+s[0]+"C")
|
s423695006
|
Accepted
| 17 | 2,940 | 51 |
s=input().split(" ")
print(s[0][0]+s[1][0]+s[2][0])
|
s841478739
|
p00292
|
u724548524
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,580 | 82 |
K 個の石から、P 人が順番に１つずつ石を取るゲームがあります。P 人目が石を取った時点で、まだ石が残っていれば、また１人目から順番に１つずつ石を取っていきます。このゲームでは、最後の石を取った人が勝ちとなります。K とP が与えられたとき、何人目が勝つか判定するプログラムを作成してください。
|
for _ in range(int(input())):
a,b = map(int,input().split())
print(a%b+1)
|
s329687840
|
Accepted
| 20 | 5,588 | 86 |
for _ in range(int(input())):
a,b = map(int,input().split())
print((a-1)%b+1)
|
s274783757
|
p03524
|
u366886346
| 2,000 | 262,144 |
Wrong Answer
| 36 | 9,704 | 147 |
Snuke has a string S consisting of three kinds of letters: `a`, `b` and `c`. He has a phobia for palindromes, and wants to permute the characters in S so that S will not contain a palindrome of length 2 or more as a substring. Determine whether this is possible.
|
s=list(input())
a=s.count("a")
b=s.count("b")
c=s.count("c")
if min(a,b,c)!=0 and max(a,b,c)-min(a,b,c)<=1:
print("Yes")
else:
print("No")
|
s550405374
|
Accepted
| 36 | 9,740 | 285 |
s=list(input())
a=s.count("a")
b=s.count("b")
c=s.count("c")
l1=[a,b,c]
l1.sort()
if l1[0]==0 and l1[1]==1 and l1[2]==1:
print("YES")
elif min(a,b,c)!=0 and max(a,b,c)-min(a,b,c)<=1:
print("YES")
elif a+b+c==max(a,b,c) and max(a,b,c)==1:
print("YES")
else:
print("NO")
|
s863567680
|
p02409
|
u150984829
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,608 | 197 |
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.
|
a=[[[0]*10]*3]*4
for _ in range(int(input())):
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])
print('#'*20)
|
s242734199
|
Accepted
| 20 | 5,620 | 199 |
a=[[10*[0]for _ in[0]*3]for _ in[0]*4]
for _ in[0]*int(input()):
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):print('',*a[i][j])
if i<3:print('#'*20)
|
s614771402
|
p03408
|
u177756077
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 260 |
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)]
dict={}
wordlist=list(set(s+t))
for i in wordlist:
cnt=0
if i in s:
cnt=cnt+1
if i in t:
cnt=cnt-1
dict[i]=cnt
print(max(dict.values()))
|
s132193809
|
Accepted
| 20 | 3,316 | 269 |
from collections import Counter
N=int(input())
s=[input() for i in range(N)]
M=int(input())
t=[input() for i in range(M)]
wordlist=list(set(s+t))
cnt=0
ss=Counter(s)
tt=Counter(t)
for i in wordlist:
buf=ss[i]-tt[i]
if cnt<buf:
cnt=buf
print(cnt)
|
s701382948
|
p03359
|
u923279197
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 70 |
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)
|
s278153330
|
Accepted
| 17 | 2,940 | 71 |
a,b=map(int,input().split())
if a<=b:
print(a)
else:
print(a-1)
|
s834984589
|
p03860
|
u265118937
| 2,000 | 262,144 |
Wrong Answer
| 27 | 9,096 | 61 |
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.
|
s = input()
print("A", end="")
print(s[0], end="")
print("C")
|
s660731489
|
Accepted
| 26 | 8,956 | 34 |
s = input()
print(s[0]+s[8]+s[-7])
|
s129347205
|
p03415
|
u608432532
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 92 |
We have a 3×3 square grid, where each square contains a lowercase English letters. The letter in the square at the i-th row from the top and j-th column from the left is c_{ij}. Print the string of length 3 that can be obtained by concatenating the letters in the squares on the diagonal connecting the top-left and bottom-right corner of the grid, from the top-left to bottom-right.
|
c = list()
for i in range(3):
c.append(input(i))
#print(c)
print(c[0][0]+c[1][1]+c[2][2])
|
s013326325
|
Accepted
| 18 | 2,940 | 91 |
c = list()
for i in range(3):
c.append(input())
#print(c)
print(c[0][0]+c[1][1]+c[2][2])
|
s907126060
|
p02393
|
u521963900
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,668 | 55 |
Write a program which reads three integers, and prints them in ascending order.
|
N = [int(i) for i in input().split()]
N.sort()
print(N)
|
s411461723
|
Accepted
| 20 | 7,752 | 74 |
N = [int(i) for i in input().split()]
N.sort()
print(" ".join(map(str,N)))
|
s733142140
|
p02388
|
u090921599
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,568 | 40 |
Write a program which calculates the cube of a given integer x.
|
x = input()
n = int(x)^3
print('n=', n)
|
s733728456
|
Accepted
| 20 | 5,576 | 35 |
x = int(input())
print(x * x * x)
|
s901948557
|
p03711
|
u882620594
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 386 |
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.
|
import sys
a=list(map(int,input().split()))
flg=0
A=[1,3,5,7,8,10,12]
B=[4,6,9,11]
C=[2]
for i in range(len(A)):
if A[i]==a[0]:
for j in range(len(A)):
if A[j]==a[1]:
flg=1
for i in range(len(B)):
if B[i]==a[0]:
for j in range(len(B)):
if B[j]==a[1]:
flg=1
if flg==1:
print("YES")
else:
print("NO")
|
s975587235
|
Accepted
| 17 | 3,064 | 386 |
import sys
a=list(map(int,input().split()))
flg=0
A=[1,3,5,7,8,10,12]
B=[4,6,9,11]
C=[2]
for i in range(len(A)):
if A[i]==a[0]:
for j in range(len(A)):
if A[j]==a[1]:
flg=1
for i in range(len(B)):
if B[i]==a[0]:
for j in range(len(B)):
if B[j]==a[1]:
flg=1
if flg==1:
print("Yes")
else:
print("No")
|
s707996624
|
p03387
|
u594244257
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 342 |
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.
|
A,B,C = map(int, input().split())
ret = 0
if not (A%2 == B%2 and B%2 == C%2):
ret += 1
if A%2 == B%2:
A += 1
B += 1
else:
B += 1
C += 1
# print(A,B,C)
max_elem = max(A,B,C)
ret += (max_elem-A)//2 + (max_elem-B)//2 + (max_elem-C)//2
print(ret)
|
s858129438
|
Accepted
| 17 | 3,064 | 455 |
A,B,C = map(int, input().split())
ret = 0
if not (A%2 == B%2 and B%2 == C%2):
ret += 1
if A%2 == B%2:
A += 1
B += 1
elif B%2 == C%2:
B += 1
C += 1
else:
A += 1
C += 1
max_elem = max(A,B,C)
ret += (max_elem-A)//2 + (max_elem-B)//2 + (max_elem-C)//2
print(ret)
|
s850203166
|
p03730
|
u576335153
| 2,000 | 262,144 |
Wrong Answer
| 32 | 9,076 | 283 |
We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`.
|
a, b, c = map(int, input().split())
ans = ''
if a % b == 0:
if c == 0:
ans = 'YES'
else:
ans = 'NO'
else:
k = a % b
for x in range(c+1):
if (k * x) % b == c:
ans = 'YES'
break
else:
ans = 'NO'
print(ans)
|
s987369256
|
Accepted
| 26 | 9,064 | 283 |
a, b, c = map(int, input().split())
ans = ''
if a % b == 0:
if c == 0:
ans = 'YES'
else:
ans = 'NO'
else:
k = a % b
for x in range(101):
if (k * x) % b == c:
ans = 'YES'
break
else:
ans = 'NO'
print(ans)
|
s712517092
|
p02281
|
u851695354
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,664 | 1,133 |
Binary trees are defined recursively. A binary tree _T_ is a structure defined on a finite set of nodes that either * contains no nodes, or * is composed of three disjoint sets of nodes: \- a root node. \- a binary tree called its left subtree. \- a binary tree called its right subtree. Your task is to write a program which perform tree walks (systematically traverse all nodes in a tree) based on the following algorithms: 1. Print the root, the left subtree and right subtree (preorder). 2. Print the left subtree, the root and right subtree (inorder). 3. Print the left subtree, right subtree and the root (postorder). Here, the given binary tree consists of _n_ nodes and evey node has a unique ID from 0 to _n_ -1.
|
class Tree:
def __init__(self,parent,left,right):
self.parent = parent
self.left = left
self.right = right
def preparse(u):
if u == -1: return
print(" ", end="")
print(u, end="")
preparse(trees[u].left)
preparse(trees[u].right)
def inparse(u):
if u == -1: return
inparse(trees[u].left)
print(" ", end="")
print(u, end="")
inparse(trees[u].right)
def postparse(u):
if u == -1: return
postparse(trees[u].left)
postparse(trees[u].right)
print(" ", end="")
print(u, end="")
N = int(input())
# ?????????
trees = [Tree(-1,-1,-1) for i in range(N)]
for i in range(N):
l = list(map(int, input().split()))
no = l[0]
left = l[1]
right = l[2]
trees[no].left = left
trees[no].right = right
if left != -1: trees[left].parent = no
if right != -1: trees[right].parent = no
print("Preorder")
preparse(0)
print("")
print("Inorder")
inparse(0)
print("")
print("Postorder")
postparse(0)
|
s998818119
|
Accepted
| 20 | 7,852 | 1,213 |
class Tree:
def __init__(self,parent,left,right):
self.parent = parent
self.left = left
self.right = right
def preparse(u):
if u == -1: return
print(" ", end="")
print(u, end="")
preparse(trees[u].left)
preparse(trees[u].right)
def inparse(u):
if u == -1: return
inparse(trees[u].left)
print(" ", end="")
print(u, end="")
inparse(trees[u].right)
def postparse(u):
if u == -1: return
postparse(trees[u].left)
postparse(trees[u].right)
print(" ", end="")
print(u, end="")
N = int(input())
# ?????????
trees = [Tree(-1,-1,-1) for i in range(N)]
for i in range(N):
l = list(map(int, input().split()))
no = l[0]
left = l[1]
right = l[2]
trees[no].left = left
trees[no].right = right
if left != -1: trees[left].parent = no
if right != -1: trees[right].parent = no
r = 0
for i in range(N):
if trees[i].parent == -1:
r = i
print("Preorder")
preparse(r)
print("")
print("Inorder")
inparse(r)
print("")
print("Postorder")
postparse(r)
print("")
|
s768340439
|
p04043
|
u632369368
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 128 |
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.
|
V = tuple([int(n) for n in input().split()])
if V in [(5, 5, 7), (5, 7, 5), (7, 5, 5)]:
print('Yes')
else:
print('No')
|
s431258015
|
Accepted
| 17 | 2,940 | 128 |
V = tuple([int(n) for n in input().split()])
if V in [(5, 5, 7), (5, 7, 5), (7, 5, 5)]:
print('YES')
else:
print('NO')
|
s099176677
|
p03711
|
u661980786
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 227 |
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())
g1 = [1,3,4,5,6,10,12]
g2 = [4,6,9,11]
g3 = [2]
if x in g1 and y in g1:
print("YES")
elif x in g2 and y in g2:
print("YES")
elif x in g3 and y in g3:
print("YES")
else:
print("NO")
|
s380951416
|
Accepted
| 17 | 3,060 | 227 |
x,y = map(int,input().split())
g1 = [1,3,5,7,8,10,12]
g2 = [4,6,9,11]
g3 = [2]
if x in g1 and y in g1:
print("Yes")
elif x in g2 and y in g2:
print("Yes")
elif x in g3 and y in g3:
print("Yes")
else:
print("No")
|
s223783587
|
p03447
|
u825528847
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,316 | 69 |
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?
|
X = int(input())
A = int(input())
B = int(input())
print((X-A) // B)
|
s183197161
|
Accepted
| 19 | 3,316 | 93 |
X = int(input())
A = int(input())
B = int(input())
tmp = (X-A) // B
print((X-A) - (tmp * B))
|
s018289636
|
p00424
|
u355726239
| 1,000 | 131,072 |
Wrong Answer
| 340 | 7,172 | 345 |
与えられた変換表にもとづき，データを変換するプログラムを作成しなさい． データに使われている文字は英字か数字で，英字は大文字と小文字を区別する．変換表に現れる文字の順序に規則性はない． 変換表は空白をはさんで前と後ろの 2 つの文字がある(文字列ではない)．変換方法は，変換表のある行の前の文字がデータに現れたら，そのたびにその文字を後ろの文字に変換し出力する．変換は 1 度だけで，変換した文字がまた変換対象の文字になっても変換しない．変換表に現れない文字は変換せず，そのまま出力する． 入力ファイルには，変換表（最初の n + 1 行）に続き変換するデータ（n + 2 行目以降）が書いてある． 1 行目に変換表の行数 n，続く n 行の各行は，空白をはさんで 2 つの文字，さらに続けて， n + 2 行目に変換するデータの行数 m，続く m 行の各行は 1 文字である． m ≤ 105 とする．出力は，出力例のように途中に空白や改行は入れず 1 行とせよ． 入力例 --- 3 A a 0 5 5 4 10 A B C 0 1 4 5 a b A 出力例 aBC5144aba
|
#!/usr/bin/env python
# -*- coding: utf-8 -*-
while True:
dic = {}
n = int(input())
if n == 0:
break
for i in range(n):
key, value = input().split()
dic[key] = value
print(dic)
ret = ''
m = int(input())
for i in range(m):
s = input()
ret += dic.get(s, s)
print(ret)
|
s552085775
|
Accepted
| 340 | 7,168 | 339 |
#!/usr/bin/env python
# -*- coding: utf-8 -*-
while True:
dic = {}
n = int(input())
if n == 0:
break
for i in range(n):
key, value = input().split()
dic[key] = value
ret = ''
m = int(input())
for i in range(m):
s = input().strip()
ret += dic.get(s, s)
print(ret)
|
s188561894
|
p03377
|
u222841610
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 68 |
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())
print('Yes' if a<=x<=a+b else 'No')
|
s916965451
|
Accepted
| 17 | 2,940 | 68 |
a,b,x = map(int,input().split())
print('YES' if a<=x<=a+b else 'NO')
|
s629721583
|
p04043
|
u919025034
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 150 |
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.
|
# coding: utf-8
# Your code here!
a = list(map(int,input().split()))
if a.count(5) == 2 and a.count(7) == 1:
print("Yes")
else:
print("No")
|
s643639194
|
Accepted
| 17 | 2,940 | 150 |
# coding: utf-8
# Your code here!
a = list(map(int,input().split()))
if a.count(5) == 2 and a.count(7) == 1:
print("YES")
else:
print("NO")
|
s802004910
|
p02694
|
u456376495
| 2,000 | 1,048,576 |
Wrong Answer
| 28 | 9,104 | 76 |
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())
i=100
j=0
while i<x:
i=int(i*1.01)
j+=1
print(j-1)
|
s956299874
|
Accepted
| 30 | 9,036 | 68 |
x=int(input())
i=100
j=0
while i<x:
i+=i//100
j+=1
print(j)
|
s750692279
|
p03129
|
u023229441
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 78 |
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>=(2*b+1):
print("YES")
else:
print("NO")
|
s272803557
|
Accepted
| 17 | 2,940 | 79 |
a,b=map(int,input().split())
if a>=(2*b-1):
print("YES")
else:
print("NO")
|
s812211870
|
p02612
|
u677253688
| 2,000 | 1,048,576 |
Wrong Answer
| 28 | 9,140 | 52 |
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 = n%1000
a = 1000-a
a%=a
print(a)
|
s484775504
|
Accepted
| 29 | 9,140 | 56 |
n = int(input())
x = (n+1000-1)//1000
x*=1000
print(x-n)
|
s829697684
|
p03493
|
u294714703
| 2,000 | 262,144 |
Wrong Answer
| 26 | 8,896 | 30 |
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()
s[0] + s[1] + s[2]
|
s335256466
|
Accepted
| 26 | 9,100 | 57 |
s = str(input())
print(int(s[0]) + int(s[1]) + int(s[2]))
|
s958501498
|
p02613
|
u254221913
| 2,000 | 1,048,576 |
Wrong Answer
| 145 | 16,168 | 216 |
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
n = int(input())
s = [0] * n
for i in range(n):
s[i] = input()
print('AC × ' + str(s.count('AC')))
print('WA × ' + str(s.count('WA')))
print('TLE × ' + str(s.count('TLE')))
print('RE × ' + str(s.count('RE')))
|
s637689440
|
Accepted
| 147 | 16,176 | 212 |
n = int(input())
s = [0] * n
for i in range(n):
s[i] = input()
print('AC x ' + str(s.count('AC')))
print('WA x ' + str(s.count('WA')))
print('TLE x ' + str(s.count('TLE')))
print('RE x ' + str(s.count('RE')))
|
s474661887
|
p02678
|
u219494936
| 2,000 | 1,048,576 |
Wrong Answer
| 2,207 | 63,356 | 616 |
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.
|
import heapq
import numpy as np
N, M = [int(x) for x in input().split(" ")]
e = {i: [] for i in range(N)}
for _ in range(M):
a, b = [int(x)-1 for x in input().split(" ")]
e[a] = e[a] + [b]
e[b] = e[b] + [a]
# roots = [[int(x) for x in input().split(" ")] for _ in range(M)]
d = np.ones(N, dtype=int) * np.inf
d[0] = 0
prev = np.ones(N, dtype=int) * -1
q = [(d[0], 0)]
heapq.heapify(q)
while len(q) != 0:
c, u = heapq.heappop(q)
for nv in e[u]:
alt = d[u] + 1
if d[nv] > alt:
d[nv] = alt
prev[nv] = u
heapq.heappush(q, (alt, nv))
print("YES")
for p in prev[1:]:
print(p+1)
|
s881030318
|
Accepted
| 1,628 | 59,708 | 481 |
import heapq
import numpy as np
N, M = [int(x) for x in input().split(" ")]
e = {i: [] for i in range(N)}
for _ in range(M):
a, b = [int(x)-1 for x in input().split(" ")]
e[a].append(b)
e[b].append(a)
d = np.ones(N, dtype=int) * -1
q = [0]
while len(q) != 0:
v = q.pop(0)
for ne in e[v]:
if d[ne] == -1:
d[ne] = v
q.append(ne)
print("Yes")
# for p in prev[1:]:
# print(p+1)
# print("\n".join([str(p+1) for p in d[1:]]))
for p in d[1:]:
print(p+1)
|
s117524590
|
p03592
|
u569272329
| 2,000 | 262,144 |
Wrong Answer
| 314 | 2,940 | 228 |
We have a grid with N rows and M columns of squares. Initially, all the squares are white. There is a button attached to each row and each column. When a button attached to a row is pressed, the colors of all the squares in that row are inverted; that is, white squares become black and vice versa. When a button attached to a column is pressed, the colors of all the squares in that column are inverted. Takahashi can freely press the buttons any number of times. Determine whether he can have exactly K black squares in the grid.
|
N, M, K = map(int, input().split())
flag = True
for i in range(0, N+1):
for j in range(0, M+1):
if (i*(M-j)+j*(N-i)) == K:
print("Yes")
flag = False
break
if flag:
print("No")
|
s558346397
|
Accepted
| 305 | 2,940 | 208 |
N, M, K = map(int, input().split())
flag = False
for i in range(0, N+1):
for j in range(0, M+1):
if (i*(M-j)+j*(N-i)) == K:
flag = True
if flag:
print("Yes")
else:
print("No")
|
s132378156
|
p02659
|
u464626513
| 2,000 | 1,048,576 |
Wrong Answer
| 23 | 9,104 | 95 |
Compute A \times B, truncate its fractional part, and print the result as an integer.
|
input = input()
split = input.split()
out = float(split[0]) * float(split[1])
print(round(out))
|
s050452298
|
Accepted
| 20 | 9,168 | 106 |
split = input().split()
num1 = int(split[0])
num2 = int(float(split[1])*1000)
print(int(num1*num2)//1000)
|
s507726056
|
p03007
|
u866769581
| 2,000 | 1,048,576 |
Wrong Answer
| 324 | 21,836 | 316 |
There are N integers, A_1, A_2, ..., A_N, written on a blackboard. We will repeat the following operation N-1 times so that we have only one integer on the blackboard. * Choose two integers x and y on the blackboard and erase these two integers. Then, write a new integer x-y. Find the maximum possible value of the final integer on the blackboard and a sequence of operations that maximizes the final integer.
|
#c.py
from collections import deque
n = int(input())
lis = list(map(int,input().split()))
lis.sort()
lis = deque(lis)
x = 0
ans_lis = []
while len(lis) != 1:
mi = lis.popleft()
ma = lis.pop()
lis.append(mi - ma)
ans_lis.append([mi,ma])
x += 1
print(lis[0])
for x in ans_lis:
print(x[0],x[1])
|
s248699856
|
Accepted
| 296 | 23,108 | 326 |
#c.py
n = int(input())
lis = list(map(int,input().split()))
lis.sort()
m = lis[0]
M = lis[-1]
ans_lis = []
for now in lis[1:-1]:
if now > 0:
ans_lis.append([m,now])
m -= now
else:
ans_lis.append([M,now])
M -= now
ans_lis.append([M,m])
print(M-m)
for x in ans_lis:
print(x[0],x[1])
|
s292780380
|
p04029
|
u180528413
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 170 |
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?
|
s = input()
s_list = list(s)
count = s.count('B')
for _ in range(count):
ind = s_list.index('B')
s_list[ind-1] = ''
s = ''.join(s_list)
s = s.replace('B','')
print(s)
|
s353625084
|
Accepted
| 355 | 21,556 | 122 |
import numpy as np
N = int(input())
all_list = [i for i in range(1, N+1)]
result = np.cumsum(all_list)
print(max(result))
|
s536901166
|
p03712
|
u535171899
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 113 |
You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}. Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1.
|
h,w = map(int,input().split())
print('#'*w)
for i in range(h):
s = input()
print('#'+s+'#')
print('#'*w)
|
s214532990
|
Accepted
| 18 | 3,060 | 189 |
h,w = map(int,input().split())
ans_map = [['#'*(w+2)]]
for i in range(h):
s = input()
ans_map.append(['#'+s+'#'])
ans_map.append(['#'*(w+2)])
for ans in ans_map:
print(ans[0])
|
s060875300
|
p03943
|
u246661425
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 112 |
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
a, b, c = map(int, input().split())
if a + b == c or a + c ==b or b + c == a:
print("YES")
else:
print("NO")
|
s973385030
|
Accepted
| 17 | 2,940 | 112 |
a, b, c = map(int, input().split())
if a + b == c or a + c ==b or b + c == a:
print("Yes")
else:
print("No")
|
s278782898
|
p03478
|
u359856428
| 2,000 | 262,144 |
Time Limit Exceeded
| 2,104 | 2,940 | 234 |
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).
|
def Fanc(x):
ans = 0
while x > 0:
ans += x % 10
x // 10
return ans
N,A,B = map(int,input().split())
sum = 0
for i in range(1,N + 1):
if A < Fanc(i) and Fanc(i) < B:
sum += i
print(sum)
|
s797296258
|
Accepted
| 31 | 3,060 | 240 |
def Fanc(x):
ans = 0
while x > 0:
ans += x % 10
x = x // 10
return ans
N,A,B = map(int,input().split())
sum = 0
for i in range(1,N + 1):
if A <= Fanc(i) and Fanc(i) <= B:
sum += i
print(sum)
|
s575871180
|
p02612
|
u626228246
| 2,000 | 1,048,576 |
Wrong Answer
| 34 | 9,144 | 93 |
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
n = int(input())
for i in range(100):
if 1000*i >= n:
print(1000*i - n)
|
s882164939
|
Accepted
| 28 | 9,152 | 103 |
import sys
n = int(input())
for i in range(100):
if 1000*i >= n:
print(1000*i - n)
sys.exit()
|
s397254078
|
p03544
|
u835482198
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 97 |
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())
ls = [2, 1]
for i in range(n - 1):
ls.append(ls[-1] + ls[-1])
print(ls[-1])
|
s420662619
|
Accepted
| 17 | 2,940 | 97 |
n = int(input())
ls = [2, 1]
for i in range(n - 1):
ls.append(ls[-1] + ls[-2])
print(ls[-1])
|
s874124288
|
p03166
|
u823885866
| 2,000 | 1,048,576 |
Wrong Answer
| 2,112 | 80,948 | 495 |
There is a directed graph G with N vertices and M edges. The vertices are numbered 1, 2, \ldots, N, and for each i (1 \leq i \leq M), the i-th directed edge goes from Vertex x_i to y_i. G **does not contain directed cycles**. Find the length of the longest directed path in G. Here, the length of a directed path is the number of edges in it.
|
import sys
import math
import itertools
import collections
import numpy as np
rl = sys.stdin.readline
N, M = map(int, rl().split())
li = [[] for _ in range(N)]
deg = [0] * N
for _ in range(M):
a, b = map(int, rl().split())
a -= 1
b -= 1
li[a].append(b)
deg[b] = 1
deg = [i for i in range(N) if deg[i] == 0]
print(deg)
dp = [0] * N
for i in deg:
for j in li[i]:
if dp[j] < dp[i] + 1:
dp[j] = dp[i] + 1
deg.append(j)
print(dp)
|
s856551136
|
Accepted
| 423 | 31,632 | 638 |
import sys
import math
import itertools
import collections
import numpy as np
rl = sys.stdin.readline
def a():
N, M = map(int, rl().split())
li = [[] for _ in range(N)]
deg = [0] * N
for _ in range(M):
a, b = map(int, rl().split())
a -= 1
b -= 1
li[a].append(b)
deg[b] += 1
d = collections.deque([i for i in range(N) if deg[i] == 0])
dp = [0] * N
while d:
i = d.popleft()
for j in li[i]:
deg[j] -= 1
if deg[j] == 0:
dp[j] = max(dp[j] ,dp[i] + 1)
d.append(j)
print(max(dp))
a()
|
s019946762
|
p03494
|
u409757418
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,060 | 162 |
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())
ais = input().split()
t = []
for a in ais:
m = 0
a = int(a)
while a % 2 == 0:
m += 1
a = a / 2
t.append(m)
print(t)
print(min(t))
|
s007600814
|
Accepted
| 19 | 2,940 | 153 |
n = int(input())
ais = input().split()
t = []
for a in ais:
m = 0
a = int(a)
while a % 2 == 0:
m += 1
a = a / 2
t.append(m)
print(min(t))
|
s949303707
|
p03433
|
u940780117
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 85 |
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())
mod = N%500
if mod<=A:
print('YES')
else:
print('NO')
|
s625084059
|
Accepted
| 17 | 2,940 | 85 |
N=int(input())
A=int(input())
mod = N%500
if mod<=A:
print('Yes')
else:
print('No')
|
s749915822
|
p02281
|
u096660561
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,456 | 1 |
Binary trees are defined recursively. A binary tree _T_ is a structure defined on a finite set of nodes that either * contains no nodes, or * is composed of three disjoint sets of nodes: \- a root node. \- a binary tree called its left subtree. \- a binary tree called its right subtree. Your task is to write a program which perform tree walks (systematically traverse all nodes in a tree) based on the following algorithms: 1. Print the root, the left subtree and right subtree (preorder). 2. Print the left subtree, the root and right subtree (inorder). 3. Print the left subtree, right subtree and the root (postorder). Here, the given binary tree consists of _n_ nodes and evey node has a unique ID from 0 to _n_ -1.
|
s240697603
|
Accepted
| 20 | 5,624 | 1,563 |
def Pre(i):
if dali[i].node != -1:
ret = dali[i].node
dali[i].node = -1
order.append(ret)
if dali[i].left != -1:
Pre(dali[i].left)
if dali[i].right != -1:
Pre(dali[i].right)
def Pre2(u):
if u == -1:
return
order.append(u)
Pre2(dali[u].left)
Pre2(dali[u].right)
def In2(u):
if u == -1:
return
In2(dali2[u].left)
order.append(u)
In2(dali2[u].right)
def Postorder(u):
if u == -1:
return None
Postorder(dali3[u].left)
Postorder(dali3[u].right)
order.append(dali3[u].node)
dali3[u].node = -1
class BiTree():
def __init__(self, node, left, right):
self.node = node
self.left = left
self.right = right
def inputdata(i):
node, left, right = list(map(int, input().split()))
dali[node] = BiTree(node, left, right)
dali2[node] = BiTree(node, left, right)
dali3[node] = BiTree(node, left, right)
# Getting Root
n = int(input())
dali = [0] * n
dali2 = [0] * n
dali3 = [0] * n
order = []
for i in range(n):
inputdata(i)
for i in range(n):
for j in range(n):
if i == dali[j].left or i == dali[j].right:
break
if j == n-1:
root = i
break
order = []
Pre2(root)
print("Preorder")
print(" ",end = "")
print(*order)
order = []
In2(root)
print("Inorder")
print(" ",end = "")
print(*order)
order = []
print("Postorder")
print(" ",end = "")
Postorder(root)
print(*order)
|
|
s742318057
|
p03549
|
u626881915
| 2,000 | 262,144 |
Wrong Answer
| 24 | 3,188 | 262 |
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).
|
import math
n, m = map(int, input().split())
s = 0
p = [0.0]
k = 1
s_old = -1.0
while s - s_old >= 0.0000001:
pr = (1 - sum(p)) / (2**m)
p.append(pr)
s_old = s
s += (1900 * m + 100 * (n - m)) * k * pr
k += 1
print(s)
print(math.ceil(s))
|
s902659726
|
Accepted
| 23 | 3,064 | 247 |
import math
n, m = map(int, input().split())
s = 0
p = [0.0]
k = 1
s_old = -1.0
while s - s_old >= 0.0000001:
pr = (1 - sum(p)) / (2**m)
p.append(pr)
s_old = s
s += (1900 * m + 100 * (n - m)) * k * pr
k += 1
print(math.ceil(s))
|
s326568746
|
p02612
|
u729008627
| 2,000 | 1,048,576 |
Wrong Answer
| 31 | 9,144 | 30 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N = int(input())
print(N%1000)
|
s096177960
|
Accepted
| 32 | 9,144 | 31 |
N = int(input())
print(-N%1000)
|
s948709210
|
p03592
|
u751663243
| 2,000 | 262,144 |
Wrong Answer
| 216 | 3,188 | 316 |
We have a grid with N rows and M columns of squares. Initially, all the squares are white. There is a button attached to each row and each column. When a button attached to a row is pressed, the colors of all the squares in that row are inverted; that is, white squares become black and vice versa. When a button attached to a column is pressed, the colors of all the squares in that column are inverted. Takahashi can freely press the buttons any number of times. Determine whether he can have exactly K black squares in the grid.
|
N,M,K = map(int,input().strip().split())
def judge(N,M,K):
if K > N * M or K < 0:
return False
for n in range(1, N + 1):
for m in range(1, M + 1):
if n * M + m * N - m * n == K:
return True
return False
if judge(N, M, K):
print("Yes")
else:
print("No")
|
s654858272
|
Accepted
| 247 | 3,060 | 318 |
N,M,K = map(int,input().strip().split())
def judge(N,M,K):
if K > N * M or K < 0:
return False
for n in range(0, N + 1):
for m in range(0, M + 1):
if n * M + m * N - 2*m * n == K:
return True
return False
if judge(N, M, K):
print("Yes")
else:
print("No")
|
s084859286
|
p03251
|
u370331385
| 2,000 | 1,048,576 |
Wrong Answer
| 19 | 3,064 | 227 |
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.
|
N,M,X,Y = map(int,input().split())
X = input().split()
Y = input().split()
for i in range(N):
X[i] = int(X[i])
for i in range(M):
Y[i] = int(Y[i])
X.sort()
Y.sort()
if((X[-1]+1)<Y[0]):
print('No war')
else:
print('war')
|
s256159016
|
Accepted
| 20 | 3,064 | 250 |
N,M,x,y = map(int,input().split())
X = input().split()
Y = input().split()
for i in range(N):
X[i] = int(X[i])
for i in range(M):
Y[i] = int(Y[i])
X.sort()
Y.sort()
Z = Y[0]
if( (X[-1]<Z)and(x<Z)and(Z<=y)):
print('No War')
else:
print('War')
|
s354463949
|
p02607
|
u084411645
| 2,000 | 1,048,576 |
Wrong Answer
| 31 | 8,972 | 53 |
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.
|
print(sum([int(i)%2 for i in input().split()[1::2]]))
|
s939173048
|
Accepted
| 26 | 8,960 | 61 |
input()
print(sum([int(i)%2 for i in input().split()[::2]]))
|
s678336141
|
p00015
|
u811773570
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,516 | 226 |
A country has a budget of more than 81 trillion yen. We want to process such data, but conventional integer type which uses signed 32 bit can represent up to 2,147,483,647. Your task is to write a program which reads two integers (more than or equal to zero), and prints a sum of these integers. If given integers or the sum have more than 80 digits, print "overflow".
|
ans = []
i = int(input())
for n in range(i):
j = int(input())
k = int(input())
x = str(j + k)
if len(x) >= 80:
ans.append("overflow")
else:
ans.append(x)
for n in range(i):
print(ans[n])
|
s142305443
|
Accepted
| 20 | 7,632 | 160 |
a = int(input())
for i in range(a):
x = int(input())
y = int(input())
if x + y >= 10 ** 80:
print("overflow")
else:
print(x + y)
|
s187062473
|
p03563
|
u257974487
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 299 |
Takahashi is a user of a site that hosts programming contests. When a user competes in a contest, the _rating_ of the user (not necessarily an integer) changes according to the _performance_ of the user, as follows: * Let the current rating of the user be a. * Suppose that the performance of the user in the contest is b. * Then, the new rating of the user will be the avarage of a and b. For example, if a user with rating 1 competes in a contest and gives performance 1000, his/her new rating will be 500.5, the average of 1 and 1000. Takahashi's current rating is R, and he wants his rating to be exactly G after the next contest. Find the performance required to achieve it.
|
import sys
s = input()
T=input()
p = 0
for i in range(len(s)-len(T)+1,0,-1):
for j in range(len(T)):
if T[j]!=s[i-1+j] and "?"!= s[i-1+j] :
break
p = 1
else:
s=s[:i-1]+T+s[i+len(T)-1:]
s = s.replace('?', 'a')
print(s)
sys.exit()
if p == 1:
print("UNRESTORABLE")
|
s757934792
|
Accepted
| 17 | 2,940 | 50 |
a = int(input())
b = int(input())
print(b*2 - a)
|
s900025888
|
p03433
|
u861141787
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 90 |
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
N = int(input())
A = int(input())
if N % 500 <= A:
print("YES")
else:
print("NO")
|
s479842987
|
Accepted
| 17 | 2,940 | 90 |
N = int(input())
A = int(input())
if N % 500 <= A:
print("Yes")
else:
print("No")
|
s137779869
|
p03352
|
u023229441
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 128 |
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.
|
z=int(input())
x=[]
for i in range(31):
for j in range(10):
if (i+1)**(j+1)<=z:
x.append((i+1)**(j+1))
print(max(x))
|
s470245675
|
Accepted
| 17 | 2,940 | 129 |
z=int(input())
x=[]
for i in range(31):
for j in range(10):
if (i+1)**(j+2)<=z:
x.append((i+1)**(j+2))
print(max(x))
|
s639873794
|
p03455
|
u911472374
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 101 |
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())
c = a * b
if c % 2 == 1:
print("Odd")
else:
print("even")
|
s497804826
|
Accepted
| 17 | 2,940 | 104 |
a, b = map(int, input().split())
c = int(a * b)
if c % 2 == 0:
print("Even")
else:
print("Odd")
|
s110528904
|
p03501
|
u426572476
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 69 |
You are parking at a parking lot. You can choose from the following two fee plans: * Plan 1: The fee will be A×T yen (the currency of Japan) when you park for T hours. * Plan 2: The fee will be B yen, regardless of the duration. Find the minimum fee when you park for N hours.
|
import math
n, a, b = map(int, input().split())
print(min(b, a + n))
|
s028730308
|
Accepted
| 17 | 2,940 | 69 |
import math
n, a, b = map(int, input().split())
print(min(b, a * n))
|
s504914822
|
p02821
|
u057668615
| 2,000 | 1,048,576 |
Wrong Answer
| 1,385 | 17,840 | 1,170 |
Takahashi has come to a party as a special guest. There are N ordinary guests at the party. The i-th ordinary guest has a _power_ of A_i. Takahashi has decided to perform M _handshakes_ to increase the _happiness_ of the party (let the current happiness be 0). A handshake will be performed as follows: * Takahashi chooses one (ordinary) guest x for his left hand and another guest y for his right hand (x and y can be the same). * Then, he shakes the left hand of Guest x and the right hand of Guest y simultaneously to increase the happiness by A_x+A_y. However, Takahashi should not perform the same handshake more than once. Formally, the following condition must hold: * Assume that, in the k-th handshake, Takahashi shakes the left hand of Guest x_k and the right hand of Guest y_k. Then, there is no pair p, q (1 \leq p < q \leq M) such that (x_p,y_p)=(x_q,y_q). What is the maximum possible happiness after M handshakes?
|
import sys, math, itertools, collections, bisect
mans = float('inf')
mod = 10 **9 +7
ans = 0
count = 0
pro = 0
N, M = map(int, input().split())
A = sorted(map(int, input().split()))
B = [0] + A[:]
for i in range(N):
B [i+1] += B[i]
print('A', A)
print('B', B)
def solve_binary(mid):
tmp = 0
for i, ai in enumerate(A):
tmp += N - bisect.bisect_left(A, mid-ai)
return tmp >= M
def binary_search(N):
ok = 0
ng = N
while abs(ok - ng ) > 1:
mid = (ok + ng) // 2
if solve_binary(mid):
ok = mid
else:
ng = mid
return ok
binresult = binary_search(2*10**5+1)
#binresult=ok
#print(binresult)
for i, ai in enumerate(A):
ans += ai*(N - bisect.bisect_left(A, binresult-ai)) + B[N] - B[bisect.bisect_left(A, binresult-ai)]
count += N - bisect.bisect_left(A, binresult-ai)
ans -= binresult * (count-M)
print(ans)
|
s176480472
|
Accepted
| 1,445 | 14,780 | 1,179 |
#Eans
import sys, math, itertools, collections, bisect
mans = float('inf')
mod = 10 **9 +7
ans = 0
count = 0
pro = 0
N, M = map(int, input().split())
A = sorted(map(int, input().split()))
B = [0] + A[:]
for i in range(N):
B [i+1] += B[i]
#print('A', A)
def solve_binary(mid):
tmp = 0
for i, ai in enumerate(A):
tmp += N - bisect.bisect_left(A, mid-ai)
return tmp >= M
def binary_search(N):
ok = 0
ng = N
while abs(ok - ng ) > 1:
mid = (ok + ng) // 2
if solve_binary(mid):
ok = mid
else:
ng = mid
return ok
binresult = binary_search(2*10**5+1)
#binresult=ok
#print(binresult)
for i, ai in enumerate(A):
ans += ai*(N - bisect.bisect_left(A, binresult-ai)) + B[N] - B[bisect.bisect_left(A, binresult-ai)]
count += N - bisect.bisect_left(A, binresult-ai)
ans -= binresult * (count-M)
print(ans)
|
s049011055
|
p03598
|
u790301364
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 356 |
There are N balls in the xy-plane. The coordinates of the i-th of them is (x_i, i). Thus, we have one ball on each of the N lines y = 1, y = 2, ..., y = N. In order to collect these balls, Snuke prepared 2N robots, N of type A and N of type B. Then, he placed the i-th type-A robot at coordinates (0, i), and the i-th type-B robot at coordinates (K, i). Thus, now we have one type-A robot and one type-B robot on each of the N lines y = 1, y = 2, ..., y = N. When activated, each type of robot will operate as follows. * When a type-A robot is activated at coordinates (0, a), it will move to the position of the ball on the line y = a, collect the ball, move back to its original position (0, a) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. * When a type-B robot is activated at coordinates (K, b), it will move to the position of the ball on the line y = b, collect the ball, move back to its original position (K, b) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. Snuke will activate some of the 2N robots to collect all of the balls. Find the minimum possible total distance covered by robots.
|
def main48():
n = int(input())
k = int(input())
data = input().split()
x = []
for i in range(n):
x.append(int(data[i]))
result = 0
for i in range(n):
if x[i] < k-x[i]:
result = result + x[i]
else:
result = result + k-x[i]
print(result)
if __name__ == "__main__":
main48()
|
s556918959
|
Accepted
| 17 | 3,064 | 366 |
def main48():
n = int(input())
k = int(input())
data = input().split()
x = []
for i in range(n):
x.append(int(data[i]))
result = 0
for i in range(n):
if x[i] < k-x[i]:
result = result + x[i] * 2
else:
result = result + (k-x[i]) * 2
print(result)
if __name__ == "__main__":
main48()
|
s941727965
|
p03455
|
u176796545
| 2,000 | 262,144 |
Wrong Answer
| 20 | 3,316 | 74 |
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())
print("Even" if a%2==0 and b%2==0 else "Odd")
|
s893562688
|
Accepted
| 19 | 2,940 | 64 |
a, b=map(int, input().split())
print("Odd" if a*b%2 else "Even")
|
s116533889
|
p03644
|
u103902792
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 74 |
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())
for i in range(n):
m = 2**i
if m > n:
print(m//2)
|
s879005688
|
Accepted
| 17 | 2,940 | 121 |
n = int(input())
if n == 1:
print(1)
exit(0)
for i in range(n):
m = 2**i
if m > n:
print(m//2)
exit(0)
|
s184192450
|
p02796
|
u997641430
| 2,000 | 1,048,576 |
Wrong Answer
| 457 | 16,984 | 227 |
In a factory, there are N robots placed on a number line. Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative. We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect. Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints. Find the maximum number of robots that we can keep.
|
n = int(input())
X = []
for i in range(n):
x, y = map(int, input().split())
X.append((x - y + 1, x + y))
X.sort()
ter = 0
cnt = 0
for left, right in X:
if ter < left:
ter = right
cnt += 1
print(cnt)
|
s562868787
|
Accepted
| 407 | 18,252 | 248 |
n = int(input())
X = []
for i in range(n):
x, y = map(int, input().split())
X.append((x - y, x + y))
X.sort(key=lambda x: x[1])
ter = -10**10
cnt = 0
for left, right in X:
if ter <= left:
ter = right
cnt += 1
print(cnt)
|
s664396099
|
p03471
|
u289288647
| 2,000 | 262,144 |
Wrong Answer
| 2,206 | 9,104 | 422 |
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.
|
import sys
N, Y = map(int, input().split())
num = [min(N, Y//10000), min(N, Y//5000), min(N, Y//1000)]
for i in reversed(range(0, num[0]+1)):
for j in reversed(range(0, N-i+1)):
if 10000*i + 5000*j > Y:
continue
for k in reversed(range(0, N-i-j+1)):
if 10000*i + 5000*j + 1000*k == Y:
print('%d %d %d' % (i, j, k))
sys.exit()
print('-1 -1 -1')
|
s571822796
|
Accepted
| 724 | 8,948 | 388 |
import sys
N, Y = map(int, input().split())
num = min(N, Y//10000)
if 10000*N < Y:
print('-1 -1 -1')
sys.exit()
for i in reversed(range(0, num+1)):
for j in reversed(range(0, N-i+1)):
if 10000*i + 5000*j > Y:
continue
if 10000*i + 5000*j + 1000*(N-i-j) == Y:
print('%d %d %d' % (i, j, N-i-j))
sys.exit()
print('-1 -1 -1')
|
s501855452
|
p03852
|
u629350026
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 66 |
Given a lowercase English letter c, determine whether it is a vowel. Here, there are five vowels in the English alphabet: `a`, `e`, `i`, `o` and `u`.
|
c=str(input())
if c in "aiueo":
print("YES")
else:
print("NO")
|
s072505949
|
Accepted
| 17 | 2,940 | 75 |
c=str(input())
if c in "aiueo":
print("vowel")
else:
print("consonant")
|
s787220098
|
p02612
|
u163971674
| 2,000 | 1,048,576 |
Wrong Answer
| 25 | 9,140 | 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)
|
s635734765
|
Accepted
| 27 | 9,152 | 70 |
n=int(input())
a=n%1000
if a==0:
print(0)
else:
print(1000-n%1000)
|
s736863778
|
p03380
|
u672475305
| 2,000 | 262,144 |
Wrong Answer
| 2,104 | 14,180 | 291 |
Let {\rm comb}(n,r) be the number of ways to choose r objects from among n objects, disregarding order. From n non-negative integers a_1, a_2, ..., a_n, select two numbers a_i > a_j so that {\rm comb}(a_i,a_j) is maximized. If there are multiple pairs that maximize the value, any of them is accepted.
|
from math import factorial
n = int(input())
lst = list(map(int,input().split()))
lst.sort()
num = lst[-1]
ans = 0
for i in range(n):
x = factorial(num) / factorial(lst[i]) / factorial(num-lst[i])
if (x > ans):
ans = x
else:
print(lst[i-1],num)
exit()
|
s572075496
|
Accepted
| 120 | 14,428 | 152 |
n = int(input())
A = list(map(int,input().split()))
A.sort()
y = max(A)
x = A[0]
for a in A:
if abs(y/2-a) < abs(y/2 - x):
x = a
print(y, x)
|
s133836884
|
p04029
|
u879693268
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 36 |
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(int(a*(a-1)/2))
|
s895126292
|
Accepted
| 17 | 2,940 | 36 |
a=int(input())
print(int(a*(a+1)/2))
|
s551040688
|
p03557
|
u368780724
| 2,000 | 262,144 |
Wrong Answer
| 1,725 | 23,328 | 758 |
The season for Snuke Festival has come again this year. First of all, Ringo will perform a ritual to summon Snuke. For the ritual, he needs an altar, which consists of three parts, one in each of the three categories: upper, middle and lower. He has N parts for each of the three categories. The size of the i-th upper part is A_i, the size of the i-th middle part is B_i, and the size of the i-th lower part is C_i. To build an altar, the size of the middle part must be strictly greater than that of the upper part, and the size of the lower part must be strictly greater than that of the middle part. On the other hand, any three parts that satisfy these conditions can be combined to form an altar. How many different altars can Ringo build? Here, two altars are considered different when at least one of the three parts used is different.
|
N = int(input())
A = list(map(int, input().split()))
#A = [int(x) for x in input().split()]
B = list(map(int, input().split()))
C = list(map(int, input().split()))
A.sort()
A = [-float('inf')] + A + [float('inf')]
B.sort()
C.sort()
C = [-float('inf')] + C + [float('inf')]
total = 0
for i in range(N):
low = 0
high = N + 1
while high - low > 1:
mid1 = (low + high) // 2
if A[mid1] < B[i]:
low = mid1
else:
high = mid1
mid1 = low
low = 0
high = N + 1
while high - low > 1:
mid2 = (low + high) // 2
if C[mid2] <= B[i]:
low = mid2
else:
high = mid2
mid2 = high
print(mid1, mid2)
total += mid1 * (N + 1 - mid2)
print(total)
|
s574556658
|
Accepted
| 1,545 | 23,328 | 736 |
N = int(input())
A = list(map(int, input().split()))
#A = [int(x) for x in input().split()]
B = list(map(int, input().split()))
C = list(map(int, input().split()))
A.sort()
A = [-float('inf')] + A + [float('inf')]
B.sort()
C.sort()
C = [-float('inf')] + C + [float('inf')]
total = 0
for i in range(N):
low = 0
high = N + 1
while high - low > 1:
mid1 = (low + high) // 2
if A[mid1] < B[i]:
low = mid1
else:
high = mid1
mid1 = low
low = 0
high = N + 1
while high - low > 1:
mid2 = (low + high) // 2
if C[mid2] <= B[i]:
low = mid2
else:
high = mid2
mid2 = high
total += mid1 * (N + 1 - mid2)
print(total)
|
s534600417
|
p03380
|
u507611905
| 2,000 | 262,144 |
Wrong Answer
| 88 | 14,052 | 423 |
Let {\rm comb}(n,r) be the number of ways to choose r objects from among n objects, disregarding order. From n non-negative integers a_1, a_2, ..., a_n, select two numbers a_i > a_j so that {\rm comb}(a_i,a_j) is maximized. If there are multiple pairs that maximize the value, any of them is accepted.
|
n = int(input())
a = [int(i) for i in input().split()]
a.sort()
if n == 2:
print(str(a[1])+" "+str(a[0]))
else:
maximum = a[-1]
a.pop()
half = maximum/2
a.append(half)
a.sort()
position = a.index(half)
large = a[position+1]
small = a[position-1]
print(position)
print(large)
print(small)
if large - (half) < (half) - small:
print(str(maximum)+" "+str(large))
else:
print(str(maximum)+" "+str(small))
|
s763521742
|
Accepted
| 91 | 14,052 | 568 |
n = int(input())
a = [int(i) for i in input().split()]
a.sort()
if n == 2:
print(str(a[1])+" "+str(a[0]))
else:
maximum = a[-1]
a.pop()
half = maximum/2
a.append(half)
a.sort()
position = a.index(half)
if position+1 >= len(a):
print(str(maximum)+" "+str(int(a[position-1])))
exit()
else:
large = int(a[position+1])
if position-1 <= 0:
print(str(maximum)+" "+str(int(a[position+1])))
exit()
else:
small = int(a[position-1])
if large - (half) < (half) - small:
print(str(maximum)+" "+str(large))
else:
print(str(maximum)+" "+str(small))
|
s359442820
|
p03760
|
u366886346
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 136 |
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=input()
e=input()
ans=""
for i in range(len(e)):
ans+=o[i//2]
ans+=e[i//2]
if len(o)!=len(e):
ans+=o[len(o)-1]
print(ans)
|
s255342086
|
Accepted
| 17 | 2,940 | 130 |
o=input()
e=input()
ans=""
for i in range(len(e)):
ans+=o[i]
ans+=e[i]
if len(o)!=len(e):
ans+=o[len(o)-1]
print(ans)
|
s502229592
|
p03860
|
u440129511
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 65 |
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.
|
s= list(map(str,input().split()))
print('A' + ''+ s[0] + '' +'C')
|
s607506939
|
Accepted
| 17 | 2,940 | 43 |
a,s,c=input().split()
print(a[0]+s[0]+c[0])
|
s521386973
|
p03170
|
u060938295
| 2,000 | 1,048,576 |
Wrong Answer
| 1,702 | 14,652 | 508 |
There is a set A = \\{ a_1, a_2, \ldots, a_N \\} consisting of N positive integers. Taro and Jiro will play the following game against each other. Initially, we have a pile consisting of K stones. The two players perform the following operation alternately, starting from Taro: * Choose an element x in A, and remove exactly x stones from the pile. A player loses when he becomes unable to play. Assuming that both players play optimally, determine the winner.
|
# -*- coding: utf-8 -*-
"""
Created on Sat Apr 25 18:20:35 2020
"""
import sys
import numpy as np
sys.setrecursionlimit(10 ** 9)
#def input():
# return sys.stdin.readline()[:-1]
mod = 10**9+7
#N = int(input())
N, K = map(int,input().split())
A = np.array(list(map(int,input().split())))
dp = np.arange(K + A[-1]+1) >= 0
for i in range(K, -1, -1):
if np.all(dp[A+i]):
dp[i] = False
# print(i, dp[i],A+i,dp[A+i])
#print(dp)
if dp[0]:
ans = 'Frist'
else:
ans = 'Second'
print(ans)
|
s100019338
|
Accepted
| 572 | 4,632 | 543 |
# -*- coding: utf-8 -*-
"""
Created on Sat Apr 25 18:20:35 2020
"""
import sys
#import numpy as np
sys.setrecursionlimit(10 ** 9)
#def input():
# return sys.stdin.readline()[:-1]
mod = 10**9+7
#N = int(input())
N, K = map(int,input().split())
A = list(map(int,input().split()))
dp = [i <= K for i in range(K + A[-1]+1)]
for i in range(K, -1, -1):
for a in A:
if dp[a+i]:
dp[i] = False
break
# print(i, dp[i],A+i,dp[A+i])
#print(dp)
if dp[0]:
ans = 'Second'
else:
ans = 'First'
print(ans)
|
s729228075
|
p03150
|
u672781822
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,064 | 426 |
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.
|
s = input()
if s.startswith('k') and s.startswith('eyence'):
print('YES')
elif s.startswith('ke') and s.startswith('yence'):
print('YES')
elif s.startswith('key') and s.startswith('ence'):
print('YES')
elif s.startswith('keye') and s.startswith('nce'):
print('YES')
elif s.startswith('keyen') and s.startswith('ce'):
print('YES')
elif s.startswith('keyenc') and s.startswith('e'):
print('YES')
else:
print('NO')
|
s673282911
|
Accepted
| 18 | 3,060 | 415 |
s = input()
if s.startswith('k') and s.endswith('eyence'):
print('YES')
elif s.startswith('ke') and s.endswith('yence'):
print('YES')
elif s.startswith('key') and s.endswith('ence'):
print('YES')
elif s.startswith('keye') and s.endswith('nce'):
print('YES')
elif s.startswith('keyen') and s.endswith('ce'):
print('YES')
elif s.startswith('keyenc') and s.endswith('e'):
print('YES')
else:
print('NO')
|
s631175138
|
p00002
|
u545973195
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,496 | 50 |
Write a program which computes the digit number of sum of two integers a and b.
|
a=map(int,input().split())
print(len(str(sum(a))))
|
s438878505
|
Accepted
| 20 | 7,464 | 97 |
while True:
try:
a=input().split()
except:
break
b=int(a[0])+int(a[1])
print(len(str(b)))
|
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