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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s904562178
|
p03351
|
u444722572
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 117 |
Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.
|
a,b,c,d=map(int,input().split())
if abs(a-c)<d or (abs(a-b)<d and abs(b-c)<d):
print("Yes")
else:
print("No")
|
s902616302
|
Accepted
| 17 | 2,940 | 130 |
a,b,c,d=map(int,input().split())
AB=abs(a-b)
BC=abs(b-c)
CA=abs(a-c)
print("Yes" if CA<=d else "Yes" if AB<=d and BC<=d else "No")
|
s143017767
|
p03549
|
u594956556
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 44 |
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())
print(2**M)
|
s950161662
|
Accepted
| 17 | 2,940 | 66 |
N, M = map(int, input().split())
print((1900*M+100*(N-M))*(2**M))
|
s812164146
|
p02619
|
u153499445
| 2,000 | 1,048,576 |
Wrong Answer
| 35 | 9,732 | 380 |
Let's first write a program to calculate the score from a pair of input and output. You can know the total score by submitting your solution, or an official program to calculate a score is often provided for local evaluation as in this contest. Nevertheless, writing a score calculator by yourself is still useful to check your understanding of the problem specification. Moreover, the source code of the score calculator can often be reused for solving the problem or debugging your solution. So it is worthwhile to write a score calculator unless it is very complicated.
|
D=int(input())
c=[int(x) for x in input().split()]
s=[[0]*i for i in range(D)]
t=[]
last_d=[0]*26
for i in range(D):
s[i]=[int(x) for x in input().split()]
for i in range(D):
t.append(int(input()))
v=0
C=sum(c)
print(C)
for i in range(D):
a=0
for j in range(26):
if j!=t[i]-1:
n=i+1-last_d[j]
a+=c[j]*n
v=v+s[i][t[i]-1]-a
print(v)
last_d[t[i]-1]=i+1
|
s071349407
|
Accepted
| 39 | 9,828 | 362 |
D=int(input())
c=[int(x) for x in input().split()]
s=[[0]*i for i in range(D)]
t=[]
last_d=[0]*26
for i in range(D):
s[i]=[int(x) for x in input().split()]
for i in range(D):
t.append(int(input()))
v=0
for i in range(D):
a=0
for j in range(26):
if j!=t[i]-1:
n=i+1-last_d[j]
a+=c[j]*n
v=v+s[i][t[i]-1]-a
print(v)
last_d[t[i]-1]=i+1
|
s293027237
|
p03645
|
u735008991
| 2,000 | 262,144 |
Wrong Answer
| 599 | 20,820 | 208 |
In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands. For convenience, we will call them Island 1, Island 2, ..., Island N. There are M kinds of regular boat services between these islands. Each service connects two islands. The i-th service connects Island a_i and Island b_i. Cat Snuke is on Island 1 now, and wants to go to Island N. However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services. Help him.
|
N, M =map(int, input().split())
sa = set()
ag = set()
for i in range(M):
a,b = map(int, input().split())
if a == 1: sa.add(b)
if b == N: sa.add(a)
print('POSSIBLE' if len(sa&ag) else 'IMPOSSIBLE')
|
s682254750
|
Accepted
| 600 | 19,024 | 208 |
N, M =map(int, input().split())
sa = set()
ag = set()
for i in range(M):
a,b = map(int, input().split())
if a == 1: sa.add(b)
if b == N: ag.add(a)
print('POSSIBLE' if len(sa&ag) else 'IMPOSSIBLE')
|
s576021917
|
p02613
|
u235499392
| 2,000 | 1,048,576 |
Wrong Answer
| 143 | 9,012 | 220 |
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())
a,b,c,d=0,0,0,0
for _ in range(n):
s=input()
if(s=="AC"):
a+=1
elif(s=="WA"):
b+=1
elif(s=="TLE"):
c+=1
else:
d+=1
print("AC X",a)
print("WA X",b)
print("TLE X",c)
print("RE X",d)
|
s066791774
|
Accepted
| 142 | 9,096 | 220 |
n=int(input())
a,b,c,d=0,0,0,0
for _ in range(n):
s=input()
if(s=="AC"):
a+=1
elif(s=="WA"):
b+=1
elif(s=="TLE"):
c+=1
else:
d+=1
print("AC x",a)
print("WA x",b)
print("TLE x",c)
print("RE x",d)
|
s815619270
|
p03836
|
u496821919
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 143 |
Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him.
|
sx,sy,tx,ty = map(int,input().split())
dx = tx-sx
dy = ty-sy
print("U"*dy+"R"*dx+"D"*dy+"L"*(dx+1)+"U"*(dy+1)+"R"*(dx+2)+"D"*(dy+2)+"L"*(dx+2))
|
s764175397
|
Accepted
| 17 | 3,060 | 152 |
sx,sy,tx,ty = map(int,input().split())
dx = tx-sx
dy = ty-sy
print("U"*dy+"R"*dx+"D"*dy+"L"*(dx+1)+"U"*(dy+1)+"R"*(dx+1)+"DR"+"D"*(dy+1)+"L"*(dx+1)+"U")
|
s134065885
|
p02936
|
u020962106
| 2,000 | 1,048,576 |
Wrong Answer
| 2,106 | 42,392 | 330 |
Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i. Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0. Now, the following Q operations will be performed: * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j. Find the value of the counter on each vertex after all operations.
|
n,q = map(int,input().split())
ki = [[] for _ in range(n)]
cnt = [0]*n
for i in range(n-1):
a,b = map(int,input().split())
ki[a-1].append(b)
for i in range(q):
p,x = map(int,input().split())
cnt[p-1]+=x
if ki[p-1]:
for j in range(ki[p-1][0],n+1):
cnt[j-1]+=x
[print(x,end=' ') for x in cnt]
|
s910542081
|
Accepted
| 1,719 | 55,856 | 501 |
n,q = map(int,input().split())
graph = [[] for _ in range(n)]
point = [0]*n
for i in range(n-1):
a,b = map(int,input().split())
graph[a-1].append(b-1)
graph[b-1].append(a-1)
for _ in range(q):
a,b = map(int,input().split())
a = a -1
point[a]+=b
stack = [];
stack.append(0)
visited = [False]*n
while stack:
x = stack.pop()
visited[x] = True
for y in graph[x]:
if visited[y] == False:
point[y] += point[x]
stack.append(y)
print(*point)
|
s992413466
|
p03478
|
u033982315
| 2,000 | 262,144 |
Wrong Answer
| 45 | 9,380 | 331 |
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).
|
s=list(map(int,input().split()))
oklist=[]
def keta(x):
ketalist=[]
for j in range(len(x)):
ketalist.append(int(x[j]))
return(sum(ketalist))
for i in range(s[0]+1):
check=keta(str(i))
if(s[1]<=check and check<=s[2]):
oklist.append(i)
print('ok')
sum_=sum(oklist)
|
s609750640
|
Accepted
| 40 | 9,100 | 320 |
s=list(map(int,input().split()))
oklist=[]
def keta(x):
ketalist=[]
for j in range(len(x)):
ketalist.append(int(x[j]))
return(sum(ketalist))
for i in range(s[0]+1):
check=keta(str(i))
if(s[1]<=check and check<=s[2]):
oklist.append(i)
sum_=sum(oklist)
print(sum_)
|
s656138008
|
p03162
|
u371467115
| 2,000 | 1,048,576 |
Wrong Answer
| 431 | 30,580 | 156 |
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())
hd=[list(map(int,input().split())) for _ in range(n)]
A,B,C=0,0,0
for a,b,c in hd:
A,B,C=A+max(b,c),B+max(a,c),C+max(a,b)
print(max(A,B,C))
|
s031928706
|
Accepted
| 392 | 3,060 | 139 |
n=int(input())
A,B,C=0,0,0
for _ in range(n):
a,b,c=map(int,input().split())
A,B,C=a+max(B,C),b+max(A,C),c+max(A,B)
print(max(A,B,C))
|
s859047086
|
p03719
|
u845937249
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 97 |
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
x , y ,z = map(int,input().split())
if z>x and y>z:
ans = 'YES'
else:
ans = 'NO'
print(ans)
|
s020800796
|
Accepted
| 17 | 2,940 | 99 |
x , y ,z = map(int,input().split())
if z>=x and y>=z:
ans = 'Yes'
else:
ans = 'No'
print(ans)
|
s124618383
|
p02388
|
u166841004
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,516 | 15 |
Write a program which calculates the cube of a given integer x.
|
x=3
print(x^3)
|
s764913883
|
Accepted
| 20 | 5,572 | 28 |
x=int(input())
print(x**3)
|
s315091526
|
p02396
|
u868716420
| 1,000 | 131,072 |
Wrong Answer
| 130 | 7,364 | 112 |
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.
|
x = input()
case = 1
while x != '0' :
print('case {}: {}'.format(case, x))
x = input()
case += 1
|
s582475432
|
Accepted
| 120 | 7,304 | 112 |
x = input()
case = 1
while x != '0' :
print('Case {}: {}'.format(case, x))
x = input()
case += 1
|
s132059832
|
p02697
|
u479719434
| 2,000 | 1,048,576 |
Wrong Answer
| 73 | 9,100 | 78 |
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.
|
N, M = map(int, input().split())
for i in range(1, M+1):
print(i, N-i+1)
|
s166459651
|
Accepted
| 85 | 9,096 | 152 |
N, M = map(int, input().split())
for i in range(M):
j = i // 2
if i % 2 == 1:
print(j+1, M-j)
else:
print(M+j+1, 2*M+1-j)
|
s221821527
|
p03599
|
u386819480
| 3,000 | 262,144 |
Wrong Answer
| 39 | 3,064 | 1,386 |
Snuke is making sugar water in a beaker. Initially, the beaker is empty. Snuke can perform the following four types of operations any number of times. He may choose not to perform some types of operations. * Operation 1: Pour 100A grams of water into the beaker. * Operation 2: Pour 100B grams of water into the beaker. * Operation 3: Put C grams of sugar into the beaker. * Operation 4: Put D grams of sugar into the beaker. In our experimental environment, E grams of sugar can dissolve into 100 grams of water. Snuke will make sugar water with the highest possible density. The beaker can contain at most F grams of substances (water and sugar combined), and there must not be any undissolved sugar in the beaker. Find the mass of the sugar water Snuke will make, and the mass of sugar dissolved in it. If there is more than one candidate, any of them will be accepted. We remind you that the sugar water that contains a grams of water and b grams of sugar is \frac{100b}{a + b} percent. Also, in this problem, pure water that does not contain any sugar is regarded as 0 percent density sugar water.
|
#!/usr/bin/env python3
import sys
from math import ceil, floor
def solve(A: int, B: int, C: int, D: int, E: int, F: int):
w = F
s = 0
for i in range(ceil(F/(100*A))):
aw = 100*A*i
for j in range(ceil(F/(100*B))):
bw = 100*B*j
for k in range(ceil((F-aw-bw)/C)):
cs = C*k
ds = min(D*floor((F-aw-bw-cs)/D), (aw*E+bw*E+cs*E)/max(1,100-E))
if(aw+bw == 0 or cs+ds == 0): continue
n = 100*s/(w+s)
mn = 100*(cs+ds)/(aw+bw+cs+ds)
if(mn <= E):
if(n < mn):
w = aw+bw
s = cs+ds
# print(i,j,k, mn)
print(w, s)
return
# Generated by 1.1.5 https://github.com/kyuridenamida/atcoder-tools (tips: You use the default template now. You can remove this line by using your custom template)
def main():
def iterate_tokens():
for line in sys.stdin:
for word in line.split():
yield word
tokens = iterate_tokens()
A = int(next(tokens)) # type: int
B = int(next(tokens)) # type: int
C = int(next(tokens)) # type: int
D = int(next(tokens)) # type: int
E = int(next(tokens)) # type: int
F = int(next(tokens)) # type: int
solve(A, B, C, D, E, F)
if __name__ == '__main__':
main()
|
s270047575
|
Accepted
| 196 | 12,464 | 1,479 |
#!/usr/bin/env python3
import sys
from math import ceil, floor
def solve(A: int, B: int, C: int, D: int, E: int, F: int):
x = []
y = []
for i in range(ceil(F/(100*A))+1):
for j in range(ceil(F/(100*B))+1):
dx = 100*A*i+100*B*j
if(min(100*A,100*B) <= dx <= F):
x.append(dx)
for i in range(ceil(F/C)+1):
for j in range(ceil(F/C)+1):
dy = C*i+D*j
if(0 <= dy <= F):
y.append(dy)
x = sorted(list(set(x)))
y = sorted(list(set(y)))
w = x[0]
s = y[0]
for i in x:
for j in y:
de = 100*s/(w+s)
ne = 100*j/(i+j)
if(i+j <= F and de < ne <= 100*E/(100+E)):
# print(i, j, ne)
w = i
s = j
print(w+s, s)
return
# Generated by 1.1.5 https://github.com/kyuridenamida/atcoder-tools (tips: You use the default template now. You can remove this line by using your custom template)
def main():
def iterate_tokens():
for line in sys.stdin:
for word in line.split():
yield word
tokens = iterate_tokens()
A = int(next(tokens)) # type: int
B = int(next(tokens)) # type: int
C = int(next(tokens)) # type: int
D = int(next(tokens)) # type: int
E = int(next(tokens)) # type: int
F = int(next(tokens)) # type: int
solve(A, B, C, D, E, F)
if __name__ == '__main__':
main()
|
s624290553
|
p03486
|
u258073778
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 241 |
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
|
s1 = input()
t1 = input()
s = list(s1)
t = list(t1)
s.sort()
t.sort(reverse=1)
i = 0
while(1):
if ord(s[i]) < ord(t[i]):
print('YES')
break
else :
if i+1 == min(len(s1),len(t1)):
print('NO')
break
else : i+=1
|
s361198380
|
Accepted
| 17 | 2,940 | 58 |
print('Yes'*(sorted(input())<sorted(input())[::-1])or'No')
|
s656156894
|
p03351
|
u496280557
| 2,000 | 1,048,576 |
Wrong Answer
| 28 | 9,160 | 152 |
Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.
|
a,b,c,d = map(int,input().split())
if abs (a - c) <= d:
print('Yes')
elif abs(a - b) <= abs(b - c) <= d:
print('Yes')
else:
print(' No ')
|
s257450908
|
Accepted
| 28 | 9,192 | 139 |
a,b,c,d = map(int,input().split())
if abs (a - c) <= d or abs (a - b) <= d and abs (b - c) <= d:
print('Yes')
else:
print(' No')
|
s387131037
|
p03999
|
u542190960
| 2,000 | 262,144 |
Wrong Answer
| 27 | 9,056 | 308 |
You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion. All strings that can be obtained in this way can be evaluated as formulas. Evaluate all possible formulas, and print the sum of the results.
|
s = str(input())
n = len(s)
ans = 0
for i in range(2**(n-1)):
s_tmp = s
nbin = str(bin(i))[2:]
nbin = nbin.zfill(n-1)
for j in range(len(nbin)-1, -1, -1):
if nbin[j] == '1':
s_tmp = s_tmp[:j+1] + '+' + s_tmp[j+1:]
print(nbin, s_tmp)
ans += eval(s_tmp)
print(ans)
|
s168363528
|
Accepted
| 25 | 9,076 | 285 |
s = str(input())
n = len(s)
ans = 0
for i in range(2**(n-1)):
s_tmp = s
nbin = str(bin(i))[2:]
nbin = nbin.zfill(n-1)
for j in range(len(nbin)-1, -1, -1):
if nbin[j] == '1':
s_tmp = s_tmp[:j+1] + '+' + s_tmp[j+1:]
ans += eval(s_tmp)
print(ans)
|
s128416917
|
p02389
|
u514745787
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,700 | 52 |
Write a program which calculates the area and perimeter of a given rectangle.
|
a, b = [int(i) for i in input().split()]
print(a, b)
|
s737720918
|
Accepted
| 30 | 7,508 | 70 |
ab = input().split()
a = int(ab[0])
b = int(ab[1])
print(a*b, (a+b)*2)
|
s954459856
|
p03478
|
u676464724
| 2,000 | 262,144 |
Wrong Answer
| 52 | 9,176 | 205 |
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
for i in range(N+1):
x = 0
for j in range(len(str(i))):
x = x + int(str(i)[j])
print(x)
if A <= x <= B:
sum = sum + i
print(sum)
|
s266426954
|
Accepted
| 48 | 9,044 | 192 |
N, A, B = map(int, input().split())
sum = 0
for i in range(N+1):
x = 0
for j in range(len(str(i))):
x = x + int(str(i)[j])
if A <= x <= B:
sum = sum + i
print(sum)
|
s176041931
|
p03556
|
u951480280
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 31 |
Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer.
|
print(int(int(input())**.5**2))
|
s414459061
|
Accepted
| 17 | 2,940 | 31 |
print(int(int(input())**.5)**2)
|
s889406493
|
p03860
|
u669382434
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 36 |
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
print("A"+input().split()[2][0]+"C")
|
s874691935
|
Accepted
| 18 | 3,064 | 36 |
print("A"+input().split()[1][0]+"C")
|
s457605103
|
p03698
|
u207075364
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 86 |
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
l=list(input())
l.sort()
if l[0]==l[len(l)-1]:
print("yes")
else:
print("no")
|
s844593192
|
Accepted
| 18 | 2,940 | 155 |
l=list(input())
l.sort()
flag = False
for i in range(len(l)-1):
if l[i]==l[i+1]:
flag = True
if flag:
print("no")
else:
print("yes")
|
s608286484
|
p03478
|
u652569315
| 2,000 | 262,144 |
Wrong Answer
| 36 | 2,940 | 139 |
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
n,a,b=map(int,input().split())
ans=0
for i in range(1,n+1):
c=sum([int(j) for j in list(str(i))])
if c>=a and c<=b:
ans+=i
print(i)
|
s454647827
|
Accepted
| 37 | 2,940 | 142 |
n,a,b=map(int,input().split())
ans=0
for i in range(1,n+1):
c=sum([int(j) for j in list(str(i))])
if c>=a and c<=b:
ans+=i
print(ans)
|
s858440565
|
p03448
|
u176796545
| 2,000 | 262,144 |
Wrong Answer
| 42 | 3,060 | 250 |
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
A=int(input())
B=int(input())
C=int(input())
total=int(input())
count=0
for a in [a*500 for a in range(A+1)]:
for b in [b*100 for b in range(B+1)]:
for c in [c*50 for c in range(C+1)]:
if a+b+c==total:
count+=1
|
s459520342
|
Accepted
| 46 | 3,064 | 263 |
A = int(input())
B = int(input())
C = int(input())
X = int(input())
count=0
for a in [a*500 for a in range(A+1)]:
for b in [b*100 for b in range(B+1)]:
for c in [c*50 for c in range(C+1)]:
if a+b+c==X:
count+=1
print(count)
|
s791517280
|
p03945
|
u871841829
| 2,000 | 262,144 |
Wrong Answer
| 45 | 3,188 | 151 |
Two foxes Jiro and Saburo are playing a game called _1D Reversi_. This game is played on a board, using black and white stones. On the board, stones are placed in a row, and each player places a new stone to either end of the row. Similarly to the original game of Reversi, when a white stone is placed, all black stones between the new white stone and another white stone, turn into white stones, and vice versa. In the middle of a game, something came up and Saburo has to leave the game. The state of the board at this point is described by a string S. There are |S| (the length of S) stones on the board, and each character in S represents the color of the i-th (1 ≦ i ≦ |S|) stone from the left. If the i-th character in S is `B`, it means that the color of the corresponding stone on the board is black. Similarly, if the i-th character in S is `W`, it means that the color of the corresponding stone is white. Jiro wants all stones on the board to be of the same color. For this purpose, he will place new stones on the board according to the rules. Find the minimum number of new stones that he needs to place.
|
import sys
s = input()
before = s[0]
ans = 0
for cx in range(1, len(s)):
if before != s[cx]:
ans += 1
before = s[cx]
print(abs)
|
s741750062
|
Accepted
| 42 | 3,188 | 151 |
import sys
s = input()
before = s[0]
ans = 0
for cx in range(1, len(s)):
if before != s[cx]:
ans += 1
before = s[cx]
print(ans)
|
s912659914
|
p02743
|
u113991073
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,060 | 141 |
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
|
import math
a,b,c = map(int, input().split())
x=math.sqrt(a)
y=math.sqrt(b)
z=math.sqrt(c)
if c>a+b:
print("Yes")
else:
print("No")
|
s488343598
|
Accepted
| 17 | 3,064 | 207 |
def Judgement(x,y,z):
if z-x-y>0 and (z-x-y)**2-4*x*y>0:
return 0
else:
return 1
a,b,c=map(int,input().split())
ans=Judgement(a,b,c)
if ans==0:
print("Yes")
else:
print("No")
|
s178555528
|
p03044
|
u059210959
| 2,000 | 1,048,576 |
Wrong Answer
| 703 | 29,512 | 791 |
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.
|
# encoding:utf-8
import copy
import random
import bisect
import fractions
import math
mod = 10**9+7
N = int(input())
u,v,w = [0 for i in range(N-1)],[0 for i in range(N-1)],[0 for i in range(N-1)]
for i in range(N-1):
u[i],v[i],w[i] = map(int,input().split())
tree = [[] for i in range(N-1)]
for i in range(N-1):
tree[i] = [u[i],v[i],w[i]]
# print(tree)
tree.sort()
color = [False for i in range(N)]
color[0] = True
for i in range(N-1):
# print(color)
if tree[i][2] % 2 == 0:
color[tree[i][1]-1] = color[tree[i][0]-1]
else:
color[tree[i][1]-1] = not(color[tree[i][0]-1])
for i in range(N):
if color[i]:
print(1)
else:
print(0)
|
s293341479
|
Accepted
| 786 | 90,192 | 863 |
# encoding:utf-8
import copy
import random
import bisect
import fractions
import math
import sys
mod = 10**9+7
sys.setrecursionlimit(mod)
N = int(input())
graph = [set() for i in range(N+1)]
for i in range(N-1):
u,v,w = map(int,input().split())
u -= 1
v -= 1
w = w % 2
graph[u].add((v,w))
graph[v].add((u,w))
res = [None for i in range(N+1)]
def dfs(node,p,color): # Depth First Serch
res[node] = color
for new_node,w in graph[node]:
if new_node == p:continue
if w == 1:dfs(new_node,node,1-color)
else:dfs(new_node,node,color)
dfs(0,-1,0)
for i in range(N):
print(res[i])
|
s982296567
|
p03672
|
u617659131
| 2,000 | 262,144 |
Wrong Answer
| 20 | 3,060 | 268 |
We will call a string that can be obtained by concatenating two equal strings an _even_ string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not. You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
|
s = list(input())
a = []
b = []
for i in range(len(s)):
s.pop()
if len(s) % 2 == 0:
for j in range(len(s)):
if j <= len(s) // 2 - 1:
a.append(s[j])
elif j <= len(s) // 2:
b.append(s[j])
if a == b:
print(len(s))
break
|
s962035334
|
Accepted
| 17 | 2,940 | 122 |
s = list(input())
for i in range(len(s)):
s.pop()
if s[:(len(s) // 2)] == s[len(s) // 2:]:
print(len(s))
break
|
s999734588
|
p02646
|
u266675845
| 2,000 | 1,048,576 |
Wrong Answer
| 21 | 9,184 | 159 |
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
a, v = map(int,input().split())
b, w = map(int,input().split())
T = int(input())
A = a + v * T
B = b + w * T
if A>= B:
print("Yes")
else:
print("NO")
|
s171482093
|
Accepted
| 23 | 9,192 | 298 |
a, v = map(int,input().split())
b, w = map(int,input().split())
T = int(input())
if a<b:
A = a + v * T
B = b + w * T
if A>= B:
print("YES")
else:
print("NO")
else:
A = a - v*T
B = b - w * T
if A <= B:
print("YES")
else:
print("NO")
|
s974882104
|
p02690
|
u919017918
| 2,000 | 1,048,576 |
Wrong Answer
| 23 | 9,188 | 379 |
Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X.
|
import sys
X = int(input())
mylist= []
for i in range(101):
mylist += [i**5]
#print(i**5)
#print(mylist)
for m in mylist:
for mm in mylist:
if m - mm == X:
print(mylist.index(m), mylist.index(mm))
sys.exit()
print()
elif m + mm == X:
print(mylist.index(m), -1*mylist.index(mm))
sys.exit()
|
s912962801
|
Accepted
| 40 | 9,180 | 253 |
x = int(input())
OP = [+1, -1]
for a in range(1000):
for b in range(a + 1):
for op1 in OP:
for op2 in OP:
if (op1 * a)**5 - (op2 * b)**5 == x:
print(op1 * a, op2 * b)
exit()
|
s605957727
|
p04043
|
u620846115
| 2,000 | 262,144 |
Wrong Answer
| 26 | 9,012 | 109 |
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 = list(map(int,input().split()))
if a.count("7")==1 and a.count("5")==2:
print("YES")
else:
print("NO")
|
s891706925
|
Accepted
| 28 | 8,876 | 94 |
a = input().split()
if a.count("7")==1 and a.count("5")==2:
print("YES")
else:
print("NO")
|
s487705628
|
p02659
|
u197868423
| 2,000 | 1,048,576 |
Wrong Answer
| 21 | 9,104 | 92 |
Compute A \times B, truncate its fractional part, and print the result as an integer.
|
import math
A, B = input().split()
A = int(A)
B = float(B)
ans = A * B
print(math.ceil(ans))
|
s029080500
|
Accepted
| 25 | 10,080 | 123 |
import math
from decimal import Decimal
A, B = input().split()
A = int(A)
B = Decimal(B)
ans = A * B
print(math.floor(ans))
|
s702376972
|
p03777
|
u811388439
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 192 |
Two deer, AtCoDeer and TopCoDeer, are playing a game called _Honest or Dishonest_. In this game, an honest player always tells the truth, and an dishonest player always tell lies. You are given two characters a and b as the input. Each of them is either `H` or `D`, and carries the following information: If a=`H`, AtCoDeer is honest; if a=`D`, AtCoDeer is dishonest. If b=`H`, AtCoDeer is saying that TopCoDeer is honest; if b=`D`, AtCoDeer is saying that TopCoDeer is dishonest. Given this information, determine whether TopCoDeer is honest.
|
l_vec = input().split()
def classifier1(l) :
if ((l[0] == 'H') and (l[1] == 'H')) or ((l[0] == 'D') and (l[1] == 'D')) :
return 'H'
else :
return 'D'
classifier1(l_vec)
|
s182730765
|
Accepted
| 17 | 2,940 | 199 |
l_vec = input().split()
def classifier1(l) :
if ((l[0] == 'H') and (l[1] == 'H')) or ((l[0] == 'D') and (l[1] == 'D')) :
return 'H'
else :
return 'D'
print(classifier1(l_vec))
|
s641285509
|
p03494
|
u123648284
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,060 | 212 |
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 = input()
l = list(map(int, input().split()))
cnt = 0
flg = True
while flg:
for i in range(len(l)):
if l[i] % 2:
l[i] /= 2
else:
flg = False
break
if flg:
cnt += 1
print(cnt)
|
s303722236
|
Accepted
| 19 | 3,060 | 217 |
n = input()
l = list(map(int, input().split()))
cnt = 0
flg = True
while flg:
for i in range(len(l)):
if l[i] % 2 == 0:
l[i] /= 2
else:
flg = False
break
if flg:
cnt += 1
print(cnt)
|
s002218429
|
p00007
|
u742178809
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,692 | 91 |
Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week. Write a program which computes the amount of the debt in n weeks.
|
x = 100000
for i in range(int(input())):
x*=1.05
x+=999
x=x//1000*1000
print(x)
|
s728090383
|
Accepted
| 30 | 7,680 | 96 |
x = 100000
for i in range(int(input())):
x*=1.05
x+=999
x=x//1000*1000
print(int(x))
|
s732224396
|
p02615
|
u081141316
| 2,000 | 1,048,576 |
Wrong Answer
| 2,269 | 72,016 | 379 |
Quickly after finishing the tutorial of the online game _ATChat_ , you have decided to visit a particular place with N-1 players who happen to be there. These N players, including you, are numbered 1 through N, and the **friendliness** of Player i is A_i. The N players will arrive at the place one by one in some order. To make sure nobody gets lost, you have set the following rule: players who have already arrived there should form a circle, and a player who has just arrived there should cut into the circle somewhere. When each player, except the first one to arrive, arrives at the place, the player gets **comfort** equal to the smaller of the friendliness of the clockwise adjacent player and that of the counter-clockwise adjacent player. The first player to arrive there gets the comfort of 0. What is the maximum total comfort the N players can get by optimally choosing the order of arrivals and the positions in the circle to cut into?
|
N = int(input())
A = [int(i) for i in input().split(' ')]
A.sort(reverse=True)
ring = []
max_value = A.pop(0)
ring.append(max_value)
ring.append(max_value)
ans = 0
for i in A:
buf = [x for x in ring]
print(ring)
n = []
for j in range(0, len(ring) - 1):
n.append(min(ring[j:j+2]))
ans = ans + max(n)
ring.insert(n.index(max(n)) + 1, i)
print(ans)
|
s271422676
|
Accepted
| 174 | 32,496 | 295 |
import sys
from collections import deque
N = int(sys.stdin.readline())
A = [int(i) for i in sys.stdin.readline().split(' ')]
A.sort(reverse=True)
max_value = A.pop(0)
diff = deque([max_value])
ans = 0
for i in A:
ans = ans + diff.popleft()
diff.append(i)
diff.append(i)
print(ans)
|
s693928354
|
p02612
|
u920204936
| 2,000 | 1,048,576 |
Wrong Answer
| 30 | 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)
|
s575032787
|
Accepted
| 30 | 9,140 | 69 |
n = int(input())
ans = 1000 - n%1000 if n%1000 != 0 else 0
print(ans)
|
s887074532
|
p03697
|
u591295155
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 55 |
You are given two integers A and B as the input. Output the value of A + B. However, if A + B is 10 or greater, output `error` instead.
|
S = input()
print(["no", "yes"][len(set(S)) == len(S)])
|
s260118490
|
Accepted
| 17 | 2,940 | 64 |
A, B = map(int, input().split())
print(["error", A+B][A+B < 10])
|
s464448010
|
p03567
|
u357751375
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 70 |
Snuke built an online judge to hold a programming contest. When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.) Determine whether the judge can return the string `AC` as the verdict to a program.
|
s = list(input())
if 'AC' in s:
print('Yes')
else:
print('No')
|
s806066352
|
Accepted
| 27 | 9,088 | 64 |
s = input()
if 'AC' in s:
print('Yes')
else:
print('No')
|
s562689108
|
p02238
|
u684325232
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,612 | 958 |
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.
|
n=int(input())
l=[[0 for i in range(n)] for j in range(n)]
d=[0 for i in range(n)]
f=[0 for i in range(n)]
stack=[]
check=[]
for i in range(n):
p=[int(x) for x in input().split()]
for j in range(n):
if j+1 in p[2:]:
l[i][j]=1
##print(l)
time=1
d[0]=1
stack+=[1]
check+=[-1]
while len(stack)!=0:
##print("----------while---------")
for i in range(n):
if l[stack[-1]-1][i]==1 and not i+1 in check:
time+=1
d[i]=time
stack+=[i+1]
## print("stack:",stack)
## print("d[",i,"]:",d[i])
break
elif i==n-1:
check+=[stack[-1]]
time+=1
f[stack.pop()-1]=time
##print("f[",i,"]:",f[i])
##print("check:",check)
for i in range(n):
print(i+1," ",d[i]," ",f[i])
|
s708326341
|
Accepted
| 30 | 5,736 | 1,222 |
n=int(input())
l=[[0 for i in range(n)] for j in range(n)]
d=[0 for i in range(n)]
f=[0 for i in range(n)]
stack=[]
check=[]
for i in range(n):
p=[int(x) for x in input().split()]
for j in range(n):
if j+1 in p[2:]:
l[i][j]=1
##print(l)
time=1
while len(check)<n:
for i in range(n):
if i+1 not in check:
stack+=[i+1]
d[i]=time
break
else: pass
while len(stack)!=0:
##print("----------while---------")
for i in range(n):
if l[stack[-1]-1][i]==1 and not i+1 in check and i+1 not in stack and stack[-1]-1!=i:
time+=1
d[i]=time
stack+=[i+1]
## print("stack:",stack)
## print("check:",check)
## print("d[",i,"]:",d[i])
break
elif i==n-1:
check+=[stack[-1]]
time+=1
f[stack.pop()-1]=time
## print("f[",i,"]:",f[i])
## print("check:",check)
time+=1
for i in range(n):
print(i+1,d[i],f[i])
|
s533296891
|
p04030
|
u024612773
| 2,000 | 262,144 |
Wrong Answer
| 41 | 3,064 | 33 |
Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?
|
print(input().replace('B', '\b'))
|
s516520304
|
Accepted
| 38 | 3,064 | 176 |
s=list(reversed(input()))
ans=""
cnt=0
for c in s:
if c == 'B':
cnt += 1
elif cnt > 0:
cnt -= 1
else:
ans += c
print(''.join(reversed(ans)))
|
s143105663
|
p03494
|
u698348858
| 2,000 | 262,144 |
Time Limit Exceeded
| 2,104 | 2,940 | 201 |
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
n = int(input())
nums = list(map(int,input().split(' ')))
cnt = 0
while True:
for num in nums:
if num % 2 != 0:
break
else:
continue
break
else:
cnt += 1
|
s270473601
|
Accepted
| 19 | 3,060 | 293 |
#coding:utf-8
n = int(input())
nums = list(map(int,input().split(' ')))
def calc():
cnt = 0
while True:
for i in range(len(nums)):
num = nums[i]
if num % 2 == 0 and num != 0:
nums[i] = num / 2
else:
return cnt
else:
cnt += 1
print(calc())
|
s760689330
|
p02972
|
u760569096
| 2,000 | 1,048,576 |
Wrong Answer
| 406 | 26,236 | 329 |
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.
|
from collections import deque
n =int(input())
a = [0]+list(map(int, input().split()))
d = deque()
cnt=0
for i in range(n,0,-1):
for j in range(i,n+1,i):
if a[j]==0:
continue
cnt+=1
b = cnt%2
a[i] = b
if b == 1:
d.appendleft(i)
cnt = 0
if len(d) == 0:
print(0)
else:
print(' '.join(map(str,d)))
|
s242503667
|
Accepted
| 404 | 26,012 | 308 |
from collections import deque
n =int(input())
a = [0]+list(map(int, input().split()))
d = deque()
cnt=0
for i in range(n,0,-1):
for j in range(i,n+1,i):
if a[j]==0:
continue
cnt+=1
b = cnt%2
a[i] = b
if b == 1:
d.appendleft(i)
cnt = 0
print(len(d))
print(' '.join(map(str,d)))
|
s113682352
|
p03623
|
u464912173
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 76 |
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
a,b,c = map(int,input().split())
print('B' if abs(a-c) > abs(b-c) else 'A')
|
s723756681
|
Accepted
| 17 | 2,940 | 76 |
x,a,b = map(int,input().split())
print('B' if abs(a-x) > abs(b-x) else 'A')
|
s812766464
|
p02399
|
u098047375
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,600 | 79 |
Write a program which reads two integers a and b, and calculates the following values: * a ÷ b: d (in integer) * remainder of a ÷ b: r (in integer) * a ÷ b: f (in real number)
|
a, b = map(int, input().split())
d = a // b
r = a % b
f = a / b
print(d, r, f)
|
s441935984
|
Accepted
| 20 | 5,604 | 96 |
a, b = map(int, input().split())
d = a // b
r = a % b
f = a / b
print(d, r, '{:.5f}'.format(f))
|
s080664651
|
p02866
|
u325282913
| 2,000 | 1,048,576 |
Wrong Answer
| 2,104 | 14,396 | 114 |
Given is an integer sequence D_1,...,D_N of N elements. Find the number, modulo 998244353, of trees with N vertices numbered 1 to N that satisfy the following condition: * For every integer i from 1 to N, the distance between Vertex 1 and Vertex i is D_i.
|
N = int(input())
array = list(map(int, input().split()))
ans = 1
for i in range(N):
ans *= array[i]
print(ans)
|
s359479255
|
Accepted
| 170 | 16,252 | 692 |
N = int(input())
array = list(map(int, input().split()))
if array[0] != 0:
print(0)
exit()
array.sort()
if 1 in array and len(set(array[1:])) == 1:
print(1)
exit()
if not (2 in array):
print(0)
exit()
if 0 in array[1:]:
print(0)
exit()
index = array.index(2)
count = 1
count_old = array.count(1)
ans = 1
first = array[index]
for i in range(index,N-1):
if first != array[i+1]:
if first != array[i+1] - 1:
print(0)
exit()
ans *= count_old ** count
count_old = count
count = 1
else:
count += 1
first = array[i+1]
ans %= 998244353
ans *= count_old ** count
ans %= 998244353
print(ans)
|
s241260238
|
p02843
|
u460009487
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 134 |
AtCoder Mart sells 1000000 of each of the six items below: * Riceballs, priced at 100 yen (the currency of Japan) each * Sandwiches, priced at 101 yen each * Cookies, priced at 102 yen each * Cakes, priced at 103 yen each * Candies, priced at 104 yen each * Computers, priced at 105 yen each Takahashi wants to buy some of them that cost exactly X yen in total. Determine whether this is possible. (Ignore consumption tax.)
|
n = int(input())
cent = n//100
under_cent = n%100
print(cent)
print(under_cent)
if under_cent > cent*5:
print(0)
else:
print(1)
|
s933277479
|
Accepted
| 17 | 2,940 | 103 |
n = int(input())
cent = n//100
under_cent = n%100
if under_cent > cent*5:
print(0)
else:
print(1)
|
s382296366
|
p03502
|
u698567423
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 103 |
An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10. Given an integer N, determine whether it is a Harshad number.
|
num = int(input())
fx = 0
for i in str(num):
fx += int(i)
print("YES" if num % fx == 0 else "NO")
|
s357957216
|
Accepted
| 17 | 2,940 | 103 |
num = int(input())
fx = 0
for i in str(num):
fx += int(i)
print("Yes" if num % fx == 0 else "No")
|
s978932353
|
p03457
|
u447528293
| 2,000 | 262,144 |
Wrong Answer
| 399 | 12,580 | 462 |
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
N = int(input().rstrip())
t = [0 for _ in range(N+1)]
x = [0 for _ in range(N+1)]
y = [0 for _ in range(N+1)]
for i in range(1, N+1):
t[i], x[i], y[i] = [int(s) for s in input().rstrip().split()]
judge = []
for i in range(N):
diff_t = t[i+1] - t[i]
diff_x = x[i+1] - x[i]
if diff_t >= diff_x and (diff_t - diff_x ) % 2 == 0:
judge.append(True)
else:
judge.append(False)
if all(judge):
print('Yes')
else:
print('No')
|
s947889662
|
Accepted
| 413 | 12,704 | 488 |
N = int(input().rstrip())
t = [0 for _ in range(N+1)]
x = [0 for _ in range(N+1)]
y = [0 for _ in range(N+1)]
for i in range(1, N+1):
t[i], x[i], y[i] = [int(s) for s in input().rstrip().split()]
judge = []
for i in range(N):
diff_t = t[i+1] - t[i]
diff_x = abs(x[i+1] - x[i]) + abs(y[i+1] - y[i])
if diff_t >= diff_x and (diff_t - diff_x ) % 2 == 0:
judge.append(True)
else:
judge.append(False)
if all(judge):
print('Yes')
else:
print('No')
|
s359671773
|
p04043
|
u040033078
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 230 |
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 -*-
a,b,c = input().split()
a = int(a)
b = int(b)
c = int(c)
if a+b+c != 17:
print('No')
else:
if (a == 5 and b == 5) or ( a ==5 and c == 5) or (b ==5 and c == 5):
print('Yes')
else:
print('No')
|
s220043130
|
Accepted
| 18 | 2,940 | 121 |
# -*- coding: utf-8 -*-
a = input().split()
a.sort()
b = ['5', '5', '7']
if a == b:
print('YES')
else:
print('NO')
|
s630098315
|
p03447
|
u705418271
| 2,000 | 262,144 |
Wrong Answer
| 25 | 9,096 | 77 |
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())
x-=a
while x>=0:
x-=b
print(x)
|
s148248151
|
Accepted
| 30 | 9,124 | 79 |
x=int(input())
a=int(input())
b=int(input())
x-=a
while x>=0:
x-=b
print(x+b)
|
s514775623
|
p03588
|
u762420987
| 2,000 | 262,144 |
Wrong Answer
| 389 | 17,808 | 125 |
A group of people played a game. All players had distinct scores, which are positive integers. Takahashi knows N facts on the players' scores. The i-th fact is as follows: the A_i-th highest score among the players is B_i. Find the maximum possible number of players in the game.
|
N = int(input())
AB = [tuple(map(int, input().split())) for _ in range(N)]
max_A = sorted(AB)[-1]
print(max_A[0] * max_A[1])
|
s069942167
|
Accepted
| 312 | 11,048 | 139 |
N = int(input())
a, b = [], []
for i in range(N):
A, B = map(int, input().split())
a.append(A)
b.append(B)
print(max(a)+min(b))
|
s850290406
|
p03479
|
u726439578
| 2,000 | 262,144 |
Wrong Answer
| 19 | 2,940 | 77 |
As a token of his gratitude, Takahashi has decided to give his mother an integer sequence. The sequence A needs to satisfy the conditions below: * A consists of integers between X and Y (inclusive). * For each 1\leq i \leq |A|-1, A_{i+1} is a multiple of A_i and strictly greater than A_i. Find the maximum possible length of the sequence.
|
x,y=map(int,input().split())
i=1
while x<=y :
i+=1
x=2*x
print(i)
|
s589849977
|
Accepted
| 18 | 2,940 | 77 |
x,y=map(int,input().split())
i=0
while x<=y :
i+=1
x=2*x
print(i)
|
s234210092
|
p02261
|
u409699893
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,832 | 816 |
Let's arrange a deck of cards. There are totally 36 cards of 4 suits(S, H, C, D) and 9 values (1, 2, ... 9). For example, 'eight of heart' is represented by H8 and 'one of diamonds' is represented by D1. Your task is to write a program which sorts a given set of cards in ascending order by their values using the Bubble Sort algorithms and the Selection Sort algorithm respectively. These algorithms should be based on the following pseudocode: BubbleSort(C) 1 for i = 0 to C.length-1 2 for j = C.length-1 downto i+1 3 if C[j].value < C[j-1].value 4 swap C[j] and C[j-1] SelectionSort(C) 1 for i = 0 to C.length-1 2 mini = i 3 for j = i to C.length-1 4 if C[j].value < C[mini].value 5 mini = j 6 swap C[i] and C[mini] Note that, indices for array elements are based on 0-origin. For each algorithm, report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).
|
def buble(n,x,count):
for i in range(1,n + 1):
for j in range(1,i):
if int(x[i - j][1]) < int(x[i - j - 1][1]):
x[i - j],x[i - j - 1] = x[i - j - 1],x[i - j]
count += 1
print (" ".join(x))
print ("Stable")
def delection(n,x,flag):
for i in range(n - 1):
minx = i
for j in range(i + 1,n):
if x[j][1] == x[i][1]:
flag = 1
if x[j][1] < x[minx][1]:
minx = j
if not i == minx:
x[i],x[minx]=x[minx],x[i]
print (" ".join(x))
if flag == 1:
print ("Stable")
else:
print ("Not stable")
n = int(input())
x = list(map(str,input().split()))
y = x
count = 0
flag = 0
buble(n,x,count)
delection(n,y,flag)
|
s776181361
|
Accepted
| 20 | 7,752 | 442 |
n = int(input())
a = input().strip().split()
b = a[:]
for i in range(n):
for j in range(n-1,i,-1):
if b[j][1] < b[j-1][1]:
b[j],b[j-1] = b[j-1],b[j]
print(' '.join(b))
print('Stable')
#Selection sort
for i in range(n):
minj = i
for j in range(i,n):
if a[j][1] < a[minj][1]:
minj = j
a[i],a[minj] = a[minj],a[i]
print(' '.join(a))
print('Stable' if b == a else 'Not stable')
|
s135004053
|
p03477
|
u129898499
| 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("Right")
else:
print("Balnced")
|
s552839705
|
Accepted
| 17 | 2,940 | 120 |
a,b,c,d=map(int, input().split())
if a+b > c+d:
print("Left")
elif a+b < c+d:
print("Right")
else:
print("Balanced")
|
s462576218
|
p03388
|
u707124227
| 2,000 | 262,144 |
Wrong Answer
| 29 | 9,364 | 310 |
10^{10^{10}} participants, including Takahashi, competed in two programming contests. In each contest, all participants had distinct ranks from first through 10^{10^{10}}-th. The _score_ of a participant is the product of his/her ranks in the two contests. Process the following Q queries: * In the i-th query, you are given two positive integers A_i and B_i. Assuming that Takahashi was ranked A_i-th in the first contest and B_i-th in the second contest, find the maximum possible number of participants whose scores are smaller than Takahashi's.
|
q=int(input())
ab=[list(map(int,input().split())) for _ in range(q)]
for a,b in ab:
if a>b:a,b=b,a
if a==b:
print(2*a-2)
elif a+1==b:
print(2*a-2)
else:
c=int((a*b)**0.5)
if c*(c+1)>=a*b:
print(2*c-2)
elif c**2<a*b:
print(2*c-1)
|
s073784356
|
Accepted
| 29 | 9,444 | 1,554 |
q=int(input())
ab=[list(map(int,input().split())) for _ in range(q)]
from math import floor
for a,b in ab:
if a==b:
print(2*a-2)
else:
t=floor((a*b)**0.5)
# (1,2*t-1),(2,2*t-2),(3,(a*b-1)//3),...,(t-1,(a*b-1)//(t-1))
if t*t>=a*b:
print(2*t-3)
elif t*(t+1)>=a*b:
print(2*t-2)
else:
print(2*t-1)
|
s861402313
|
p03943
|
u637175065
| 2,000 | 262,144 |
Wrong Answer
| 50 | 5,420 | 356 |
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.
|
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time
sys.setrecursionlimit(10**7)
inf = 10**20
mod = 10**9 + 7
def LI(): return list(map(int, input().split()))
def II(): return int(input())
def S(): return input()
def main():
a = LI()
if sum(a) % 3 == 0:
return 'YES'
return 'NO'
print(main())
|
s230274569
|
Accepted
| 73 | 6,976 | 360 |
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time
sys.setrecursionlimit(10**7)
inf = 10**20
mod = 10**9 + 7
def LI(): return list(map(int, input().split()))
def II(): return int(input())
def S(): return input()
def main():
a = LI()
if sum(a) == max(a)*2:
return 'Yes'
return 'No'
print(main())
|
s812386041
|
p03433
|
u243535639
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 89 |
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")
|
s839523584
|
Accepted
| 17 | 2,940 | 89 |
n = int(input())
a = int(input())
if n % 500 <= a:
print("Yes")
else:
print("No")
|
s513386354
|
p03719
|
u710907926
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 105 |
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
a, b, c = [int(s) for s in input().split()]
if c >= a and c <= b:
print("YES")
else:
print("NO")
|
s427094396
|
Accepted
| 18 | 2,940 | 105 |
a, b, c = [int(s) for s in input().split()]
if c >= a and c <= b:
print("Yes")
else:
print("No")
|
s880593823
|
p03636
|
u031146664
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 64 |
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.
|
s = input()
lis = [s[:1],str(len(s)),s[-1]]
print("".join(lis))
|
s788224508
|
Accepted
| 17 | 2,940 | 66 |
s = input()
lis = [s[:1],str(len(s)-2),s[-1]]
print("".join(lis))
|
s238939680
|
p02388
|
u745214779
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,504 | 11 |
Write a program which calculates the cube of a given integer x.
|
x = 3
x**3
|
s169071276
|
Accepted
| 20 | 5,572 | 35 |
x = input()
x = int(x)
print(x**3)
|
s006487252
|
p03457
|
u985376351
| 2,000 | 262,144 |
Wrong Answer
| 535 | 28,068 | 418 |
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
n = int(input())
p = [None]*n
for i in range(n):
p[i] = list(map(int,input().split()))
plan = [[0,0,0]]
plan.extend(p)
for i in range(n):
diff = [plan[i+1][k] - plan[i][k] for k in range(len(plan[0]))]
if abs(diff[1])+abs(diff[2])>diff[0]:
print('No')
break
elif diff[0]-(abs(diff[1])+abs(diff[2]))%2==1:
print('No')
break
elif i==n-1:
print('YES')
|
s331426231
|
Accepted
| 528 | 28,068 | 352 |
n = int(input())
p = [None]*n
for i in range(n):
p[i] = list(map(int,input().split()))
plan = [[0,0,0]]
plan.extend(p)
for i in range(n):
diff = [plan[i+1][k] - plan[i][k] for k in range(len(plan[0]))]
if (abs(diff[1])+abs(diff[2]))>diff[0] or (diff[0]-(abs(diff[1])+abs(diff[2])))%2==1:
print('No')
exit()
print('Yes')
|
s166293425
|
p03129
|
u112318601
| 2,000 | 1,048,576 |
Wrong Answer
| 28 | 9,136 | 65 |
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())
print("Yes" if (n+1)/2>=k else "No")
|
s170313872
|
Accepted
| 28 | 9,072 | 66 |
n,k=map(int,input().split())
print("YES" if (n+1)/2>=k else "NO")
|
s870315659
|
p04011
|
u000770457
| 2,000 | 262,144 |
Wrong Answer
| 20 | 2,940 | 139 |
There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee.
|
x=[]
for i in range(0,4):
x.append(int(input()))
s=0
for j in range(0,x[0]):
if j+1<x[1]:
s+=x[2]
else:
s+=x[3]
print(s)
|
s655677980
|
Accepted
| 21 | 3,064 | 141 |
x=[]
for i in range(0,4):
x.append(int(input()))
s=0
for j in range(0,x[0]):
if j+1<=x[1]:
s+=x[2]
else:
s+=x[3]
print(s)
|
s739162806
|
p00511
|
u847467233
| 8,000 | 131,072 |
Wrong Answer
| 20 | 5,584 | 265 |
加減乗除の計算をする電卓プログラムを作りなさい． 入力データの各行には数と記号 +, -, *, /, = のどれか1つが交互に書いてある． １行目は数である． 演算 +, -, *, / の優先順位（乗除算 *, / を加減算 +, - よりも先に計算すること）は考えず，入力の順序で計算し，= の行になったら，計算結果を出力する． 入力データの数値は108以下の正の整数とする． 計算中および計算結果は，0または負の数になることもあるが -108〜108 の範囲は超えない． 割り算は切り捨てとする． したがって，100/3*3= は 99 になる． 出力ファイルにおいては， 出力の最後の行にも改行コードを入れること．
|
# AOJ 0588: Simple Calculator
# Python3 2018.6.30 bal4u
a = int(input())
while True:
op = input().strip()
print(op)
if op == '=': break
b = int(input())
if op == '+': a += b
elif s == '-': a -= b
elif s == '*': a *= b
else: a //= b
print(a)
|
s956897530
|
Accepted
| 20 | 5,592 | 253 |
# AOJ 0588: Simple Calculator
# Python3 2018.6.30 bal4u
a = int(input())
while True:
op = input().strip()
if op == '=': break
b = int(input())
if op == '+': a += b
elif op == '-': a -= b
elif op == '*': a *= b
else: a //= b
print(a)
|
s800354868
|
p03485
|
u045270305
| 2,000 | 262,144 |
Wrong Answer
| 175 | 13,556 | 231 |
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
import numpy as np
from math import modf
l = input().split()
l = list(map(int, l))
mean = np.mean(l)
decimal, integer = modf(mean)
j = decimal * 10
if j >= 5 :
res = integer + 1
else :
res = integer
print(res)
|
s323872978
|
Accepted
| 155 | 12,504 | 218 |
import numpy as np
l = input().split()
l = list(map(int, l))
mean = np.mean(l)
integer = int(mean)
decimal = mean - integer
j = decimal * 10
if j >= 5 :
res = integer + 1
else :
res = integer
print(res)
|
s951191061
|
p04043
|
u962170423
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 155 |
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.
|
l,m,n = map(int,input().split())
cnt =0
if l == 5:
cnt += 1
if m == 7:
cnt += 1
if l == 5:
cnt += 1
if cnt == 3:
print("YES")
else:
print("NO")
|
s752196460
|
Accepted
| 17 | 2,940 | 97 |
l,m,n = map(int,input().split())
cnt = l+m+n
if cnt == 17:
print("YES")
else:
print("NO")
|
s258239201
|
p03387
|
u501643136
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 197 |
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())
if max(A,B,C)*3%2 == (A+B+C)%2:
print((max(A,B,C)*3-(A+B+C))/2)
else:
print((max(A,B,C)*3+3-(A+B+C))/2)
|
s717105797
|
Accepted
| 17 | 3,064 | 207 |
A, B, C = map(int, input().split())
if max(A,B,C)*3%2 == (A+B+C)%2:
print(int((max(A,B,C)*3-(A+B+C))/2))
else:
print(int((max(A,B,C)*3+3-(A+B+C))/2))
|
s306886419
|
p00173
|
u814278309
| 1,000 | 131,072 |
Wrong Answer
| 30 | 5,588 | 76 |
会津学園高等学校では、毎年学園祭をおこなっています。その中でも一番人気はお化け屋敷です。一番人気の理由は、お化け屋敷をおこなうクラスが 1クラスや 2クラスではなく、9クラスがお化け屋敷をおこなうことです。それぞれが工夫することより、それぞれが個性的なお化け屋敷になっています。そのため、最近では近隣から多くの来場者が訪れます。 そこで、学園祭実行委員会では、お化け屋敷の入場料金を下表のように校内で統一し、これにもとづき各クラスごとに入場者総数と収入の集計をおこなうことにしました。 入場料金表(入場者 1人あたりの入場料) 午前 午後 200円 300円 各クラス毎の午前と午後の入場者数を入力とし、各クラス毎の入場者総数及び収入の一覧表を作成するプログラムを作成してください。
|
a,b,c=map(str,input().split())
print(a,int(b)+int(c),int(b)*200+int(c)*300)
|
s611636566
|
Accepted
| 20 | 5,608 | 127 |
for i in range(9):
n,a,b = list(input().split())
x = int(a) + int(b)
y = int(a)*200 + int(b)*300
print(f"{n} {x} {y}")
|
s198424210
|
p03673
|
u439396449
| 2,000 | 262,144 |
Wrong Answer
| 2,104 | 26,032 | 157 |
You are given an integer sequence of length n, a_1, ..., a_n. Let us consider performing the following n operations on an empty sequence b. The i-th operation is as follows: 1. Append a_i to the end of b. 2. Reverse the order of the elements in b. Find the sequence b obtained after these n operations.
|
import sys
input = sys.stdin.readline
n = int(input())
a = [int(x) for x in input().split()]
b = []
for ai in a:
b.append(ai)
b = b[::-1]
print(b)
|
s907288767
|
Accepted
| 101 | 28,112 | 278 |
import sys
from collections import deque
input = sys.stdin.readline
n = int(input())
a = input().split()
b = deque([])
for i in range(n):
if i % 2 == 0:
b.append(a[i])
else:
b.appendleft(a[i])
if n % 2 == 1:
b = list(b)[::-1]
print(' '.join(b))
|
s581867060
|
p03645
|
u953110527
| 2,000 | 262,144 |
Wrong Answer
| 584 | 6,320 | 327 |
In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands. For convenience, we will call them Island 1, Island 2, ..., Island N. There are M kinds of regular boat services between these islands. Each service connects two islands. The i-th service connects Island a_i and Island b_i. Cat Snuke is on Island 1 now, and wants to go to Island N. However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services. Help him.
|
n,m = map(int,input().split())
c = [False for i in range(200000)]
d = [False for i in range(200000)]
for i in range(m):
a,b = map(int,input().split())
if a == 1:
c[b] == True
if b == n:
d[a] = True
for i in range(n):
if c[i] and d[i]:
print("POSSIBLE")
exit()
print("IMPOSSIBLE")
|
s850540463
|
Accepted
| 602 | 6,320 | 326 |
n,m = map(int,input().split())
c = [False for i in range(200000)]
d = [False for i in range(200000)]
for i in range(m):
a,b = map(int,input().split())
if a == 1:
c[b] = True
if b == n:
d[a] = True
for i in range(n):
if c[i] and d[i]:
print("POSSIBLE")
exit()
print("IMPOSSIBLE")
|
s840084945
|
p03730
|
u646336933
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 155 |
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 = [int(i) for i in input().split()]
for i in range(1, b+1):
if a*i % b == c:
print("YES")
break
else:
print("NO")
|
s207844417
|
Accepted
| 19 | 2,940 | 143 |
a, b, c = [int(i) for i in input().split()]
for i in range(1, b+1):
if a*i % b == c:
print("YES")
break
else: print("NO")
|
s962604580
|
p02615
|
u524534026
| 2,000 | 1,048,576 |
Wrong Answer
| 151 | 31,388 | 143 |
Quickly after finishing the tutorial of the online game _ATChat_ , you have decided to visit a particular place with N-1 players who happen to be there. These N players, including you, are numbered 1 through N, and the **friendliness** of Player i is A_i. The N players will arrive at the place one by one in some order. To make sure nobody gets lost, you have set the following rule: players who have already arrived there should form a circle, and a player who has just arrived there should cut into the circle somewhere. When each player, except the first one to arrive, arrives at the place, the player gets **comfort** equal to the smaller of the friendliness of the clockwise adjacent player and that of the counter-clockwise adjacent player. The first player to arrive there gets the comfort of 0. What is the maximum total comfort the N players can get by optimally choosing the order of arrivals and the positions in the circle to cut into?
|
n=int(input())
A=list(map(int,input().split()))
A.sort()
A.reverse()
print(A)
ans=0
for i in range(len(A)-1):
ans+=A[i]
print(ans)
|
s457671633
|
Accepted
| 213 | 32,296 | 246 |
from collections import deque
n=int(input())
A=list(map(int,input().split()))
A.sort()
A.reverse()
conf=[A[0]]
ans=0
con=deque(conf)
A.pop(0)
for elem in A:
ans+=con.popleft()
for _ in range(2):
con.append(elem)
print(ans)
|
s288654701
|
p02388
|
u597483031
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,516 | 19 |
Write a program which calculates the cube of a given integer x.
|
x=2
print("x**2")
|
s141666041
|
Accepted
| 20 | 5,576 | 23 |
print(int(input())**3)
|
s434668295
|
p03565
|
u978783125
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 412 |
E869120 found a chest which is likely to contain treasure. However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters. He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`. One more thing he found is a sheet of paper with the following facts written on it: * Condition 1: The string S contains a string T as a contiguous substring. * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1. Print the string S. If such a string does not exist, print `UNRESTORABLE`.
|
S = input()
T = input()
result = "UNRESTORABLE"
b = True
for i in range(len(S) - len(T) + 1):
for j in range(len(T)):
print("%d, %d" % (i, j))
if S[i + j] != T[j] and S[i + j] != "?":
break
if j == len(T) - 1:
s = (S[:i] + T + S[i + len(T):]).replace("?", "a")
if b or s < result:
result = s
b = False
print(result)
|
s970640699
|
Accepted
| 17 | 3,060 | 379 |
S = input()
T = input()
result = "UNRESTORABLE"
b = True
for i in range(len(S) - len(T) + 1):
for j in range(len(T)):
if S[i + j] != T[j] and S[i + j] != "?":
break
if j == len(T) - 1:
s = (S[:i] + T + S[i + len(T):]).replace("?", "a")
if b or s < result:
result = s
b = False
print(result)
|
s488079175
|
p03550
|
u143509139
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,188 | 150 |
We have a deck consisting of N cards. Each card has an integer written on it. The integer on the i-th card from the top is a_i. Two people X and Y will play a game using this deck. Initially, X has a card with Z written on it in his hand, and Y has a card with W written on it in his hand. Then, starting from X, they will alternately perform the following action: * Draw some number of cards from the top of the deck. Then, discard the card in his hand and keep the last drawn card instead. Here, at least one card must be drawn. The game ends when there is no more card in the deck. The score of the game is the absolute difference of the integers written on the cards in the two players' hand. X will play the game so that the score will be maximized, and Y will play the game so that the score will be minimized. What will be the score of the game?
|
n, z, w = map(int, input().split())
a = list(map(int, input().split()))
if n == 1:
print(abs(a[0] - w))
else:
print(max(a[-1] - w, a[-1] - a[-2]))
|
s204192645
|
Accepted
| 18 | 3,188 | 161 |
n, z, w = map(int, input().split())
a = list(map(int, input().split()))
if n == 1:
print(abs(a[0] - w))
else:
print(max(abs(a[-1] - w), abs(a[-1] - a[-2])))
|
s081219188
|
p02841
|
u553824105
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 74 |
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.
|
a = input().split()
c = input().split
if a != c:
print(1)
else:
print(0)
|
s532069485
|
Accepted
| 19 | 2,940 | 82 |
a = input().split()
c = input().split()
if a[0] != c[0]:
print(1)
else:
print(0)
|
s107846575
|
p02261
|
u447562175
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,604 | 727 |
Let's arrange a deck of cards. There are totally 36 cards of 4 suits(S, H, C, D) and 9 values (1, 2, ... 9). For example, 'eight of heart' is represented by H8 and 'one of diamonds' is represented by D1. Your task is to write a program which sorts a given set of cards in ascending order by their values using the Bubble Sort algorithms and the Selection Sort algorithm respectively. These algorithms should be based on the following pseudocode: BubbleSort(C) 1 for i = 0 to C.length-1 2 for j = C.length-1 downto i+1 3 if C[j].value < C[j-1].value 4 swap C[j] and C[j-1] SelectionSort(C) 1 for i = 0 to C.length-1 2 mini = i 3 for j = i to C.length-1 4 if C[j].value < C[mini].value 5 mini = j 6 swap C[i] and C[mini] Note that, indices for array elements are based on 0-origin. For each algorithm, report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).
|
def BubbleSort(C, N):
for i in range(N):
for j in range(N-1, i, -1):
if C[j][1] < C[j-1][1]:
C[j], C[j-1] = C[j-1], C[j]
return C
def SelectionSort(C, N):
for i in range(N):
minj = i
for j in range(i, N):
if C[j][1] < C[minj][1]:
minj = j
C[i], C[minj] = C[minj], C[i]
return C
def isStable(C1, C2, N):
if C1 != C2:
return "Stable"
return "Not stable"
Num = int(input())
Cards = [x for x in input().split()]
card1 = BubbleSort(list(Cards), Num)
card2 = SelectionSort(list(Cards), Num)
print(" ".join(card1))
print(isStable(card1, card1, Num))
print(" ".join(card2))
print(isStable(card1, card2, Num))
|
s461690980
|
Accepted
| 20 | 5,612 | 727 |
def BubbleSort(C, N):
for i in range(N):
for j in range(N-1, i, -1):
if C[j][1] < C[j-1][1]:
C[j], C[j-1] = C[j-1], C[j]
return C
def SelectionSort(C, N):
for i in range(N):
minj = i
for j in range(i, N):
if C[j][1] < C[minj][1]:
minj = j
C[i], C[minj] = C[minj], C[i]
return C
def isStable(C1, C2, N):
if C1 != C2:
return "Not stable"
return "Stable"
Num = int(input())
Cards = [x for x in input().split()]
card1 = BubbleSort(list(Cards), Num)
card2 = SelectionSort(list(Cards), Num)
print(" ".join(card1))
print(isStable(card1, card1, Num))
print(" ".join(card2))
print(isStable(card1, card2, Num))
|
s979410750
|
p03377
|
u011212399
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 84 |
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 + b >= x) and (x >= a) else "No")
|
s637308688
|
Accepted
| 18 | 2,940 | 84 |
a,b,x = map(int,input().split())
print("YES" if (a + b >= x) and (x >= a) else "NO")
|
s161837250
|
p03623
|
u869728296
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 129 |
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
a=input().strip().split(" ")
b=[int(i) for i in a]
A=abs(b[0]-b[1])
B=abs(b[0]-b[2])
if(A>B):
print(B)
else:
print(A)
|
s572164783
|
Accepted
| 18 | 2,940 | 133 |
a=input().strip().split(" ")
b=[int(i) for i in a]
A=abs(b[0]-b[1])
B=abs(b[0]-b[2])
if(A>B):
print("B")
else:
print("A")
|
s560379269
|
p03946
|
u545368057
| 2,000 | 262,144 |
Wrong Answer
| 333 | 29,964 | 473 |
There are N towns located in a line, conveniently numbered 1 through N. Takahashi the merchant is going on a travel from town 1 to town N, buying and selling apples. Takahashi will begin the travel at town 1, with no apple in his possession. The actions that can be performed during the travel are as follows: * _Move_ : When at town i (i < N), move to town i + 1. * _Merchandise_ : Buy or sell an arbitrary number of apples at the current town. Here, it is assumed that one apple can always be bought and sold for A_i yen (the currency of Japan) at town i (1 ≦ i ≦ N), where A_i are distinct integers. Also, you can assume that he has an infinite supply of money. For some reason, there is a constraint on merchandising apple during the travel: the sum of the number of apples bought and the number of apples sold during the whole travel, must be at most T. (Note that a single apple can be counted in both.) During the travel, Takahashi will perform actions so that the _profit_ of the travel is maximized. Here, the profit of the travel is the amount of money that is gained by selling apples, minus the amount of money that is spent on buying apples. Note that we are not interested in apples in his possession at the end of the travel. Aoki, a business rival of Takahashi, wants to trouble Takahashi by manipulating the market price of apples. Prior to the beginning of Takahashi's travel, Aoki can change A_i into another arbitrary non-negative integer A_i' for any town i, any number of times. The cost of performing this operation is |A_i - A_i'|. After performing this operation, different towns may have equal values of A_i. Aoki's objective is to decrease Takahashi's expected profit by at least 1 yen. Find the minimum total cost to achieve it. You may assume that Takahashi's expected profit is initially at least 1 yen.
|
N,T = map(int, input().split())
As = list(map(int, input().split()))
AsR = As[::-1]
mxs = []
mx = 0
ind = 0
for i, a in enumerate(AsR):
if a > mx:
mx = a
ind = N-i-1
mxs.append((mx,ind))
pairs = []
for a,mx in zip(As,mxs[::-1]):
print(mx[0]-a,mx[1])
pairs.append((mx[0]-a,mx[1]))
pairs.sort(reverse=True)
mx = pairs[0][0]
from collections import defaultdict
ans = set()
for x, i in pairs:
if x != mx:break
ans.add(i)
print(len(ans))
|
s714675438
|
Accepted
| 204 | 28,260 | 476 |
N,T = map(int, input().split())
As = list(map(int, input().split()))
AsR = As[::-1]
mxs = []
mx = 0
ind = 0
for i, a in enumerate(AsR):
if a > mx:
mx = a
ind = N-i-1
mxs.append((mx,ind))
pairs = []
for a,mx in zip(As,mxs[::-1]):
# print(mx[0]-a,mx[1])
pairs.append((mx[0]-a,mx[1]))
pairs.sort(reverse=True)
mx = pairs[0][0]
from collections import defaultdict
ans = set()
for x, i in pairs:
if x != mx:break
ans.add(i)
print(len(ans))
|
s556859390
|
p02259
|
u424041287
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,596 | 328 |
Write a program of the Bubble Sort algorithm which sorts a sequence _A_ in ascending order. The algorithm should be based on the following pseudocode: BubbleSort(A) 1 for i = 0 to A.length-1 2 for j = A.length-1 downto i+1 3 if A[j] < A[j-1] 4 swap A[j] and A[j-1] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 4 of the pseudocode.
|
def bubbleSort(A, N):
t = 0
flag = 0
while flag == 0:
flag = 1
for j in range(N - 1, 0, -1):
if A[j] < A[j - 1]:
A[j], A[j - 1] = A[j - 1], A[j]
flag = 0
t += 1
n = int(input())
a = [int(i) for i in input().split()]
print(bubbleSort(a, n))
|
s907385549
|
Accepted
| 20 | 5,604 | 423 |
def bubbleSort(A, N):
s = 0
flag = 0
while flag == 0:
flag = 1
for j in range(N - 1, 0, -1):
if A[j] < A[j - 1]:
A[j], A[j - 1] = A[j - 1], A[j]
flag = 0
s += 1
t = str(A[0])
for i in range(1,N):
t = t + " " + str(A[i])
print(t)
print(s)
n = int(input())
a = [int(i) for i in input().split()]
bubbleSort(a, n)
|
s302180528
|
p03814
|
u896741788
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,512 | 55 |
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()
a=s.index("A")
b=s.rfind("z")
print(s[a:b+1])
|
s550150623
|
Accepted
| 18 | 3,500 | 60 |
s=input()
a=s.index("A")
b=s.rfind("Z")
print(len(s[a:b+1]))
|
s483821464
|
p02613
|
u820205438
| 2,000 | 1,048,576 |
Wrong Answer
| 157 | 16,556 | 246 |
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())
res=[]
for i in range(n):
res.append(input())
from collections import defaultdict
cnt=defaultdict(int)
cnt["AC"]=0
cnt["TLE"]=0
cnt["WA"]=0
cnt["RE"]=0
for i in res:
cnt[i]+=1
for i in cnt.items():
print(i[0],"x",i[1])
|
s910733664
|
Accepted
| 164 | 16,544 | 256 |
n=int(input())
res=[]
for i in range(n):
res.append(input())
from collections import defaultdict
cnt=defaultdict(int)
cnt["AC"]=0
cnt["TLE"]=0
cnt["WA"]=0
cnt["RE"]=0
for i in res:
cnt[i]+=1
for i in ["AC","WA","TLE","RE"]:
print(i,"x",cnt[i])
|
s313781518
|
p03760
|
u736729525
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 123 |
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()
A = [E, O]
for i in range(len(O)+len(E)):
j, r = divmod(i, 2)
print(A[1-r],end="")
print()
|
s688484816
|
Accepted
| 17 | 2,940 | 124 |
O = list(input())
E = list(input())
if len(O) > len(E):
E.append("")
print("".join(o+e for o, e in zip(O, E)))
|
s278494577
|
p03501
|
u769411997
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 87 |
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.
|
s = input()
cnt = 0
for i in range(3):
if s[i] == '1':
cnt += 1
print(cnt)
|
s463752559
|
Accepted
| 17 | 2,940 | 82 |
n, a, b = map(int, input().split())
if n*a > b:
print(b)
else:
print(n*a)
|
s064466774
|
p03478
|
u452015170
| 2,000 | 262,144 |
Wrong Answer
| 38 | 3,424 | 268 |
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())
sumdict = dict()
for i in range(1, 37) :
sumdict[i] = []
for i in range(1, n) :
sumofi = sum([int(k) for k in list(str(i))])
sumdict[sumofi] += [i]
ans = 0
for i in range(a, b + 1) :
ans += sum(sumdict[i])
print(ans)
|
s997929159
|
Accepted
| 37 | 3,424 | 273 |
n, a, b = map(int, input().split())
sumdict = dict()
for i in range(1, 37) :
sumdict[i] = [0]
for i in range(1, n + 1) :
sumofi = sum([int(k) for k in list(str(i))])
sumdict[sumofi] += [i]
ans = 0
for i in range(a, b + 1) :
ans += sum(sumdict[i])
print(ans)
|
s801898236
|
p02261
|
u019011517
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,604 | 645 |
Let's arrange a deck of cards. There are totally 36 cards of 4 suits(S, H, C, D) and 9 values (1, 2, ... 9). For example, 'eight of heart' is represented by H8 and 'one of diamonds' is represented by D1. Your task is to write a program which sorts a given set of cards in ascending order by their values using the Bubble Sort algorithms and the Selection Sort algorithm respectively. These algorithms should be based on the following pseudocode: BubbleSort(C) 1 for i = 0 to C.length-1 2 for j = C.length-1 downto i+1 3 if C[j].value < C[j-1].value 4 swap C[j] and C[j-1] SelectionSort(C) 1 for i = 0 to C.length-1 2 mini = i 3 for j = i to C.length-1 4 if C[j].value < C[mini].value 5 mini = j 6 swap C[i] and C[mini] Note that, indices for array elements are based on 0-origin. For each algorithm, report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).
|
n = int(input())
*s1, = input().split()
s2 = s1[:]
def bubbleSort(s):
flag = True
while flag:
flag = False
for i in range(n-1):
if int(s[i][1]) > int(s[i+1][1]):
s[i],s[i+1] = s[i+1],s[i]
flag = True
return s
def selectionSort(s):
for i in range(n):
minj = i
for j in range(i+1,n):
if int(s[minj][1]) > int(s[j][1]):
minj = j
s[i],s[minj] = s[minj],s[i]
return s
print(' '.join(bubbleSort(s1)))
print("Stable")
print(' '.join(selectionSort(s2)))
if s1 == s2:
print("Stable")
else:
print("Not Stable")
|
s856362107
|
Accepted
| 20 | 5,608 | 625 |
n = int(input())
a = [i for i in input().split()]
b = [a[i] for i in range(n)]
def bubbleSort(A):
for i in range(n):
for j in range(i+1,n)[::-1]:
if int(A[j][1]) < int(A[j-1][1]):
A[j], A[j-1] = A[j-1], A[j]
return A
def selectionSort(A):
for i in range(n):
minj = i
for j in range(i,n):
if int(A[j][1]) < int(A[minj][1]):
minj = j
A[i], A[minj] = A[minj], A[i]
return A
bubbleSort(a)
selectionSort(b)
print(' '.join(a))
print("Stable")
print(' '.join(b))
if a == b:
print("Stable")
else:
print("Not stable")
|
s665208869
|
p03795
|
u993642190
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 48 |
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())
print(N * 800 - (N % 15) * 200)
|
s264172009
|
Accepted
| 18 | 2,940 | 49 |
N = int(input())
print(N * 800 - (N // 15) * 200)
|
s792894762
|
p02694
|
u536642030
| 2,000 | 1,048,576 |
Wrong Answer
| 23 | 9,168 | 136 |
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())
count = 0
a = 100
while True:
a *= 1.01
a = int(a)
if int(a) >= x:
break
else:
count += 1
print(count)
|
s097707403
|
Accepted
| 21 | 9,096 | 136 |
x = int(input())
count = 1
a = 100
while True:
a *= 1.01
a = int(a)
if int(a) >= x:
break
else:
count += 1
print(count)
|
s541692968
|
p03351
|
u306142032
| 2,000 | 1,048,576 |
Wrong Answer
| 21 | 3,316 | 197 |
Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.
|
a, b, c, d = map(int, input().split())
lst = []
lst.append(a)
lst.append(b)
lst.append(c)
lst.sort()
print(lst)
if lst[2]-lst[1] <= d and lst[1]-lst[0] <= d:
print("Yes")
else:
print("No")
|
s024461681
|
Accepted
| 17 | 2,940 | 130 |
a, b, c, d = map(int, input().split())
if abs(c-a) <= d or abs(c-b)<=d and abs(b-a) <= d:
print("Yes")
else:
print("No")
|
s609623060
|
p03251
|
u525090128
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,064 | 300 |
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.
|
# B
N,M,X,Y = map(int,input().split())
x = list(map(int,input().split()))
y = list(map(int,input().split()))
x.append(X)
y.append(Y)
ans = "War"
Xmax = max(x)
Ymin = min(y)
# print("X",X)
# print("Xmax",Xmax)
# print("Ymin",Ymin)
# 1
if Xmax < Ymin:
ans = "No War"
print("Ans",ans)
|
s409906543
|
Accepted
| 17 | 3,064 | 294 |
# B
N,M,X,Y = map(int,input().split())
x = list(map(int,input().split()))
y = list(map(int,input().split()))
x.append(X)
y.append(Y)
ans = "War"
Xmax = max(x)
Ymin = min(y)
# print("X",X)
# print("Xmax",Xmax)
# print("Ymin",Ymin)
# 1
if Xmax < Ymin:
ans = "No War"
print(ans)
|
s936844537
|
p03385
|
u503901534
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 160 |
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
n = str(input())
nl = list(n)
nlset = set(nl)
def check(onelist):
if nlset == {'a','b','c'}:
print('YES')
else:
print('NO')
check(nlset)
|
s716969550
|
Accepted
| 17 | 2,940 | 160 |
n = str(input())
nl = list(n)
nlset = set(nl)
def check(onelist):
if nlset == {'a','b','c'}:
print('Yes')
else:
print('No')
check(nlset)
|
s036726960
|
p03457
|
u635272634
| 2,000 | 262,144 |
Wrong Answer
| 348 | 27,380 | 321 |
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
N = int(input())
txys = [list(map(int, input().split())) for _ in range(N) ]
txys.insert(0, [0, 0, 0])
for i in range(N):
dt = txys[i][0] - txys[i+1][0]
dist = abs(txys[i][1] - txys[i+1][1]) + abs(txys[i][2] - txys[i+1][2])
if dt % 2 != dist % 2 or dist > dt:
print("No")
exit()
print("Yes")
|
s793203541
|
Accepted
| 403 | 27,324 | 320 |
N = int(input())
txys = [list(map(int, input().split())) for _ in range(N) ]
txys.insert(0, [0, 0, 0])
for i in range(N):
t1, x1, y1 = txys[i]
t2, x2, y2 = txys[i+1]
dt = t2 - t1
dist = abs(x1 - x2) + abs(y1 - y2)
if dt % 2 != dist % 2 or dist > dt:
print("No")
exit()
print("Yes")
|
s766967283
|
p02612
|
u196940531
| 2,000 | 1,048,576 |
Wrong Answer
| 30 | 9,096 | 87 |
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("What does it cost?"))
if N%1000==0:
print(0)
else:
print(1000-N%1000)
|
s311794047
|
Accepted
| 28 | 9,112 | 68 |
N = int(input())
if N%1000==0:
print(0)
else:
print(1000-N%1000)
|
s477465553
|
p03469
|
u917859896
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 155 |
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
|
while True:
try:
days = input().split("/")
answer = days[0].replace("7", "8")
print(answer)
except EOFError:
break
|
s404840581
|
Accepted
| 18 | 2,940 | 164 |
while True:
try:
days = input().split("/")
days[0] = days[0].replace("7", "8")
print("/".join(days))
except EOFError:
break
|
s297093694
|
p03160
|
u957100216
| 2,000 | 1,048,576 |
Wrong Answer
| 136 | 13,980 | 209 |
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.
|
N = int(input())
h = list(map(int, input().split()))
dp = [0] * (N)
for i, _h in enumerate(h):
if i > 1:
dp[i] = min(dp[i - 1] + abs(_h - h[i - 1]), dp[i - 2] + abs(_h - h[i - 2]))
print(dp[-1])
|
s803013044
|
Accepted
| 136 | 13,980 | 255 |
N = int(input())
h = list(map(int, input().split()))
dp = [0] * (N)
for i in range(1, N):
if i > 1:
dp[i] = min(dp[i-1] + abs(h[i] - h[i-1]), dp[i-2] + abs(h[i] - h[i-2]))
else:
dp[i] = dp[i-1] + abs(h[i] - h[i-1])
print(dp[-1])
|
s186769474
|
p03377
|
u534953209
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 90 |
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
A, B, X = map(int, input().split())
if A <= X <= A + B:
print("Yes")
else:
print("No")
|
s018573873
|
Accepted
| 17 | 2,940 | 91 |
A, B, X = map(int, input().split())
if A <= X <= A + B:
print("YES")
else:
print("NO")
|
s106782073
|
p03477
|
u580236524
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 136 |
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 = list(map(int, input().split()))
l= a+b
r= c+d
if l<r:
print('Right')
elif l==r:
print('Balanced')
else:
print('left')
|
s515483126
|
Accepted
| 17 | 3,060 | 129 |
a,b,c,d = map(int, input().split())
l= a+b
r= c+d
if l<r:
print('Right')
elif l==r:
print('Balanced')
else:
print('Left')
|
s451851639
|
p02743
|
u867848444
| 2,000 | 1,048,576 |
Wrong Answer
| 19 | 2,940 | 107 |
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
|
a,b,c=map(int,input().split())
if pow(a,0.5)+pow(a,0.5)<pow(c,0.5):
print('Yes')
else:
print('No')
|
s936245123
|
Accepted
| 17 | 2,940 | 106 |
a,b,c=map(int,input().split())
if c-(a+b)>=0 and (a+b-c)**2>4*a*b:
print('Yes')
else:
print('No')
|
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