QOJ.ac
QOJ
| ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
|---|---|---|---|---|---|---|---|---|---|
| #595419 | #5141. Identical Parity | louhao088# | WA | 11ms | 3824kb | C++23 | 660b | 2024-09-28 13:40:55 | 2024-09-28 13:40:58 |
Judging History
answer
#include<bits/stdc++.h>
using namespace std;
int T,n,K;
int main()
{
scanf("%d",&T);
while(T--)
{
scanf("%d %d",&n,&K);
int p=(n/2-1)/(n/K+1)+1;
int q=((n+1)/2-1)/(n/K+1)+1;
if((p+q)*(n/K+1)>n+K-n%K) {printf("No\n");continue;}
int s=((n/2)/(n/K)+n/2)/(n/K+1);
int t=(((n+1)/2)/(n/K)+((n+1)/2))/(n/K+1);
if(p>s || q>t) {printf("No\n");continue;}
if((p+q)*(n/K+1)-n>K-n%K) {printf("No\n");continue;}
if((s+t)*(n/K+1)-n<K-n%K) {printf("No\n");continue;}
if((n+K-n%K)%(n/K+1)==0) printf("Yes\n");
else printf("No\n");
}
return 0;
}
详细
Test #1:
score: 100
Accepted
time: 0ms
memory: 3816kb
input:
3 3 1 4 2 5 3
output:
No Yes Yes
result:
ok 3 token(s): yes count is 2, no count is 1
Test #2:
score: -100
Wrong Answer
time: 11ms
memory: 3824kb
input:
100000 1 1 2 1 2 2 3 1 3 2 3 3 4 1 4 2 4 3 4 4 5 1 5 2 5 3 5 4 5 5 6 1 6 2 6 3 6 4 6 5 6 6 7 1 7 2 7 3 7 4 7 5 7 6 7 7 8 1 8 2 8 3 8 4 8 5 8 6 8 7 8 8 9 1 9 2 9 3 9 4 9 5 9 6 9 7 9 8 9 9 10 1 10 2 10 3 10 4 10 5 10 6 10 7 10 8 10 9 10 10 11 1 11 2 11 3 11 4 11 5 11 6 11 7 11 8 11 9 11 10 11 11 12 1 ...
output:
No No Yes No Yes Yes No Yes Yes Yes No Yes Yes Yes Yes No Yes No Yes Yes Yes No Yes Yes Yes Yes Yes Yes No Yes No Yes Yes Yes Yes Yes No Yes No Yes Yes Yes Yes Yes Yes No Yes No Yes No Yes Yes Yes Yes Yes No Yes No Yes Yes Yes Yes Yes Yes Yes Yes No Yes No Yes Yes Yes Yes Yes Yes Yes Yes Yes No Yes ...
result:
wrong answer expected YES, found NO [1st token]