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 |
|---|---|---|---|---|---|---|---|---|---|
| #750146 | #5141. Identical Parity | ckq328 | WA | 72ms | 3648kb | C++20 | 1.0kb | 2024-11-15 13:02:20 | 2024-11-15 13:02:21 |
Judging History
answer
#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N=1e6+10,M=200+10,inf=1e18,mod=998244353;
int n,k;
int a[N],f[N];
void solve(){
cin>>n>>k;
if(k==1){
if(n==1) cout<<"yes"<<endl;
else cout<<"No"<<endl;
}
else if(k%2==0) cout<<"yes"<<endl;
else{
if(k==n) cout<<"yes"<<endl;
else{
vector<int> vc;
int sum=0;
for(int i=1;i<=k;i++){
int cnt=n/k;
if(n%k>=i) cnt++;
vc.push_back(cnt);
if(cnt!=1) sum+=cnt;
}
if(sum<n) cout<<"yes"<<endl;
else{
if(n%k==0) cout<<"no"<<endl;
else if((n/k)*(k-(n%k))==((n/k)+1)*(n%k)) cout<<"yes"<<endl;
else if(n/k+1==n/k*2&&n%2==0) cout<<"yes"<<endl;
else if(2==n/k&&n%2==1) cout<<"yes"<<endl;
else cout<<"no"<<endl;
}
}
}
}
signed main(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int tt;
cin>>tt;
// tt=1;
while(tt--){
solve();
}
}
/*
1
10
598196518 640274071 983359971 71550121 96204862 799843967 446173607 796619138 402690754 223219513
*/
详细
Test #1:
score: 100
Accepted
time: 1ms
memory: 3508kb
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: 72ms
memory: 3648kb
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:
yes 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 [127th token]