| ||||||||||
| Online Judge | Problem Set | Authors | Online Contests | User | ||||||
|---|---|---|---|---|---|---|---|---|---|---|
| Web Board Home Page F.A.Qs Statistical Charts | Current Contest Past Contests Scheduled Contests Award Contest | |||||||||
有个定理,用了就很简单的。#include <iostream>
using namespace std;
int main ()
{
int n,k;
while(cin>>n>>k)
cout<<(((n-k)&k)>0?0:1)<<endl;
return 0;
}
Followed by: Post your reply here: |
All Rights Reserved 2003-2013 Ying Fuchen,Xu Pengcheng,Xie Di
Any problem, Please Contact Administrator