Information
- ID
- 47
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 8
- Tags
- # Submissions
- 1548
- Accepted
- 244
- Uploaded By
#include<stdio.h>
int main(){
long long num;//要判断的数
int m;
int count;
scanf("%d",&m);
for(int i=0; i<m; i++){//m行数
scanf("%lld",&num);
//是2的幂次方只有两个条件,1:大于0 2:二进制只有一个1,因此一旦减1比为0
if((num&(num-1))== 0 && num>0)
printf("YES\n");
else
printf("NO\n");
}
}
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