poj 3372(找规律)

Posted 2020-08-14 AC菜鸟机

Candy Distribution

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6033		Accepted: 3351

Description

N children standing in circle who are numbered 1 through N clockwise are waiting their candies. Their teacher distributes the candies by in the following way:

First the teacher gives child No.1 and No.2 a candy each. Then he walks clockwise along the circle, skipping one child (child No.3) and giving the next one (child No.4) a candy. And then he goes on his walk, skipping two children (child No.5 and No.6) and giving the next one (child No.7) a candy. And so on.

Now you have to tell the teacher whether all the children will get at least one candy?

Input

The input consists of several data sets, each containing a positive integer N (2 ≤ N ≤ 1,000,000,000).

Output

For each data set the output should be either "YES" or "NO".

Sample Input

2
3 
4

Sample Output

YES
NO
YES


老师给糖给 1 2号学生,然后接下来隔着一个学生给糖,隔着两个学生给糖,问一直这样下去,是否所有学生都会拿到糖?

找规律，判断 n是否为 2^k...快速判断 n是否为 2的指数幂的方法 n&n-1...

#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
typedef long long LL;
const int N = 1005;

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF){
        if((n&(n-1))==0) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}