POJ 3372 Candy Distribution

最新推荐文章于 2017-02-22 20:02:07 发布

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解决一个有趣的问题：一群孩子围成一圈，老师以特定的模式分发糖果。通过算法判断每个孩子是否都能得到至少一颗糖果。

Candy Distribution

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

题目大意：有N个小孩围成一圈，老师先发给第一个小孩一颗糖，然后跳过0个人，给2号小孩发一颗糖，接着跳过1个人，给4号小孩发糖，再跳过2个人，给7号小孩发糖……以此类推，问N个小孩能否每个人得到一颗糖。

解题思路：刚开始读错了题意，以为是跳过的人数是0,1,2的循环……后来发现题意理解错了，是跳过依次0,1,2,3……个人。这道题是有规律的，打表发现只要N == 2 ^ k，所有人都可以得到糖果，所以关键在于如何判断N是否等于2 ^ k。一种方法是判断N & (N - 1)是否为0，若为0，则N是2 ^ k，否则不是。原理如下：

设N = 16(10) = 10000(2)，则N - 1 = 15 = 01111(2)，N & (N - 1) = 0

代码如下：

#include <cstdio>

typedef long long ll;

int main()
{
	ll n; 
	while(scanf("%lld",&n) != EOF){
		printf("%s\n",n & (n - 1) ? "NO" : "YES");
	}
	return 0;
}