poj3372-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

代码实现：

#include<iostream>

#include<cstdio>

using namespace std;

int main()

{

    int n;

    while(cin>>n)

    {

        printf("%s",n&(n-1)?"NO":"YES");

        cout<<endl;

    }

    return 0;

}

1.求解的过程中便能发现，手推这道题目的规律非常难推，也尝试过猜测，but WA，既然不好推，程序是干嘛的呢，不就是解决实际问题的，写一段程序来跑一下，代码如下，善于应用取余

#include<iostream>

#include<cstring>

using namespace std;

int main()

{

    int n,a[1000],i,s,sum;//确保输入的n要比数组的数据范围小

    while(cin>>n)

    {

        memset(a,0,sizeof(a));

        a[1]=a[2]=1;

        s=2;

        for(i=1;i<1000000;i++)//这个地方要注意，有的n需要很多次之后才能所有的孩子都拿到糖果，所以i的范围尽量大一点

        {

            s=s+i+1;

            if(s%n==0)

            {

                a
=1;

            }

            else

            {

                a[s%n]=1;

            }

        }

        for(i=1;i<=n;i++)

        {

            cout<<a[i]<<" ";

        }

        cout<<endl;

    }

    return 0;

}

2.推理发现，所有2的倍数都输出YES

3.解释为什么n&(n-1)==0,n便为2的倍数，设n为2的x方

如果n为2的倍数，n除了第x位为1，其他位都为0，而n-1第x位一定为0，所以与的结果为0

其他的数我觉得至少最高非0位相同，这样结果不为0