Graph（2014辽宁ACM省赛）[通俗易懂]

大家好，又见面了，我是全栈君。

问题 F: Graph

时间限制: 1 Sec 内存限制: 128 MB 提交: 30 解决: 5

题目描写叙述

Your task is to judge whether a regular polygon can be drawn only by straightedge and compass.

The length of the straightedge is infinite.

The width of the compass is infinite.

The straightedge does not have scale.

输入

There are several test cases. Each test case contains a positive integer n (3<=n<=10^9). The input will be ended by the End Of File.

输出

If the regular polygon with n sides can be drawn only by straightedge and compass, output YES in one line, otherwise, output NO in one line.

例子输入

34567

例子输出

YESYESYESYESNO

坑大爹的一题。该死的费马数。。。。。

。

p=2^n;

或 p=(2^n)*m; m为若干个不同样的费马数的积

//满足要求的边为 (2^n)*p p为费马素数
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        while(n%2==0)
        {
            n/=2;
        }
        if(n==1)
        {
            printf("YES\n");
            continue;
        }
        if(n%3==0)
            n/=3;
        if(n%5==0)
            n/=5;
        if(n%17==0)
            n/=17;
        if(n%257==0)
            n/=257;
        if(n%65537==0)
            n/=65537;
        if(n==1)
        {
            printf("YES\n");
        }
        else
            printf("NO\n");
    }
    return 0;
}

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