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问题 F: Graph
时间限制: 1 Sec
内存限制: 128 MB
提交: 30
解决: 5
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论坛]
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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