問題 F: Graph

時間限制:?1 Sec??內存限制:?128 MB
提交:?30??解決:?5
[提交][狀態][論壇]

題目描寫敘述

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.

輸入

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.

輸出

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.

例子輸入

34567

例子輸出

YESYESYESYESNO


坑大爹的一題。該死的費馬數。。。。。

。

p=2^n; 或 p=(2^n)*m; m為若干個不同樣的費馬數的積 //滿足要求的邊為 (2^n)*p p為費馬素數 #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");}elseprintf("NO\n");}return 0; }

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