POJ 2154 color 波利亚群置换

最新推荐文章于 2018-10-24 21:02:41 发布

最新推荐文章于 2018-10-24 21:02:41 发布 · 79 阅读

本文介绍了如何使用波利亚原理解决计算不同颜色圆珠手链种类数量的问题，包括输入测试案例和输出答案，以及代码实现细节。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Color

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4071		Accepted: 1371

Description

Beads of N colors are connected together into a circular necklace of N beads (N<=1000000000). Your job is to calculate how many different kinds of the necklace can be produced. You should know that the necklace might not use up all the N colors, and the repetitions that are produced by rotation around the center of the circular necklace are all neglected.

You only need to output the answer module a given number P.

Input

The first line of the input is an integer X (X <= 3500) representing the number of test cases. The following X lines each contains two numbers N and P (1 <= N <= 1000000000, 1 <= P <= 30000), representing a test case.

Output

For each test case, output one line containing the answer.

Sample Input

Sample Output

Source

POJ Monthly,Lou Tiancheng

解法就是用波利亚原理

不知道为什么，我用__int64就会超时，但是用int就会WA

我考虑到可能是二分幂的时候会超过int，于是换了很多个写法的二分幂模板。最后只有一个可以过

我的代码：

#include<stdio.h> #include<string.h> typedef int ll; ll prime[35000],m; bool flag[35000]; ll eular(ll n) { ll ret=1,i; for(i=2;i*i<=n;i++) { if(n%i==0) { ret=ret*(i-1); n=n/i; while(n%i==0) { n=n/i; ret=ret*i; } } if(n==1) break; } if(n>1) ret=ret*(n-1); return ret%m; } ll exmod(ll p,ll n) { ll sq=1; while(n>0) { if(n%2==1) sq=sq*p%m; p=(p%m)*(p%m)%m; n/=2; } return sq%m; } int main() { ll n,t,ans,i; scanf("%d",&t); while(t--) { ans=0; scanf("%d%d",&n,&m); for(i=1;i*i<n;i++) { if(n%i==0) { ans=(ans+eular(n/i)*exmod(n,i-1)%m)%m; ans=(ans+eular(i)*exmod(n,n/i-1)%m)%m; } } if(i*i==n) ans=(ans+eular(i)*exmod(n,i-1)%m)%m; printf("%d\n",ans%m); } return 0; }