POJ 2154-Color（Polya定理-旋转 串项链）

Color

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 10352 Accepted: 3368

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

51 300002 300003 300004 300005 30000

Sample Output

131170629

Source

POJ Monthly,Lou Tiancheng

题目意思：

给定m种颜色的珠子，每种颜色的珠子个数不限，将它们串成长度为n的项链，计算一共能做成多少种不重复的项链。仅考虑旋转，不考虑翻转。

解题思路：

旋转，将项链顺时针旋转i格之后，其循环节数是gcd(n,i)，计算出所有不同的着色方案。

枚举i会TLE，所以用欧拉函数优化，每个循环的长度L=N/gcd(i,N)，结果是∑(Euler(L)×N^(L-1))%P。

#include<cstdio>#include<cmath>#include<vector>#include<algorithm>using namespace std;int quick(int a,int b,int m )//a^b%m 快速幂取模{    int ans=1;    a%=m;    while(b)    {        if(b&1)        {            ans=(ans*a)%m;            --b;        }        b/=2;        a=a*a%m;    }    return ans;}int Euler(int n)//欧拉函数{    int i,ans=n;    for(i=2; i*i<=n; ++i)        if(n%i==0)        {            ans-=ans/i;            while(n%i==0)                n/=i;        }    if(n>1)        ans-=ans/n;    return ans;}int main(){#ifdef ONLINE_JUDGE#else    freopen("G:/cbx/read.txt","r",stdin);    //freopen("G:/cbx/out.txt","w",stdout);#endif    int t;    scanf("%d",&t);    while(t--)    {        int n,p,ans=0;        scanf("%d%d",&n,&p);        int i;        for(i=1; i*i<n; i++)            if(n%i==0) //如果i是n的约数，则n/i也是n的约数，也要做同样的处理            {                ans=(ans+(Euler(n/i)%p)*quick(n,i-1,p))%p ;                ans=(ans+(Euler(i)%p)*quick(n,n/i-1,p))%p ;            }        if(i*i==n) ans=(ans+(Euler(i)%p)*quick(n,i-1,p))%p ;        printf("%d\n",ans);    }    return 0;}