2154 Color//POLYA-优快云博客

本文链接：https://blog.youkuaiyun.com/hqd_acm/article/details/6181992

Color

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 2967		Accepted: 1016

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

/*
题目求N色N个不等价置换个数(mod p) N<= 10^9
置换只有旋转一种方式，那么共有n个置换
基本知识：环的个数为gcd(n , i) , 长度L=n / gcd(n , i)   其中 i 为转的位子数
普通求法: ∑n^( gcd(n,i) )  0<=i<n  复杂度过高
优化:枚举环的长度L
枚举优化: L可以从1取到sqrt(n) ,因为L|n , n/L | n
对于每个L,我们再看有几个i满足条件
n/L = gcd(n , i)
那么令 a=n/L = gcd(n , i) , 再设i = at
那么当且仅当gcd(L,t)=1时候，才有gcd(n,i) = a
显然满足条件的i的个数就是t的个数也就是phi(L)
那么最后统计一下就是 ∑(phi(L) * N^(L-1) ) % p  (L即枚举值)
*/
#include<cstdio>
#include<cstring>
const int N=40000;
int prime_num;
int prime[20000];
int n,p;
bool isprime[N];
void init()
{
    memset(isprime,true,sizeof(isprime));
    for(int i=2;i*i<N;i++)
    if(isprime[i])
    {
        for(int j=i*i;j<N;j+=i)
        isprime[j]=false;
    }
    prime_num=0;
    for(int i=2;i<N;i++)
    if(isprime[i])  prime[++prime_num]=i;
}
int phi(int x)//欧拉函数
{
    int num=x;
    for(int i=1;prime[i]*prime[i]<=x;i++)
    {
        if(x%prime[i]==0)
        {
            num = (num/prime[i])*(prime[i]-1);
            while(x % prime[i] == 0)
            {
                x = x / prime[i];
            }
        }
    }
    if(x != 1) num = (num/x)*(x-1);
    return num % p;
}
int exp(int x, int y)
{
    long long i;
    if(y == 0) return 1;
    if(y == 1) return x%p;
    i = exp(x,y/2);
    i = (i*i)%p;
    if(y&1) i=i*x;
    return i % p;
}
int main()
{
    int cas;
    scanf("%d",&cas);
    init();
    while(cas--)
    {
         int ans=0;
         scanf("%d%d",&n,&p);
         for(int i=1;i*i<=n;i++)
         {
             if(i*i==n)
             {
                 ans+=phi(i)*exp(n,i-1);
                 ans%=p;
                 break;
             }
             else if(n%i==0)//i,n/i都是质因子
             {
                 ans+=phi(i)*exp(n,n/i-1);
                 ans+=phi(n/i)*exp(n,i-1);
                 ans%=p;
             }
         }
         printf("%d/n",ans);
    }
    return 0;
}