【POJ 2154】 Color （置换、burnside引理）

Color

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

5
1 30000
2 30000
3 30000
4 30000
5 30000

Sample Output

1
3
11
70
629

Source

 

 

【分析】　　

　　经典的置换和burnside引理应用。只有旋转。

　　考虑旋转i个单位，那么循环节有gcd(n,i)个

　　可以化出求 sigma(n^gcd(i,n))/n

　　根据莫比乌斯反演得 sigma(n^d*phi[n/d])/n   (d|n) 【表示我莫比乌斯反演白学了ORZ

　　即sigma(n^(d-1)*phi[n/d]) (d|n)

　　就算一算就好了。【不用欧拉筛，筛不到n的，先求出n的质因子，那么d的质因子也在里面，然后根据定义分解质因数求phi

　　最后要除以n，不能求逆元哦，每次少乘一个n就好了。

 

　　然而我还WA着！！！！【AC了再放代码吧

　　好了AC了：

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define Maxn 35000

int n,p;
int pri[Maxn],phi[Maxn],pl;
int d[Maxn],dl;
bool vis[Maxn];

void init()
{
    pl=0;
    memset(vis,0,sizeof(vis));
    for(int i=2;i<=Maxn-110;i++)
    {
        if(vis[i]==0)
        {
            pri[++pl]=i;
            phi[i]=i-1;
        }
        for(int j=1;j<=pl;j++)
        {
            if(i*pri[j]>Maxn-110) break;
            vis[i*pri[j]]=1;
            if(i%pri[j]!=0) phi[i*pri[j]]=phi[i]*(pri[j]-1);
            else phi[i*pri[j]]=phi[i]*pri[j];
            if(i%pri[j]==0) break;
        }
    }
    phi[1]=1;
    // for(int i=2;i<=10;i++) printf("%d ",phi[i]);
    // printf("
");
}

int qpow(int a,int b,int p)
{
    a%=p;
    int ans=1;
    while(b)
    {
        if(b&1) ans=(ans*a)%p;
        a=(a*a)%p;
        b>>=1;
    }
    return ans;
}

void get_d(int n)
{
    dl=0;
    for(int i=1;i<=pl;i++) if(n%pri[i]==0)
    {
        while(n%pri[i]==0) n/=pri[i];
        d[++dl]=pri[i];
    }
    if(n!=1) d[++dl]=n;
}

int gphi(int x)
{
    int ans=x;
    for(int i=1;i<=dl;i++) if(x%d[i]==0)
    {
        ans=ans/d[i]*(d[i]-1);
    }
    return ans;
}

int main()
{
    int T;
    scanf("%d",&T);
    init();
    while(T--)
    {
        scanf("%d%d",&n,&p);
        get_d(n);
        int ans=0;
        for(int i=1;i*i<=n;i++) if(n%i==0)
        {
            ans=(ans+(qpow(n,i-1,p)*(gphi(n/i)%p)))%p;
            if(i*i!=n) ans=(ans+(qpow(n,n/i-1,p)*(phi[i]%p)))%p;
        }
        printf("%d
",ans);
    }
    return 0;
}

 

2017-01-13 11:48:09