poj-2154-polya+euler函数

Color

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 11758		Accepted: 3783

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

POJ Monthly,Lou Tiancheng

　　n种颜色的珠子组成一个长度为n的圆环，旋转能够到达的属于重复状态，问一共有多少种不同的方案。

　　由polya定理可知 ans=SUM{ n^gcd(k,n) |  0<=k<n} / n ，n<1e9  如果直接算的话肯定会T。由于所有的gcd(k,n)都是n的因子，可以考虑对

相同的因子分成一组，这样找到每一组的个数就好办了， ans=1/n * SUM{ s(d)*n^d |  d|n ,s(d)为使得gcd(k,n)=d成立的k的个数 }，

gcd(k,n)==d   -->  gcd(k/d,n/d)==1  --> phi(n/d)  。 所以得出s(d)=phi(n/d) ,枚举一下n的因子统计答案就好了。

　　在计算phi的时候先把sqrt(1e9)之内的素数筛出来再计算会更快，否则会超时。

　　

#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<cmath> 
using namespace std;
#define LL long long 
#define PI acos(-1.0)
#define N 33333
int mod;
bool isp[N+10];
vector<int>prime;
void init(){
    int m=sqrt(N+0.5);
    for(int i=2;i<=N;i++){
        if(!isp[i]){
            prime.push_back(i);
            for(int j=i*i;j<=N;j+=i)isp[j]=1;
        }
    }
}
int qpow(int a,int b,int m){
    a%=m;
    int r=1;
    while(b){
        if(b&1) r=r*a%m;
        b>>=1;
        a=a*a%m;
    }
    return r%m;
}
int euler(int n){
    int m=sqrt(n+0.5);
    int ans=n;
    for(int i=0;prime[i]<=m;++i){
        if(n%prime[i]==0){
        ans=ans/prime[i]*(prime[i]-1);
        while(n%prime[i]==0) n/=prime[i];
        }
    }
    if(n>1) ans=ans/n*(n-1);
    return ans%mod;
}
int main()
{
    int t,n,i,j,k,d;
    init();
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&mod);
        int ans=0;
        for(i=1;i*i<n;++i){
            if(n%i==0){
                (ans+=(euler(n/i)*qpow(n,i-1,mod)%mod))%=mod;
                
                (ans+=(euler(i)*qpow(n,n/i-1,mod)%mod))%=mod;
            }
        }
        if(i*i==n) (ans+=(euler(i)*qpow(n,i-1,mod)%mod))%=mod;
        cout<<ans<<endl;
    }
    return 0;
}