Problem Description
Declare:
k=∑mi=1φ(i∗n)mod1000000007

n is a square-free number.

φ is the Euler’s totient function.

find:
ans=kkkk...k mod p

There are infinite number of k

Input
Multiple test cases(test cases ≤100), one line per case.

Each line contains three integers, n,m and p.

1≤n,m,p≤107

Output
For each case, output a single line with one integer, ans.

Sample Input

1 2 6
1 100 9

Sample Output

4
7

Author
HIT

Source
2016 Multi-University Training Contest 1

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
using namespace std
 
;

#define LL long long

const int mod=1000000007;
const int maxn=10000005;
bool check[maxn];          //用於打表記錄的中間量
LL sumPhi[maxn];           //前i個的尤拉函式和
int cnt,phi[maxn],prime[maxn];  //素數個數，尤拉表，素數表
//素數表是第幾個素數是什麼，尤拉表是i的尤拉是phi[i];
void init(){               //素數+尤拉表
    phi[1]=1;
    cnt=0;
    for(int i=2;i<maxn;i++){
        if
 
(!check[i]){
            phi[i]=i-1;
            prime[cnt++]=i;
        }
        for(int j=0;j<cnt;j++){
            if(i*prime[j]>=maxn)break;
            check[i*prime[j]]=true;
            if(i%prime[j]==0){
                phi[i*prime[j]]=phi[i]*prime[j];
                break;
            }
            else{
                phi[i*prime[j]]=phi[i]*(prime[j]-1);
            }
        }
    }
    sumPhi[0]=0;
    for(int i=1;i<maxn;i++) sumPhi[i]=(sumPhi[i-1]+phi[i])%mod;
}
LL phiz(int n,int m){       //計算k，
    if(n==1) return sumPhi[m];
    if(m==1) return phi[n];
    if(m<1) return 0;

    for(int i=0;i<cnt;i++){
        if(n%prime[i]==0){  //n的最小質數prime[i]
            LL temp1=(prime[i]-1)*phiz(n/prime[i],m)%mod;
            LL temp2=phiz(n,m/prime[i])%mod;
            return (temp1+temp2)%mod;
        }
    }
    return 0;
}
LL pow(LL a,LL b,int p){  //快速冪
    LL ret=1;
    a%=p;
    while(b){
        if(b&1) ret=(ret*a)%p;
        a=(a*a)%p;
        b/=2;
    }
    return ret;
}
LL powe(LL k,int p){  //k^b%p
    if(p==1) return 0;
    LL temp=powe(k,phi[p]);
    return pow(k,temp+phi[p],p);
}
int main(){
    init();
    int n,m,p;
    while(scanf("%d %d %d",&n,&m,&p)!=EOF){
        LL k=phiz(n,m);
        printf("%I64d\n",powe(k,p));
    }
    return 0;
}