超級冪的弱化版本這裡

然後就是把底數k求出來,假設p是n的一個質因數，因為n無平方因子，所以gcd(n,pn)=1,所以可以得到f(n,m)=∑i=1mϕ(i×n)=ϕ(p)∑i=1&&i%p≠0mϕ(i×np)+∑i=1mpϕ(i×p×n)=ϕ(p)∑i=1&&i%p≠0mϕ(i×np)+p∑i=1mpϕ(i×n)=ϕ(p)∑i=1&&i%p≠0mϕ(i×np)+(ϕ(p)+1)∑i=1mpϕ(i×n)=ϕ(p)∑i=1mϕ(i×np)+∑i=1mpϕ(i×n)=ϕ(p)f(np,m)+f(n,mp)

接下來就暴力遞迴。

#include <cstdio>
 

#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <map>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
#define maxn 10050000
#define mod 1000000007

bool vis[maxn];
int phi[maxn], prime[maxn], cnt, sum[maxn];
int fac[maxn];//每一個數的一個素因子
 


void init () {
    cnt = 0;
    memset (vis, 0, sizeof vis);
    phi[1] = 1;
    for (int i = 2; i < maxn; i++) {
        if (!vis[i]) {
            prime[cnt++] = i;
            fac[i] = i;
            phi[i] = i-1;
        }
        for (int j = 0; j < cnt; j++) {
            if (1LL*i*prime[j] >= maxn) break
 
;
            vis[i*prime[j]] = 1;
            fac[i*prime[j]] = prime[j];
            if (i%prime[j] == 0) {
                phi[i*prime[j]] = phi[i] * prime[j];
                break;
            }
            else {
                phi[i*prime[j]] = phi[i] * (prime[j]-1);
            }
        }
    }
    sum[0] = 0;
    for (int i = 1; i < maxn; i++) {
        sum[i] = sum[i-1]+phi[i];
        sum[i] %= mod;
    }
}

long long n, m, p, k;

long long qpow (long long a, long long b, int p) {
    if (b == 0)
        return 1%p;
    long long ans = qpow (a, b>>1, p);
    ans = ans*ans%p;
    if (b&1)
        ans = ans*a%p;
    return ans;
}

long long cal (int n, int m) {
    if (!n || !m)
        return 0;
    if (n == 1) 
        return sum[m];
    if (m == 1) 
        return phi[n];
    return (1LL*phi[fac[n]] * cal (n/fac[n], m) % mod + cal (n, m/fac[n])) % mod;
}

long long solve (int p) {
    if (p == 1)
        return 0;
    long long ans = solve (phi[p]);
    ans += phi[p];
    return qpow (k, ans, p);
}

int main () {
    init (); 
    while (scanf ("%lld%lld%lld", &n, &m, &p) == 3) {
        k = cal (n, m);
        printf ("%lld\n", solve (p));
    }
    return 0;
}