【hdu 5728 PowMod】【数论】【欧拉函数】【欧拉降幂递归取模】【欧拉积性函数】

【链接】

http://acm.hdu.edu.cn/showproblem.php?pid=5728

【题意】

n是无平方因子的数

定义k=∑mi=1φ(i∗n) mod 1000000007，求K^k^k^k......%p

【思路】

先欧拉性质求出k,再用欧拉降幂，A^B=A^B%phi(C)+phi(C)  (mod C)求出答案

∑(i=1~m)phi(i*n)=sum[n][m]

∑(i=1~m)phi[i*n]=phi[p]*∑(i=1~m)phi(i*n/p)+∑(i=1~m/p)phi(i*n)

对于i%p!=0，那么显然——∑(i=1~m,i不为p的倍数)phi[i*n]=phi[p]*∑(i=1~m,i不为p的倍数)phi(i*n/p)

对于i%p==0，那么则有——∑(i=1~m,i为p的倍数）phi[i*n]=(phi[p]+1)*∑(i=1~m,i为p的倍数)phi(i*n/p)

【代码】

#include<bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int maxn = 1e7 + 5;
using ll = long long;
int phi[maxn];
int v[maxn];
int prime[maxn/5];
int phisum[maxn];

void euler_all() {
	int m = 0;
	phi[1] = 1; phisum[1] = 1;
	for (int i = 2; i < maxn; ++i) {
		if (!v[i]) {
			prime[++m] = i;
			phi[i] = i - 1;
		}
		for (int j = 1; j <= m && prime[j] * i < maxn; ++j) {
			v[prime[j] * i] = 1;
			if (i % prime[j] == 0) {
				phi[prime[j] * i] = phi[i] * prime[j];
				break;
			}
			else phi[prime[j] * i] = phi[i] * (prime[j] - 1);
		}
		phisum[i] = (phisum[i - 1] + phi[i]) % mod;
	}
}

//sum(n,m)=phi[p]*sum(n/p,m)+sum(n,m/p);

ll  sum(int n, int m){
	if (n == 1)return phisum[m];
	if (m == 1)return phi[n];
	if (m < 1)return 0;
	for (int i = 1;; ++i)if (n%prime[i] == 0){
		int pp = prime[i];
		return (phi[pp] * sum(n / pp, m) + sum(n, m / pp)) % mod;
	}
}

ll qpow(ll x, int p, int Z) {
	ll y = 1;
	while (p) {
		if (p & 1)y = y * x%Z;
		x = x * x%Z;
		p >>= 1;
	}
	return y;
}

int cal(int k, int p) {
	if (p == 1)return 0;
	int tmp = cal(k, phi[p]);
	return qpow(k, tmp+phi[p], p);
}

int main() {
	int n, m, p;
	euler_all();
	while (~scanf("%d%d%d", &n, &m, &p)) {
		int k = sum(n, m);
		printf("%d\n", cal(k, p));
	}
}