【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));
}
}