给定 \(n,\ m,\ p\) ，令 \(k=\displaystyle\sum_{i=1}^m\varphi(i\times n)\pmod{10^9+7}\)
求 \(k^{k^{k^{\cdots^{k}}}}\pmod{p}\)
共 \(T\) 组询问， \(n\) 无平方因子
\(T\leq100,\ n,\ m,\ p\leq10^7\)
数论，计数
令 \(f(n,\ m)=\displaystyle\sum_{i=1}^m\varphi(i\times n)\) 。假设 \(p\) 是 \(n\) 的一个质因子，若 \((i,\ n)=1\) ，则 \(\varphi(i\times n)=\varphi(i)\times\varphi(n)\) ，否则若 \(p\ |\ i\) ，可以将 \(i\) 看作 \(k\times p\) ，否则 \(p\not |\;i\) ，于是分类讨论
\[\begin{aligned}f(n,\ m)&=\displaystyle\sum_{i=1}^m{[p\ |\ i](\varphi(p)\times\varphi(i\times\frac{n}{p}))}+\sum_{i=1}^{\frac{m}{p}}\varphi(i\times n\times p)\\&=\varphi(p)\times\sum_{i=1}^m{[p\not|\;i]\varphi(i\times\frac{n}{p})+p\times\sum_{i=1}^{\frac{m}{p}}\varphi(i\times n)}\\&=\varphi(p)\times\sum_{i=1}^m{[p\not|\;i]\varphi(i\times\frac{n}{p})+(\varphi(p)+1)\times\sum_{i=1}^{\frac{m}{p}}\varphi(i\times n)}\\&=\varphi(p)\times\sum_{i=1}^m{[p\not|\;i]\varphi(i\times\frac{n}{p})}+\varphi(p)\times\sum_{i=1}^{\frac{m}{p}}\varphi(i\times n)+\sum_{i=1}^{\frac{m}{p}}\varphi(i\times n)\end{aligned}\]
但前两项是可以合并的，第二项恰好将第一项补全了，于是
\[f(n,\ m)=\varphi(p)\times\sum_{i=1}^m\varphi(i\times\frac{n}{p})+\sum_{i=1}^{\frac{m}{p}}\varphi(i\times n)\]
所以 \[f(n,\ m)=\varphi(p)\times f(\frac{n}{p},\ m)+f(n,\ \frac{m}{p})\]
递归时枚举一个质因子就够了
而求 \(k^{k^{k^{\cdots^{k}}}}\pmod{p}\) 直接用欧拉定理就可以了
时间复杂度 \(O(\) 能过 \()\)
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e7 + 10, P = 1e9 + 7;
int tot, p[maxn], phi[maxn], sum[maxn];
inline int inc(int x, int y) {
return x + y < P ? x + y : x + y - P;
}
inline int qp(int a, int k, int P) {
int res = 1;
for (; k; k >>= 1, a = 1ll * a * a % P) {
if (k & 1) res = 1ll * res * a % P;
}
return res;
}
void sieve() {
int N = 10000000;
for (int i = 2; i <= N; i++) {
if (!p[i]) p[++tot] = i, phi[i] = i - 1;
for (int j = 1; j <= tot && i * p[j] <= N; j++) {
p[i * p[j]] = 1;
if (i % p[j] == 0) {
phi[i * p[j]] = phi[i] * p[j]; break;
}
phi[i * p[j]] = phi[i] * (p[j] - 1);
}
}
phi[1] = 1;
for (int i = 1; i <= N; i++) {
sum[i] = inc(sum[i - 1], phi[i]);
}
}
int dfs(int a, int P) {
int t = phi[P];
return t == 1 ? 0 : qp(a, dfs(a, t) % t + t, P);
}
int calc(int n, int m) {
if (!m) return 0;
if (n == 1) return sum[m];
int tmp = sqrt(n);
for (int i = 2; i <= tmp; i++) {
if (n % i == 0) {
return (calc(n, m / i) + 1ll * phi[i] * calc(n / i, m)) % P;
}
}
return (calc(n, m / n) + 1ll * phi[n] * sum[m]) % P;
}
int main() {
sieve();
int n, m, p;
while (~scanf("%d %d %d", &n, &m, &p)) {
printf("%d\n", dfs(calc(n, m), p));
}
return 0;
}原文地址：https://www.cnblogs.com/Juanzhang/p/11044555.html