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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;
}

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原文地址：https://www.cnblogs.com/Juanzhang/p/11044555.html