hdu5728 PowMod

hdu5728 PowMod

给定 (n, m, p) ，令 (k=displaystylesum_{i=1}^mvarphi(i imes n)pmod{10^9+7})

求 (k^{k^{k^{cdots^{k}}}}pmod{p})

共 (T) 组询问， (n) 无平方因子

(Tleq100, n, m, pleq10^7)

数论，计数

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令 (f(n, m)=displaystylesum_{i=1}^mvarphi(i imes n)) 。假设 (p) 是 (n) 的一个质因子，若 ((i, n)=1) ，则 (varphi(i imes n)=varphi(i) imesvarphi(n)) ，否则若 (p | i) ，可以将 (i) 看作 (k imes p) ，否则 (p ot |;i) ，于是分类讨论

[egin{aligned}f(n, m)&=displaystylesum_{i=1}^m{[p | i](varphi(p) imesvarphi(i imesfrac{n}{p}))}+sum_{i=1}^{frac{m}{p}}varphi(i imes n imes p)\&=varphi(p) imessum_{i=1}^m{[p ot|;i]varphi(i imesfrac{n}{p})+p imessum_{i=1}^{frac{m}{p}}varphi(i imes n)}\&=varphi(p) imessum_{i=1}^m{[p ot|;i]varphi(i imesfrac{n}{p})+(varphi(p)+1) imessum_{i=1}^{frac{m}{p}}varphi(i imes n)}\&=varphi(p) imessum_{i=1}^m{[p ot|;i]varphi(i imesfrac{n}{p})}+varphi(p) imessum_{i=1}^{frac{m}{p}}varphi(i imes n)+sum_{i=1}^{frac{m}{p}}varphi(i imes 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) imessum_{i=1}^mvarphi(i imesfrac{n}{p})+sum_{i=1}^{frac{m}{p}}varphi(i imes n) ]

所以 $$f(n, m)=varphi(p) imes f(frac{n}{p}, m)+f(n, frac{m}{p})$$

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
", dfs(calc(n, m), p));
  }
  return 0;
}