HDU-2138-How many prime numbers(Miller-Rabin新解法)

题目传送门

sol1：普通判到sqrt(n)的素数判定，不多说了。

• 素数判定

  #include "bits/stdc++.h"
  using namespace std;
  bool is_prime(int n) {
      for (int i = 2; 1LL * i * i <= n; i++)
      {
          if (n % i == 0) return false;
      }
      return true;
  }
  int main() {
      int n, m;
      while (~scanf("%d", &n)) {
          int cnt = 0; 
          for (int i = 1; i <= n; i++) {
              scanf("%d", &m);
              if (is_prime(m)) cnt++;
          }
          printf("%d\n", cnt);
      }
      return 0;
  }

  复杂度sqrt(m),判断n次就是n * sqrt(m);

sol2：新get到的技巧Miller-Rabin素数测试，结合了费马小定理和二次探测定理，可以更高效的判断素数，存在误判可能，不过误判可能非常小，可以忽略不计；

• Miller-Rabin素数测试

  #include "bits/stdc++.h"
  using namespace std;
  int quick_pow(int n, int k, int p) {
      int ans = 1;
      while (k) {
          if (k & 1) ans = 1LL * ans * n % p;
          n = 1LL * n * n % p;
          k >>= 1;
      }
      return ans;
  }
  bool is_prime(int n) {
  //    if (n < 2) return false;
      int s = 0, t = n - 1;
      while (!(t & 1)) s++, t >>= 1;
      for (int i = 1; i <= 5; i++) {
          int a = rand() % (n - 1) + 1;
          int k = quick_pow(a, t, n);
          for (int j = 1; j <= s; j++) {
              int x = 1LL * k * k % n;
              if (x == 1 && k != 1 && k != n - 1) return false;
              k = x;
          }
          if (k != 1) return false;
      }
      return true;
  }
  int main() {
      int n, m;
      srand(time(NULL));
      while (~scanf("%d", &n)) {
          int cnt = 0;
          for (int i = 1; i <= n; i++) {
              scanf("%d", &m);
              if (is_prime(m)) cnt++;
          }
          printf("%d\n", cnt);
      }
      return 0;
  }

  复杂度logm，判断n次就是n * log(m);

附加一个用于 LL 范围素数测试的模板：s

• Miller-Rabin素数测试 LL 范围模板

  typedef long long LL;
  LL quick_mul(LL n, LL k, LL p) {
      LL ans = 0;
      while (k) {
          if (k & 1) ans = (ans + n) % p;
          n = (n + n) % p;
          k >>= 1;
      }
      return ans;
  }
  LL quick_pow(LL n, LL k, LL p) {
      LL ans = 1;
      while (k) {
          if (k & 1) ans = quick_mul(ans, n, p);
          n = quick_mul(n, n, p);
          k >>= 1;
      }
      return ans;
  }
  bool is_prime(LL n) {
      if (n < 2) return false;
      LL s = 0, t = n - 1;
      while (!(t & 1)) s++, t >>= 1;
      for (int i = 1; i <= 5; i++) {
          LL a = rand() % (n - 1) + 1;
          LL k = quick_pow(a, t, n);
          for (int j = 1; j <= s; j++) {
              LL x = quick_mul(k, k, n);
              if (x == 1 && k != 1 && k != n - 1) return false;
              k = x;
          }
          if (k != 1) return false;
      }
      return true;
  }

  大致差不多，为了防止爆 LL 加了一个快速积用于乘， 所以复杂度变成了log(m) * log(m)