【LOJ143】质数判定

【题目链接】

• 点击打开链接

【思路要点】

• MillerRabin M i l l e r R a b i n 素数测试模板题。
• 时间复杂度 O(TLog2N) O ( T L o g 2 N ) 。

【代码】

#include

using namespace std;
const int MAXN = 100005;
const int p[9] = {2, 3, 5, 7, 11, 13, 17, 19, 23};
typedef long long ll;
typedef long double ld;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
  x = 0; int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
template <typename T> void write(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
  write(x);
  puts("");
}
ll times(ll a, ll b, ll P) {
  ll tmp = (ld) a * b / P;
  return ((a * b - tmp * P) % P + P) % P;
}
ll power(ll a, ll b, ll P) {
  if (b == 0) return 1;
  ll tmp = power(a, b / 2, P);
  if (b % 2 == 0) return times(tmp, tmp, P);
  else return times(a, times(tmp, tmp, P), P);
}
bool prime(ll n) {
  for (int i = 0; i <= 8; i++) {
      if (p[i] == n) return true;
      else if (p[i] > n) return false;
      ll tmp = power(p[i], n - 1, n), tnp = n - 1;
      if (tmp != 1) return false;
      while (tmp == 1 && tnp % 2 == 0) {
          tnp /= 2;
          tmp = power(p[i], tnp, n);
          if (tmp != 1 && tmp != n - 1) return false;
      }
  }
  return true;
}
int main() {
  ll n;
  while (scanf("%lld", &n) != EOF) {
      if (n == 1) {
          puts("N");
          continue;
      }
      if (prime(n)) puts("Y");
      else puts("N");
  }
  return 0;
}