soj 1454. Pseudoprime numbers

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描述：

费马小定理指出 for any prime number p and for any integer a > 1, a^p == a (mod p)。

对于合数p，如果存在1 < a < p满足上式，那么就称 p 为以 a 为底的伪素数。

给出p和a，问p是不是以a为底的伪素数。2 < p ≤ 1,000,000,000 and 1 < a < p

思路：

先用Miller-Rabin测试p是不是素数，再测试a^p%p是否等于a。a^p%p要用快速指数幂和二分相乘防止溢出。

代码：

#include <cstdio>#include <ctime>#include <cstdlib>using namespace std;#define Times 12typedef long long LL;LL random(LL n){return (LL)((double)rand()/RAND_MAX * n + 0.5);}LL multi(LL a, LL b, LL &m) // (a * b) % m{LL ret = 0;while (b){if (b & 1) ret = (ret + a) % m;b >>= 1;a = (a<<1) % m;}return ret;}LL pow(LL a, LL b, LL &m) // (a ^ b) % m{LL ans = 1;a %= m;while (b){if (b & 1) ans = multi(ans, a, m);b >>= 1;a = multi(a,a,m);}return ans;}bool Witness(LL a, LL n) // if n is base-a pseudoprime{LL m = n-1;int j = 0;while (! (m & 1)){j ++;m >>= 1;}LL x = pow(a, m, n);if (x == 1 || x == n-1) return false;while (j --){x = multi(x, x, n);if (x == n-1) return false;}return true;}bool Miller_Rabin(LL n) // if n is a prime{if (n < 2) return false;if (n == 2) return true;if (! (n&1)) return false;for (int i = 1; i <= Times; ++ i){LL a = random(n-2) + 1;if (Witness(a, n)) return false;}return true;}bool PseudoPrime(LL a, LL n){return pow(a, n, n) == a;}int main(){LL n, a;while (scanf("%lld %lld", &n, &a)){if (n == 0 && a == 0) break;if (!Miller_Rabin(n) && PseudoPrime(a, n)) printf("yes\n");else printf("no\n");}}