POJ1811 Prime Test(miller素数判断&&pollar_rho大数分解)

http://blog.csdn.net/shiyuankongbu/article/details/9202373

发现自己原来的那份模板是有问题的，而且竟然找不出是哪里的问题，所以就用了上面的链接上的一份代码，下面只是寄存一下这份代码，以后打印出来当模板好了。

#pragma warning(disable:4996)
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <cstdlib>
#include <cstdio>
using namespace std;

#define Times 10
map<long long, int>m;
long long mi;
long long random(long long n)
{
	return ((double)rand() / RAND_MAX*n + 0.5);
}

long long multi(long long a, long long b, long long mod)
{
	long long ans = 0;
	while (b){
		if (b & 1) ans = (ans + a) % mod;
		b >>= 1;
		a = (a << 1) % mod;
	}
	return ans;
}
long long Pow(long long a, long long b, long long mod)
{
	long long ans = 1;
	while (b){
		if (b & 1) ans = multi(ans, a, mod);
		b >>= 1;
		a = multi(a, a, mod);
	}
	return ans;
}
bool witness(long long a, long long n)
{
	long long d = n - 1;
	while (!(d & 1))
		d >>= 1;
	long long t = Pow(a, d, n);
	while (d != n - 1 && t != 1 && t != n - 1)
	{
		t = multi(t, t, n);
		d <<= 1;
	}
	return t == n - 1 || d & 1;
}
bool miller_rabin(long long n)
{
	if (n == 2)
		return true;
	if (n<2 || !(n & 1))
		return false;
	for (int i = 1; i <= Times; i++)
	{
		long long a = random(n - 2) + 1;
		if (!witness(a, n))
			return false;
	}
	return true;
}
long long gcd(long long a, long long b)
{
	return a&&b ? gcd(b, a%b) : a + b;
}
long long pollard_rho(long long n, int c)
{
	long long x, y, d, i = 1, k = 2;
	x = random(n - 2) + 1;
	y = x;
	while (1){
		i++;
		x = (multi(x, x, n) + c) % n;
		d = gcd(y - x, n);
		if (1<d&&d<n)
			return d;
		if (y == x)
			return n;
		if (i == k){
			y = x;
			k <<= 1;
		}
	}
}
void find(long long n, int c)
{
	if (n == 1)
		return;
	if (miller_rabin(n)){
		m[n]++;
		mi = min(mi, n);
		return;
	}
	long long p = n;
	while (p >= n)
		p = pollard_rho(p, c--);
	find(p, c);
	find(n / p, c);
}
int main()
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		long long n;
		scanf("%lld", &n);
		mi = n;
		if (miller_rabin(n))
			cout << "Prime" << endl;
		else
		{
			find(n, 12312);
			cout << mi << endl;
		}
	}
	return 0;
}