POJ1811 (Prime Test Pollard rho整数分解,Miller-Rabin素数测试)

最新推荐文章于 2020-09-09 09:51:22 发布

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最新推荐文章于 2020-09-09 09:51:22 发布

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分类专栏： poj(北大)OJ题目数论文章标签： Miller-Rabin素数测试 Pollard rho整数分解

本文链接：https://blog.youkuaiyun.com/u013021513/article/details/47361811

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本文介绍了使用Miller-Rabin素数测试和Pollard rho方法来判断大整数是否为素数，并在不是素数的情况下找到最小的素因子。通过算法实现，解决了大数分解的问题。

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Prime Test

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 30648		Accepted: 7880
Case Time Limit: 4000MS

Description

Given a big integer number, you are required to find out whether it's a prime number.

Input

The first line contains the number of test cases T (1 <= T <= 20 ), then the following T lines each contains an integer number N (2 <= N < 2⁵⁴).

Output

For each test case, if N is a prime number, output a line containing the word "Prime", otherwise, output a line containing the smallest prime factor of N.

Sample Input

2
5
10

Sample Output

Prime
2

题目大意:

给一个数N(2 <= N < 2 ^ 54),判断N是否为素数.

解题思路:

由于N非常大，用普通的素数判断方法不行,所以需要采用Miller-Rabin素数测试,如果是素数直接输出"Prime",如果不是素数,用Pollard rho方法进行整数分解,求出所有的质因数后取其最小的即为所求.

AC代码:

#include<iostream>
#include<ctime>
#include<cstdlib>
#include<cmath> 
using namespace std;

const int N = 501;
const int C = 201;
const int Times = 11;
#define MAX ((long long)1<<61)
typedef long long ll;

ll mini;
int ct;
ll res[N];

ll gcd(ll a,ll b)
{
	return b == 0 ? a : gcd(b,a % b);
}

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 > 0)
	{
		if(b & 1)
			ret = (ret + a) % m;
		b >>= 1;
		a = (a << 1) % m;
	}
	return ret;
}

ll quick_mod(ll a,ll b,ll m)
{
	ll ans = 1;
	a %= m;
	while(b)
	{
		if(b & 1)
		{
			ans = multi(ans,a,m);
			b--;
		}
		b /= 2;
		a = multi(a,a,m);
	}
	return ans;
}

bool Witness(ll a,ll n) //判断大的素数 
{
	ll m = n - 1;
	int j = 0;
	while(!(m & 1))
	{
		j++;
		m >>= 1;
	}
	ll x = quick_mod(a,m,n);
	if(x == 1 || x == n - 1)
		return false;
	while(j--)
	{
		x = x * x % n;
		if(x == n - 1)
			return false;
	}
	return true;
}

bool miller_rabin(ll n)
{
	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;
}


ll pollard_rho(ll n,int c) //大数分解因数 
{
	ll x,y,d,i = 1,k = 2;
	x = random(n-1) + 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(ll n,int k)
{
	if(n == 1)
		return;
	//cout<<"1111"<<endl;
	if(miller_rabin(n)) //没进去 
	{
		res[++ct] = n;
		//cout<<"1111"<<endl;
		if(n < mini)
		{
			mini = n;
			//cout<<mini<<endl;
		}
		return;
	}
	ll p = n;
	while(p >= n)
	{
		p = pollard_rho(p,k--);
	}
	find(p,k);
	find(n / p,k);
}

int main()
{
	int total;
	ll n;
	scanf("%d",&total);
	while(total--)
	{
		scanf("%lld",&n);
		ct = 0;
		mini = MAX;
		if(miller_rabin(n))
		{
			printf("Prime\n");
		}
		else
		{
			find(n,C);
			printf("%lld\n",mini);
		}
	}
	return 0;	
}