Prime Test
Time Limit: 6000MS Memory Limit: 65536K
Total Submissions: 29925 Accepted: 7631
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
Source

POJ Monthly

直接套bin神的板子。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<time.h>
#include<iostream>
#include<string.h>
#include<math.h>
#include<algorithm>

using namespace std;

//****************************************************************
// Miller_Rabin 算法进行素数测试
//速度快，而且可以判断 < 2^63的数
//****************************************************************
const int S = 10;//随机算法判定次数，S越大，判错概率越小 一般8-10足够


//计算 (a*b)%c.   a,b都是long long的数，直接相乘可能溢出的
//  a,b,c <2^63
long long mult_mod(long long a, long long b, long long c)
{
    a %= c;
    b %= c;
    long long ret = 0;
    while (b)
    {
        if (b & 1){ ret += a; ret %= c; }
        a <<= 1;
        if (a >= c)a %= c;
        b >>= 1;
    }
    return ret;
}



//计算 ret =  x^n %c
long long pow_mod(long long x, long long n, long long mod)//x^n%c
{
    if (n == 1)return x%mod;
    x %= mod;
    long long tmp = x;
    long long ret = 1;
    while (n)
    {
        if (n & 1) ret = mult_mod(ret, tmp, mod);
        tmp = mult_mod(tmp, tmp, mod);
        n >>= 1;
    }
    return ret;
}


//通过费马小定理，即 a^(n-1)=1(mod n)  验证n是不是合数
//一定是合数返回true,不一定返回false
bool check(long long a, long long n, long long x, long long t)
{
    long long ret = pow_mod(a, x, n);
    long long last = ret;
    for (int i = 1; i <= t; i++)
    {
        ret = mult_mod(ret, ret, n);
        if (ret == 1 && last != 1 && last != n - 1) return true;//合数
        last = ret;
    }
    if (ret != 1) return true;
    return false;
}

// Miller_Rabin()算法素数判定
//是素数返回true.(可能是伪素数，但概率极小)
//合数返回false;

bool Miller_Rabin(long long n)
{
    if (n<2)return false;
    if (n == 2)return true;
    if ((n & 1) == 0) return false;//偶数
    long long x = n - 1;
    long long t = 0;
    while ((x & 1) == 0){ x >>= 1; t++; }
    for (int i = 0; i<S; i++)
    {
        long long a = rand() % (n - 1) + 1;//rand()需要stdlib.h头文件
        if (check(a, n, x, t))
            return false;//合数
    }
    return true;
}


//************************************************
//pollard_rho 算法进行质因数分解
//************************************************
long long factor[100];//质因数分解结果（刚返回时是无序的）
int tol;//质因数的个数。数组下标从0~tol-1开始

long long gcd(long long a, long long b)
{
    if (a == 0)return 1;//???????
    if (a<0) return gcd(-a, b);
    while (b)
    {
        long long t = a%b;
        a = b;
        b = t;
    }
    return a;
}
//找出一个因子
long long Pollard_rho(long long x, long long c)
{
    long long i = 1, k = 2;
    long long x0 = rand() % x;
    long long y = x0;
    while (1)
    {
        i++;
        x0 = (mult_mod(x0, x0, x) + c) % x;
        long long d = gcd(y - x0, x);
        if (d != 1 && d != x) return d;
        if (y == x0) return x;
        if (i == k){ y = x0; k += k; }
    }
}
//对n进行素因子分解 存入factor数组
void findfac(long long n)
{
    if (Miller_Rabin(n))//素数
    {
        factor[tol++] = n;
        return;
    }
    long long p = n;
    while (p >= n)p = Pollard_rho(p, rand() % (n - 1) + 1);
    findfac(p);
    findfac(n / p);
}
int main()
{
    // srand(time(NULL));//需要time.h头文件  //POJ上G++要去掉这句话
    int T;
    long long n;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%I64d", &n);
        if (Miller_Rabin(n))
        {
            printf("Prime\n");
            continue;
        }
        tol = 0;
        findfac(n);
        long long ans = factor[0];
        for (int i = 1; i<tol; i++)
            if (factor[i]<ans)
                ans = factor[i];
        printf("%I64d\n", ans);
    }
    return 0;
}