poj 1811 Prime Test【 随机素数测试与大数分解】

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 <2^63long 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 %clong 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falsebool 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;}