C++实现的大整数分解Pollard's rho算法程序

最新推荐文章于 2023-11-11 20:09:43 发布

weixin_34087307

最新推荐文章于 2023-11-11 20:09:43 发布

阅读量411

点赞数

CC 4.0 BY-SA版权

文章标签： c/c++

原文链接：http://www.cnblogs.com/tigerisland/p/7564849.html

本文介绍了一种用于合数质因数分解的Pollard's Rho算法，并提供了一个使用C++实现的具体例子。该算法通过迭代计算来寻找非平凡因子。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

代码来自GeeksforGeeks的Pollard’s Rho Algorithm for Prime Factorization。

C++语言程序代码如下：

/* C++ program to find a prime factor of composite using
   Pollard's Rho algorithm */
#include<bits/stdc++.h>
using namespace std;
 
/* Function to calculate (base^exponent)%modulus */
long long int modular_pow(long long int base, int exponent,
                          long long int modulus)
{
    /* initialize result */
    long long int result = 1;
 
    while (exponent > 0)
    {
        /* if y is odd, multiply base with result */
        if (exponent & 1)
            result = (result * base) % modulus;
 
        /* exponent = exponent/2 */
        exponent = exponent >> 1;
 
        /* base = base * base */
        base = (base * base) % modulus;
    }
    return result;
}
 
/* method to return prime divisor for n */
long long int PollardRho(long long int n)
{
    /* initialize random seed */
    srand (time(NULL));
 
    /* no prime divisor for 1 */
    if (n==1) return n;
 
    /* even number means one of the divisors is 2 */
    if (n % 2 == 0) return 2;
 
    /* we will pick from the range [2, N) */
    long long int x = (rand()%(n-2))+2;
    long long int y = x;
 
    /* the constant in f(x).
     * Algorithm can be re-run with a different c
     * if it throws failure for a composite. */
    long long int c = (rand()%(n-1))+1;
 
    /* Initialize candidate divisor (or result) */
    long long int d = 1;  
 
    /* until the prime factor isn't obtained.
       If n is prime, return n */
    while (d==1)
    {
        /* Tortoise Move: x(i+1) = f(x(i)) */
        x = (modular_pow(x, 2, n) + c + n)%n;
 
        /* Hare Move: y(i+1) = f(f(y(i))) */
        y = (modular_pow(y, 2, n) + c + n)%n;
        y = (modular_pow(y, 2, n) + c + n)%n;
 
        /* check gcd of |x-y| and n */
        d = __gcd(abs(x-y), n);
 
        /* retry if the algorithm fails to find prime factor
         * with chosen x and c */
        if (d==n) return PollardRho(n);
    }
 
    return d;
}
 
/* driver function */
int main()
{
    long long int n = 10967535067;
    printf("One of the divisors for %lld is %lld.",
          n, PollardRho(n));
    return 0;
}

转载于:https://www.cnblogs.com/tigerisland/p/7564849.html