使用Pollard_rho算法分解质因数

分解质因数的朴素算法

最简单的算法即为从 [2, sqrt（N）] 进行遍历。

vector<int> breakdown(int N) {vector<int> result;for (int i = 2; i * i <= N; i++) {if (N % i == 0) {  // 如果 i 能够整除 N，说明 i 为 N 的一个质因子。while (N % i == 0) N /= i;result.push_back(i);}}if (N != 1) {  // 说明再经过操作之后 N 留下了一个素数result.push_back(N);}return result;
}

这个算法当 n 是质数时拥有最差时间复杂度 O(n) ，不过可以先用米勒-拉宾特判一下 n 是不是质数，这样的话最差时间复杂度就是当 n 是质数的平方时的 O(sqrt (n)) 了

Pollard_rho将一个数分解为质因数相乘的形式

如：200 = 22255
Pollard_rho算法是找到一个数的最大质因数，我参考此算法进行修改，实现了将一个数分解为质因数相乘形式的程序。

#include <iostream>
#include <stdlib.h>
#include <vector>int t;
long long max_factor, n;
std::vector<long long> factor; // 存储质因数long long gcd(long long a, long long b) {if (b == 0) return a;return gcd(b, a % b);
}long long quick_pow(long long x, long long p, long long mod) {  // 快速幂long long ans = 1;while (p) {if (p & 1) ans = (__int128)ans * x % mod;x = (__int128)x * x % mod;p >>= 1;}return ans;
}bool Miller_Rabin(long long p) {  // 判断素数if (p < 2) return 0;if (p == 2) return 1;if (p == 3) return 1;long long d = p - 1, r = 0;while (!(d & 1)) ++r, d >>= 1;  // 将d处理为奇数std::vector<long long> ud = {2, 325, 9375, 28178, 450775, 9780504, 1795265022};for (long long a:ud) {long long x = quick_pow(a, d, p);if (x == 1 || x == p - 1) continue;for (int i = 0; i < r - 1; ++i) {x = (__int128)x * x % p;if (x == p - 1) break;}if (x != p - 1) return 0;}return 1;
}long long Pollard_Rho(long long x) {long long s = 0, t = 0;long long c = (long long)rand() % (x - 1) + 1;int step = 0, goal = 1;long long val = 1;for (goal = 1;; goal *= 2, s = t, val = 1) {  // 倍增优化for (step = 1; step <= goal; ++step) {t = ((__int128)t * t + c) % x;val = (__int128)val * abs(t - s) % x;if ((step % 127) == 0) {long long d = gcd(val, x);if (d > 1) return d;}}long long d = gcd(val, x);if (d > 1) return d;}
}void fac(long long x) {if (x <= max_factor || x < 2) return;if (Miller_Rabin(x)) {              // 如果x为质数max_factor = std::max(max_factor, x);  // 更新答案return;}long long p = x;while (p >= x) p = Pollard_Rho(x);  // 使用该算法while ((x % p) == 0) x /= p;fac(x), fac(p);  // 继续向下分解x和p
}void decompose(long long n) { // 将n分解为质因数相乘的形式srand((unsigned)time(NULL));max_factor = 0;fac(n);if(max_factor == n) { // 最大的质因数是自己factor.push_back(max_factor);}else {factor.push_back(max_factor);n /= max_factor;decompose(n);}
}int main() {std::cout << "----start decompose n: [exit please input number 0 !]----" << std::endl;while(1) {std::cout << "input number: ";std::cin >> n;if(n == 0) {std::cout << "stop and exit program!" << std::endl;break;}decompose(n);std::cout << n << " = ";for(std::vector<long long>::iterator it = factor.begin(); it != factor.end()-1; ++it) {std::cout << *it << "*";}std::cout << *(factor.end()-1) << std::endl;std::vector<long long>().swap(factor);}return 0;
}