素数分解-优快云博客

本文链接：https://blog.youkuaiyun.com/xing_mo/article/details/87455410

本文介绍了两种质因数分解算法的实现，包括基础的分解方法和更高效的Pollard's rho算法。通过C++代码详细展示了如何将一个整数分解为其质因数的乘积，适用于数学和计算机科学领域的学习和研究。

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Eddy’s research I

hdu-1164
Eddy’s interest is very extensive, recently he is interested in prime number. Eddy discover the all number owned can be divided into the multiply of prime number, but he can’t write program, so Eddy has to ask intelligent you to help him, he asks you to write a program which can do the number to divided into the multiply of prime number factor .

Input
The input will contain a number 1 < x<= 65535 per line representing the number of elements of the set.
Output
You have to print a line in the output for each entry with the answer to the previous question.
Sample Input
11
9412
Sample Output
11
2213*181

代码1：

#include<iostream>
#include<cstdio>
using namespace std;
long long n,factor[50],cnt;
int main()
{
	while(~scanf("%lld",&n))
	{
		cnt = 0;
		long long x = 2;
		while(x < n)
		{
			if(n % x)
				++x;
			else
			{
				n /= x;
				factor[cnt++] = x;
			}
		}
		factor[cnt++] = x;
		for(int i = 0;i < cnt-1;++i)
			printf("%lld*",factor[i]);
		printf("%lld\n",factor[cnt-1]);
	}
	return 0;
}

代码2：(pollard_rho算法)

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
 
using namespace std;
typedef long long LL;
 
long long factor[110], cnt;
long long qmul(long long a, long long b, long long m)//?a*b%m
{
    long long ans = 0;
    a %= m;
    while(b)
    {
        if(b & 1)ans = (ans + a) % m;
        b >>= 1;
        a = (a + a) % m;
    }
    return ans;
}
long long qpow(long long a, long long b, long long m)//?a^b % m
{
    long long ans = 1;
    a %= m;
    while(b)
    {
        if(b & 1)ans = qmul(a, ans, m);
        b >>= 1;
        a = qmul(a, a, m);
    }
    return ans;
}
bool Miller_rabin(long long n, int timenum = 10) {
    if (n < 2)
        return false;
    if (n == 2)
        return true;
    while (timenum--) {
        if (qpow(rand() % (n - 2) + 2, n - 1, n) != 1)
            return false;
    }
    return true;
}
long long Gcd(long long a, long long b) {
    long long t;
    while (b) {
        t = a;
        a = b;
        b = t % b;
    }
    return a;
}
void Pollard(long long n);
 
void Factor(long long n) {
    long long d = 2;
    while (true) {
        if (n % d == 0) {
            Pollard(d);
            Pollard(n / d);
            return;
        }
        d++;
    }
}
void Pollard(long long n) {
    if (n <= 0)
        printf("error\n");
    if (n == 1)
        return;
    if (Miller_rabin(n)) {
        factor[cnt++] = n;
        return;
    }
    long long i = 0, k = 2, x, y, d;
    x = y = rand() % (n - 1) + 1;
    while (i++ < 123456) {
        x = (qmul(x, x, n) + n - 1) % n;
        d = Gcd((y - x + n) % n, n);
        if (d != 1) {
            Pollard(d);
            Pollard(n / d);
            return;
        }
        if (i == k) {
            y = x;
            k *= 2;
        }
    }
    Factor(n);
}
 
 
int main() {
    long long n;
    while (~scanf("%lld",&n)) 
	{
        if (Miller_rabin(n))
            printf("%lld\n",n);
        else 
		{
            cnt = 0;
            Pollard(n);
            sort(factor, factor + cnt);
            for(int i=0;i<cnt-1;i++) 
            	printf("%lld*", factor[i]);
            printf("%lld\n", factor[cnt-1]);
        }
    }
    return 0;
}