POJ 2429 GCD & LCM Inverse（Pollard-rho 大整数分解+DFS）

GCD & LCM Inverse Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 16795   Accepted: 3107 Description Given two positive integers a and b, we can easily calculate the greatest common divisor (GCD) and the least common multiple (LCM) of a and b. But what about the inverse? That is: given GCD and LCM, finding a and b. Input The input contains multiple test cases, each of which contains two positive integers, the GCD and the LCM. You can assume that these two numbers are both less than 2^63. Output For each test case, output a and b in ascending order. If there are multiple solutions, output the pair with smallest a + b. Sample Input 3 60 Sample Output 12 15 Source POJ Achilles

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题意：给你两个数的gcd和lcm，让你求这两个数

题解：我们考虑lcm/gcd，这样得出的两个数一定是互质的。

然后我们考虑分解质因数，两个互质的数一定没有公因数，因此dfs暴力其中一个数的因数即可。

数很大，无法直接暴力分解因数，这里采用Pollard-rho算法分解（不知道的自行百度）

然后怎么保证a+b最小呢，其实两个数最接近时，这两个数的和一定最小（可以证明的）。

呢我们直接dfs即可。

#include<set>      
#include<map>         
#include<stack>                
#include<queue>                
#include<vector>        
#include<string>     
#include<time.h>    
#include<math.h>                
#include<stdio.h>                
#include<iostream>                
#include<string.h>                
#include<stdlib.h>        
#include<algorithm>       
#include<functional>        
using namespace std;                
#define ll long long          
#define inf 1000000000           
#define Mod 10007                
#define maxn  50500    
#define lowbit(x) (x&-x)                
#define eps 1e-9   
ll cnt, fat[101],mx,ans;
ll pri[2550005], a[2550005] = {1,1};
ll Multi(ll a, ll b, ll mod)        //和上面一样，但比上面慢很多 
{ 
    ll ans = 0; 
    a %= mod; 
    while(b) 
    { 
        if(b%2==1)  ans = (ans+a)%mod, b--; 
        else  a = (a+a)%mod, b /= 2; 
    } 
    return ans; 
} 
ll Pow(ll a, ll b, ll mod)  
{  
    ll ans = 1;  
    a %= mod;  
    while(b)  
    {  
        if(b&1)  ans = Multi(ans, a, mod), b--;  
        else  a = Multi(a, a, mod), b /= 2;  
    }  
    return ans;  
}  
ll Gcd(ll a, ll b)  
{ 
	if(b==0)
		return a;
	return Gcd(b, a%b);
}  
int Miller_Rabin(ll n)  
{  
    int i, j, k;  
    ll a, x, y, mod;  
    if(n==2)  return 1;  
    if(n<2 || n%2==0)  return 0;  
    k = 0, mod = n-1;  
    while(mod%2==0)  
    {  
        k++;  
        mod /= 2;  
    }  
    for(i=1;i<=15;i++)  
    {  
        a = rand()%(n-1)+1;  
        x = Pow(a, mod, n);  
        y = 0;  
        for(j=1;j<=k;j++)  
        {  
            y = Multi(x, x, n);  
            if(y==1 && x!=1 && x!=n-1)  
                return 0;  
            x = y;  
        }  
        if(y!=1)  
            return 0;  
    }  
    return 1;  
}  
ll Divi(ll n)  
{  
    ll i, k, x, y, p, c;  
    if(n==1)  
        return 1;  
    k = 2, p = 1;  
    y = x = rand()%n, c = rand()%(n-1)+1;  
    for(i=1;p==1;i++)
    {
        x = (Multi(x, x, n)+c)%n;
		p = x-y;
		if(p<0)
			p = -p;
        p = Gcd(n, p);  
        if(i==k)  
            y = x, k *= 2;  
    }
    return p;  
}  
void Pollard_rho(ll n)  
{  
    ll p;  
    if(n==1)  
        return;  
    if(Miller_Rabin(n))  
        fat[++cnt] = n;  
    else  
    {  
        p = Divi(n);
		Pollard_rho(p);  
        Pollard_rho(n/p);  
    }  
}  
void dfs(ll x,ll y)
{
	if(x>cnt)
	{
		if(y>ans && y<=mx)
			ans=y;
		return;
	}
	dfs(x+1,y);
	dfs(x+1,y*fat[x]);
}
int main(void)
{
	ll a,b;
	while(scanf("%lld%lld",&a,&b)!=EOF)
	{
		cnt=0;b=b/a;
		Pollard_rho(b);
		sort(fat+1,fat+cnt+1);
		int i,j=1;
		for(i=2;i<=cnt;i++)
		{
			while(fat[i]==fat[i-1] && i<=cnt)
				fat[j]*=fat[i],i++;     
			if(i<=cnt)
				fat[++j]=fat[i];
		}   
		cnt=j;ans=1;
		mx=(ll)sqrt((double)b);
		dfs(1ll,1ll);
		printf("%lld %lld\n",ans*a,b/ans*a);
	}
	return 0;
}