poj 2429 GCD & LCM Inverse（Pollard_rho整数分解+dfs枚举）

题目链接：http://poj.org/problem?id=2429

题意：给你两个数a和b的最大公约数和最小公倍数，求a和b，使得a+b的和尽量小

思路:因为a*b/gcd=lcm,所以a/gcd*b/gcd=lcm/gcd,令p为a/gcd,q为b/gcd,所以p，q互质，所以只要把lcm/gcd分解为两个互质的数即可，分解后再用dfs枚举，a+b>=2ab,所以a越和b接近，a+b越小。

代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <map>
#include <vector>
#include <math.h>
using namespace std;
#define ll long long
const int N=1e6+10;
const ll inf=0x3f3f3f3f;
ll mul(ll a,ll b,ll n)
{
    ll ans=0;
    while(b)
    {
        if(b&1)ans=(ans+a)%n;
        a=(a+a)%n;
        b>>=1;
    }
    return ans;
}
ll poww(ll a,ll b,ll n)
{
    ll ans=1;
    while(b)
    {
        if(b&1)ans=mul(ans,a,n);
        a=mul(a,a,n);
        b>>=1;
    }
    return ans;
}
ll yz[N],yz1[N];
int cnt,z;
ll ans;
bool miller_rabin(ll n)
{
    ll t,u,a,x,y;
    if(n==2)return true;
    if(n==1||!(n&1))return false;
    for(t=0,u=n-1;!(u&1);t++,u>>=1);
    for(int i=0;i<10;i++)
    {
        ll a=rand()%(n-1)+1;
        x=poww(a,u,n);
        for(int j=0;j<t;j++)
        {
            y=mul(x,x,n);
            if(y==1&&x!=1&&x!=n-1)
                return false;
            x=y;
        }
        if(x!=1)return false;
    }
    return true;
}
ll gcd(ll a,ll b)
{
    if(a<0)return gcd(-a,b);
    if(a==0)return 1;
    while(b)
    {
        ll t=a%b;
        a=b;
        b=t;
    }
    return a;
}
ll Pollard_rho(ll n,ll c)
{
    int i=1,k=2;
    ll x=rand()%(n-1)+1;
    ll y=x;
    while(1)
    {
        i++;
        x=(mul(x,x,n)+c)%n;
        ll p=gcd(y-x,n);
        if(p!=1&&p!=n)return p;
        if(y==x)return n;
        if(i==k)
        {
            y=x;
            k+=k;
        }
    }
}
void findx(ll n)
{
    if(n==1)return ;
    if(miller_rabin(n))
    {
        yz[cnt++]=n;
        //cout<<n<<endl;
        return;
    }
    ll p=n;
    while(p>=n)
        p=Pollard_rho(p,rand()%(n-1)+1);
    findx(p);
    findx(n/p);
}
void dfs(ll i,ll x,ll k)
{
    if(i>z)return;
    if(x>ans&&x<=k)ans=x;
    dfs(i+1,x,k);
    x*=yz1[i];
    if(x>ans&&x<=k)ans=x;
    dfs(i+1,x,k);
}
int main()
{
    ll x,y;
    while(~scanf("%lld %lld",&x,&y))
    {
        //cout<<gcd(x,y)<<endl;
        ll k=y/x;
        if(x==y)
        {
            printf("%lld %lld\n",x,y);
        }
        else if(miller_rabin(k))
        {
            ll aa=1LL;
            ll bb=k;
            printf("%lld %lld\n",aa*x,bb*x);
        }
        else
        {
             cnt=0;
            findx(k);
            sort(yz,yz+cnt);
            yz1[0]=yz[0];
            z=0;
            for(int i=0;i<cnt-1;i++)
            {
                if(yz[i]==yz[i+1])yz1[z]*=yz[i+1];
                else
                {
                    z++;
                    yz1[z]=yz[i+1];
                }
            }
            ll temp=(ll)sqrt(k*1.0);
            ans=0;
            dfs(0,1,temp);
            printf("%lld %lld\n",x*ans,k/ans*x);
        }
    }
    return 0;
}

import java.util.Scanner;

public class Main {
      public static void main(String []args)
      {
    	  Scanner cin=new Scanner(System.in);
    	  while(cin.hasNext())
    	  {
    		  long x=cin.nextLong();
    		  long y=cin.nextLong();
    		  long a,b;
    		  if(x==y)
    		  {
    			  System.out.println(x+" "+y);
    		  }
    		  else
    		  {
    			  y/=x;
    			  a=b=0;
    			  for(long i=(long)Math.sqrt(y);i>0;i--)
    			  {
    				  if(y%i==0&&gcd(i,y/i)==1)
    				  {
    					  a=i*x;
    					  b=y/i*x;
    					  break;
    				  }
    			  }
    			  System.out.println(a+" "+b);
    		  }
    	  }
      }
      public static long gcd(long a,long b)
      {
    	  if(a<b)
    	  {
    		  long t=a;
    		  a=b;
    		  b=t;
    	  }
    	  if(b==0) return a;
    	  return gcd(b,a%b);
      }
}