学步园

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.

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.

For each test case, output a and b in ascending order. If there are multiple solutions, output the pair with smallest a + b.

3 60

12 15

 

大数质因数分解

然后DFS搜索出每一个因子

 

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<iostream>
#include<time.h>

using namespace std;

typedef unsigned __int64 ll;
const ll MAX=100;
const ll inf=(ll)1<<62;
ll f0[100],ff,n,tmp,ret,ret1;

ll myrandom()
{
    ll a;
    a=rand();
    a*=rand();
    a*=rand();
    a*=rand();
    return a;
}

ll mulmod(ll a,ll b,ll c)
{
    ll ret=0;
    while(b)
	{
        if(b&1)
		{
            ret+=a;
            if(ret>c)
				ret=ret-c;
        }
		a=a<<1;
        if(a>c)
            a=a-c;
       b=b>>1;
    }
    return ret;
}

ll powmod(ll a,ll b,ll c)
{
    ll ret=1;
    while(b)
	{
        if(b&1)
            ret=mulmod(ret,a,c);
        a=mulmod(a,a,c);
        b=b>>1;
    }
    return ret;
}

int miller(ll base,ll n)
{
    ll m=n-1,k=0;
	int i;
    while(m%2==0)
	{
        m=m>>1;
        k++;
    }
    if(powmod(base,m,n)==1)
        return 1;
    for(i=0;i<k;i++)
	{
        if(powmod(base,m<<i,n)==n-1)
            return 1;
    }
    return 0;
}

int miller_rabin(ll n)
{
	int i;
    for( i=2;i<100;++i)
        if(n%i==0)
            if(n==i)
                return 1;
            else return 0;
    for(i=0;i<MAX;++i)
	{
        ll base=myrandom()%(n-1)+1;
        if(!miller(base,n))
            return 0;
    }
    return 1;
}

ll gcd(ll a,ll b)
{
    if(b==0) 
		return a;
	else
		return gcd(b,a%b);
}

ll f(ll a,ll b)
{
    return (mulmod(a,a,b)+1)%b;
}

ll pollard_rho(ll n)
{
	int i;
    if(n<=2) 
		return 0;
    for( i=2;i<100;++i)
        if(n%i==0)
            return i;
    ll p,x,xx;
    for( i=1;i<MAX;i ++)
	{
        x=myrandom()%n;
        xx=f(x,n);
        p=gcd((xx+n-x)%n,n);
        while(p==1)
		{
            x=f(x,n);
            xx=f(f(xx,n),n);
            p=gcd((xx+n-x)%n,n)%n;
        }
        if(p) 
			return p;
    }
    return 0;
}

ll prime(ll a)
{
    if(miller_rabin(a)) 
		return 0;
    ll t=pollard_rho(a);
    ll p=prime(t);
    if(p) 
		return p;
    return t;
}

ll ans,ansa,ansb;

ll min(ll a,ll b)
{
	if(a>b)
		return b;
	else
		return a;
}

void dfs(ll n,ll now,ll id)
{
	if(now*now>n)//剪哈枝
		return;
	if(id==ff)
	{
		ll a,b;
		a=now;
		b=n/now;
		if(a+b<ans)
		{
			ans=a+b;
			ansa=a;
			ansb=b;
		}
		return;
	}
	dfs(n,now*f0[id],id+1);
	dfs(n,now,id+1);
}

void swap(ll &a,ll &b)
{
	ll temp;
	temp=a;
	a=b;
	b=temp;
}

int main()
{
	ll a,b,key,tmp;
	srand(time(NULL));
	while(scanf("%I64u%I64u",&a,&b)!=EOF)
	{
		if(a>b)
			swap(a,b);
		if(a==b||miller_rabin(b))
		{
			printf("%I64u %I64u\n",a,b);
			continue;
		}
		key=b/a;
		ff=0,n=key;
		while(key>1)
		{
			if(miller_rabin(key))
				break;
			tmp=prime(key);
			f0[ff]=tmp;
			key=key/tmp;
			while(key%tmp==0)
			{
				key=key/tmp;
				f0[ff]=f0[ff]*tmp;
			}
			ff++;
		}
		if(key>1)
		{
			f0[ff++]=key;
		}
		ans=inf;
		dfs(n,1,0);
		printf("%I64u %I64u\n",ansa*a,ansb*a);
	}
	return 0;
}