poj 2429

因子分解很简单 直接套用模版即可。。然后之后的处理，这里一个讲解不错。。。。http://blog.sina.com.cn/s/blog_69c3f0410100uac0.html  可以参考一下的。。

Problem 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

#include<iostream>
#include<ctime>
#include<algorithm>
#include<cmath>
using namespace std;
long long p,f[1000],ans;
long long f2[1000],q;
long long mod2(long long a,long long b,long long n)// a*b%n
{
    long long  exp = a%n, res = 0;
    while(b){
        if(b&1){
            res += exp;
            if(res>n) res -= n;
        }
        exp <<= 1;
        if(exp>n)  exp -= n;
        b>>=1;
    }
    return res;
}
 
long long mod(long long a,long long b,long long m) //(a^b)%n
{
    long long exp=a%m, res=1;
    while(b>=1){
        if(b&1)
            res=mod2(res,exp,m);
        exp = mod2(exp,exp,m);
        b>>=1;
    }
    return res;
}
//miller-rabin法测试素数
bool miller_rabin(long long n)
{
    if(n==2)return 1;
    if(n<2||!(n&1))return 0;
    long long a, u=n-1, x, y;
    int t=0;
    while(u%2==0){
        t++;
        u/=2;
    }
    srand(time(0));
    for(int i=0;i<10;i++) {
        a = rand() % (n-1) + 1;
        x = mod(a, u, n);
        for(int j=0;j<t;j++){
            y = mod2(x, x, n);
            if ( y == 1 && x != 1 && x != n-1 )
                return false;
            x = y;
        }
        if( y!=1) return false;
    }
    return true;
}
// 因子分解
long long gcd(long long a,long long b){
    long long c;
    while(b){
        c=b; b=a%b; a=c;
    }
    return a;
}
long long pollard_rho(long long n,long long k){
    srand(time(0));
    long long x=rand()%(n-1)+1;
    long long i=1,t=2,y=x,d;
    while(1){
        i++;
        x=(mod2(x,x,n)+k)%n;
        d=gcd(y-x,n);
        if(d>1 && d<n) return d;
        if(x==y) return n;
        if(t==i) y=x,t<<=1;
    }
}
void get_factor(long long n,long long k){
    if(miller_rabin(n)){
        f[p++]=n;
       return ;
    }
    long long h=n;
    while(h>=n)
        h=pollard_rho(h,k--);
    get_factor(h,k);
    get_factor(n/h,k);
}
void dfs(long long i,long long x,long long k){
    if(i>q) return ;
    if(x>ans && x<=k) ans=x;
    dfs(i+1,x,k);
    x*=f2[i];
    if(x>ans && x<=k) ans=x;
    dfs(i+1,x,k);
}
int main(int argc, char** argv) {
    long long m,n,k;
    while(scanf("%lld%lld",&m,&n)!=-1){
        if(m==n){
            printf("%lld %lld\n",m,n); continue;
        }
        k=n/m; p=0; q=0; ans=1;
        get_factor(k,180);
        sort(f,f+p);
        f2[0]=f[0];
        for(int i=0;i<p-1;i++)
            if(f[i]==f[i+1]) f2[q]*=f[i+1];
            else{
                q++; f2[q]=f[i+1];
            }
        long long tmp=(long long)sqrt(k*1.0);
        dfs(0,1,tmp);
        printf("%lld %lld\n",m*ans,k/ans*m);
    }
    return 0;
}