题目：
GCD & LCM Inverse
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 17978 | Accepted: 3331 |
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
题意：输入两个数的最大公约数和最小公倍数，求这两个数。如果有多组，输出和最小的一对。
思路：
设这两个数分别为：x ， y。
那么 必然有 gcd(x,y)*k1 = x ； gcd(x,y)*k2 = y。
由 ：x*y / gcd（x,y)= lcm(x,y) 变形得=>> x*y = lcm*gcd;
则：(gcd*k1) * (gcd*k2) = lcm*gcd.
化简得：k1*k2 = lcm/gcd
若我们找到一组k1,k2,满足：k1*k2 = lcm/gcd 的关系，那么k1*gcd就是可能的x，k2*gcd就是可能的y。
我们还应该注意到由于 k1 = x/gcd , k2 = y/gcd 。那么一定有gcd（k1,k2）=1 。
所以我们可以将lcm/gcd进行质因数分解，之后将质因数分别再凑为两堆，一堆相乘结果就是x ，令一堆就是y。特别要注意的是，由于gcd（k1,k2）=1 ,那么为了保证这个性质，分解出的质因数若存在多个，要再乘回去，否则gcd（k1,k2）就会等于那个质因数了。
题设还有最后一个条件:输出和最小的一对。由于x*y =gcd*lcm , 那么求和最小的一对也就是求差最小的一对解。而要使，差最小，无非就是使分解出的两个数都接近sqrt（lcm/gcd），所以我们将所有质因数的组合进行dfs，找到最优解即可。
代码：
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<stack>
#include<set>typedef long long ll;using namespace std;const int INF = 0x3f3f3f3f;
const int N = 5500;
const int Times = 10; ll ct,cnt;
ll fac[N];
ll f[N];
ll ans;ll gcd(ll a , ll b){return b==0? a:gcd(b,a%b);
}ll multi(ll a , ll b , ll m){ll ans = 0;a %= m;while(b){if(b&1){ans = (ans+a)%m;b--;}b >>= 1;a = (a+a)%m;}return ans;
}ll quick_mod(ll a , ll b , ll m){ll ans = 1;a %= m;while(b){if(b&1){ans = multi(ans , a , m);b--;}b>>=1;a = multi(a,a,m);}return ans;
}bool Miller_Rabin(ll n){if(n==2) return true;if(n<2 || !(n&1)) return false;ll m = n-1;int k = 0;while((m&1)==0){k++;m>>=1;}for(int i = 0 ; i<Times ; i++){ll a = rand()%(n-1)+1;ll x = quick_mod(a,m,n);ll y = 0;for(int j = 0 ; j<k ; j++){y = multi(x,x,n);if(y==1 && x!=1 && x!=n-1) return false;x = y;}if(y!=1) return false;}return true;
}ll pollard_rho(ll n , ll c){ll i = 1 , k=2;ll x = rand()%(n-1)+1;ll y = x;while(true){i++;x = (multi(x,x,n)+c)%n;ll d = gcd((y-x+n)%n , n);if(1<d && d<n) return d;if(y == x) return n;if(i == k){y = x;k <<= 1;}}
}void find(ll n , int c){if(n == 1) return;if(Miller_Rabin(n)){fac[ct++] = n;return;}ll p = n;ll k = c;while(p >= n) p = pollard_rho(p , c--);find(p , k);find(n/p , k);
}void findx(ll i , ll x , ll q){if(i==cnt) return;if(x>ans && x<=q) ans = x;findx(i+1 , x , q);x *= f[i];if(x>ans && x<=q) ans = x;findx(i+1 , x , q);
}int main(){ll a,b;while(scanf("%lld %lld",&a,&b)!=EOF){if(a == b){printf("%lld %lld\n",a,b);continue;}ct = 0;ll n = b/a;find(n,120);sort(fac,fac+ct);f[0] = fac[0]; cnt = 0;for(int i = 1 ; i<ct ; i++){if(fac[i] == fac[i-1]){f[cnt] = f[cnt] * fac[i];}else{cnt++;f[cnt] = fac[i];}}cnt++;ll goal = (ll)sqrt(n*1.0);ans = 1;findx(0,1,goal);printf("%lld %lld\n" , ans*a , n/ans*a);}return 0;
}