A Simple Math Problem HDU - 5974

Sample Input

6 8
798 10780

Sample Output

No Solution
308 490

题意：给出a，b两个数字，问是否存在x，y，使得x+y=a，lcm（x，y）=b。找不到输出“No Solution”
思路：大佬队友写出来的。x+y=a，x*y/gcd(x,y)=b，令gcd(x,y)=c，x=i*c,y=j*c，i*c+j*c=a，c*i*j=b，c*(i+j)=a，c*i*j=b。
（转载自：https://www.cnblogs.com/kimsimple/p/6792395.html）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
#define PI 3.14159
#define ll long long
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
int main()
{
    ll a, b, aa, bb, cnt, x, y;
    while (~scanf("%lld%lld", &a, &b))
    {
        cnt = gcd(a, b);
        aa = a / cnt, bb = b / cnt;
        if (aa * aa < 4 * bb) { printf("No Solution\n"); continue; }
        cnt = sqrt(aa * aa - 4 * bb);
        if (cnt * cnt != aa * aa - 4 * bb || (cnt + aa) % 2) { printf("No Solution\n"); continue; }
        x = (cnt + aa) / 2;
        y = aa - x;
        if (b % x || b % y || b / x + b / y != a) { printf("No Solution\n"); continue; }
        printf("%lld %lld\n", b / x, b / y);
    }
}