A Simple Math Problem（hdu 5974 数论题

#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll gcd(ll n,ll m)
{
    return m==0?n:gcd(m,n%m);
}
int main()
{
    int i,j,t,n,m,a,b;
    while(~scanf("%d%d",&a,&b))
    {
        ll g=gcd(a,b);
        ll A=a/g;
        ll B=b/g;
        ll deita=A*A-4*B;
        if(deita<0)cout<<"No Solution\n";
        else
        {
            ll z1=(-A+sqrt(deita));
            ll z2=(-A-sqrt(deita));
            ll de=sqrt(deita);
            if(z1%-2==1||z2%-2==1||de*de!=deita)cout<<"No Solution\n";//非整数情况
            else cout<<min(z1/(-2),z2/(-2))*g<<" "<<max(z1/(-2),z2/(-2))*g<<endl;
        }//因为都是整数解，不可能有负数解；所以两个解一个是p一个是q，再*g就是x，y了
    }
    return 0;
}