2016ACM/ICPC亚洲区大连站 D - A Simple Math Problem（尚未通过）

Given two positive integers a and b,find suitable X and Y to meet the conditions:
X+Y=a
Least Common Multiple (X, Y) =b

Input

Output
For each set of input data,output a line of two integers,representing X, Y.If you cannot find such X and Y,output one line of “No Solution”(without quotation).

Sample Input
6 8
798 10780
Sample Output
No Solution
308 490

#include<iostream>
#include<cmath>
#include<cstdio>

using namespace std;
long long gcd(long long a,long long b);
long long lcm(long long a,long long b);

int main(){
    long long a,b;

    while(cin >> a >> b){
        int temp = gcd(a,b);
        long long x,y;
        if(a*a > 4*b*temp){
            x = (a + sqrt(a*a - 4*b*temp))/2;
            y = a-x;
            if(x+y == a && lcm(x,y) == b)
                printf("%d %d\n",x,y);
            else
                cout << "No Solution" << endl;
        }
        else{
            cout << "No Solution" << endl;
        }
    }

    return 0;
}

long long gcd(long long a,long long b){
    return b == 0 ? a : gcd(b,a%b);
}

long long lcm(long long a,long long b){
    return a*b/gcd(a,b);
}