# HDU 5974 A Simple Math Problem （数论+解方程组）

## Problem Description

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

Input includes multiple sets of test data.Each test data occupies one line,including two positive integers a(1≤a≤2×10^4),b(1≤b≤10^9),and their meanings are shown in the description.Contains most of the 12W test cases.

## 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


## AC 代码

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<cmath>
#include<iostream>
using namespace std;
#define MAXX 110000
typedef __int64 LL;

LL gcd(LL a,LL b)
{
if(b==0)return a;
return gcd(b,a%b);
}

void solve(int a,int b)
{
LL gc=gcd(a,b);
LL xy=b*gc;
LL x_y=sqrt(a*a-4.0*xy);
LL x=(a+x_y)/2;
LL y=a-x;
if(x/gcd(x,y)*y!=b)
{
printf("No Solution\n");
return;
}
if(x>y)swap(x,y);
printf("%I64d %I64d\n",x,y);
}

int main()
{
LL a,b;
while(~scanf("%I64d%I64d",&a,&b))
solve(a,b);
return 0;
}