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
InputInput 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.OutputFor 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 设gcd(x,y)=u,x=m*u,y=n*u
a/b=(mu+nu)/(m*n*u)
化简一下可以求出m,n，从而求出X,Y
#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<queue>
#include<vector>
#include<cmath>
using namespace std;
const double pi=acos(-1.0);
int n,d,x;
long long a,b,u;
long long gcd(long long a,long long b)
{
if (b==) return a;
else return gcd(b,a%b);
}
int main()
{
while(~scanf("%lld%lld",&a,&b))
{
u=gcd(a,b);
a=a/u;
b=b/u;
if ( (long long)sqrt(a*a-*b)*(long long)sqrt(a*a-*b)==a*a-*b )
{
long long t=(long long)sqrt(a*a-*b);
if ((a+t)%== && a>t) printf("%lld %lld\n",u*(a-t)/,u*(a+t)/);
else printf("No Solution\n");
} else printf("No Solution\n");
}
return ;
}