2016ICPC-大连 A Simple Math Problem （数学）

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 ;
 }