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
1 #include <iostream> 2 #include<cstdio> 3 #include<algorithm> 4 #include<cstring> 5 #include<string> 6 #include<queue> 7 #include<vector> 8 #include<cmath> 9 using namespace std; 10 const double pi=acos(-1.0); 11 int n,d,x; 12 long long a,b,u; 13 long long gcd(long long a,long long b) 14 { 15 if (b==0) return a; 16 else return gcd(b,a%b); 17 } 18 int main() 19 { 20 while(~scanf("%lld%lld",&a,&b)) 21 { 22 u=gcd(a,b); 23 a=a/u; 24 b=b/u; 25 if ( (long long)sqrt(a*a-4*b)*(long long)sqrt(a*a-4*b)==a*a-4*b ) 26 { 27 long long t=(long long)sqrt(a*a-4*b); 28 if ((a+t)%2==0 && a>t) printf("%lld %lld ",u*(a-t)/2,u*(a+t)/2); 29 else printf("No Solution "); 30 } else printf("No Solution "); 31 } 32 return 0; 33 }