A Simple Math Problem HDU - 5974 （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

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

题意：找到两个整数x和y，使得 x+y=a  且  lcm（x,y）=b，如果能输出x，y，否者，输出No Solution

思路：

令gcd(x,y)=g;

那么

g   *   k1   =   x

g   *   k2   =   y

由于g是x，y的gcd，则k1和k2互质

>>>>>>   g * k1 * k2 = b; 

>>>>>>   g * k1  +  g * k2 = a                         

>>>>>>   k1  *  k2 =  b/g     k1 +  k2 =  a/g

因为k1,k2互质，则 k1  *  k2   与    k1 +  k2   互质

则 b/g  与 a/g  互质，则  g  =  gcd(a,b)

所以  gcd（x,y） =  gcd(a,b)

化解 x+y=a  且  lcm（x,y）=b

为一元二次方程：  x^2  -  ax  + b*gcd(a,b)=0;

求解x

代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int Gcd(int x,int y){
	while(y){
		int z=x%y;
		x=y;
		y=z;
	}
	return x;
}
int main(){
	int a,b;
	while(~scanf("%d%d",&a,&b)){
		int gcd=Gcd(a,b);
		int det=a*a-4*b*gcd;
		if(det<0) printf("No Solution\n");
		else{
			int p=sqrt(det);
			if(p*p!=det) printf("No Solution\n");
			else{
				int x1=(a+p)/2;
				int x2=(a-p)/2;
				int x=min(x1,x2);
				printf("%d %d\n",x,a-x);
			}
		}
	}
	return 0;
}