A Simple Math Problem HDU - 5974

A Simple Math Problem HDU - 5974

转自http://www.cnblogs.com/kimsimple/p/6792395.html

题意

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

分析

x*y = b*gcd(x,y);
设 gcd(x,y) = d;
x =i * d;
y = j * d;
a = (i + j)*d;
b = i*j*d;
gcd(a,b) = d;
x*y = b*gcd(a,b);
x + y = a;
解一元二次方程可得解

参考代码

int gcd(int a,int b)
{
    return b==0?a:gcd(b,a%b);
}
int main(void)
{
    int a,b;
    while(~scanf("%d %d",&a,&b))
    {
        int tmp = a*a-4*b*gcd(a,b);
        if(tmp < 0)
        {
            printf("No Solution\n");
            continue;
        }
        double t = sqrt(tmp);
        t = (a-t)/2;
        if(t-(int)t>1e-6)
        {
             printf("No Solution\n");
              continue;
        }
        tmp = t;

        printf("%d %d\n",tmp,a-tmp);

    }
}