生成具有给定的和与根的乘积的二次方程式

给定两个整数S和M ，任务是找到二次方程的系数，使得根的总和和乘积分别为S和M。

例子：

Input: S = 5, M = 6
Output: 1 -5 6
Explanation:
For the quadratic equation x² – 5x + 6 = 0. The root of the equation are 2 and 3. Therefore, the sum of roots is 2 + 3 = 5, and the product of roots is 2*3 = 6.

Input: S = -2, M = 1
Output: 1 2 1

方法：可以通过使用二次方程的属性来解决给定的问题，如下所示：

$aX^2 + bX + c = 0$

For the above quadratic equation the roots are given by:

$X = \frac{-b + \sqrt{b^2 - 4ac}}{2*a}$ and $X = \frac{-b - \sqrt{b^2 - 4ac}}{2*a}$

The sum of roots is given by:

=> $S = \frac{-b + \sqrt{b^2 - 4ac}}{2*a} + \frac{-b - \sqrt{b^2 - 4ac}}{2*a}$
=> $S = \frac{-2b}{2a}$
=> $S = \frac{-b}{a}$

The product of roots is given by:

$P = \frac{-b + \sqrt{b^2 - 4ac}}{2*a} * \frac{-b - \sqrt{b^2 - 4ac}}{2*a}$
=> $P = \frac{{-b}^2 - {(\sqrt{b^2 - 4ac})}^2}{4{a}^2}$
=> $P = \frac{c}{a}$

根据以上两个方程，如果a的值为1，则b的值为(-1)* S ， c为P。因此，该等式由下式给出：

$X^2 + (-1)*S*X + P = 0$

下面是上述方法的实现：

C++

// C++ program for the above approach
  
#include 
using namespace std;
  
// Function to find the quadratic
// equation from the given sum and
// products of roots
void findEquation(int S, int M)
{
    // Print the coefficients
    cout << "1 " << (-1) * S << " "
         << M << endl;
}
  
// Driver Code
int main()
{
    int S = 5, M = 6;
    findEquation(S, M);
  
    return 0;
}

输出：

1 -5 6

时间复杂度： O(1)
辅助空间： O(1)