生成给定根的和与积的二次方程

原文:https://www . geeksforgeeks . org/generate-二次方程-具有给定的根式和与积/

给定两个整数 S 和 M ，任务是求二次方程的系数，使得根的和与积分别为 S 和 M 。

示例:

输入: S = 5，M = 6 输出: 1 -5 6 说明: 对于二次方程 x2–5x+6 = 0。方程的根是 2 和 3。所以根的和是 2 + 3 = 5，根的积是 2*3 = 6。

输入: S = -2，M = 1 T3】输出: 1 2 1

方法:给定的问题可以通过使用如下所示的二次方程的性质来解决:

aX^2 + bX + c = 0

对于上述二次方程，根由下式给出:

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

根的总和由下式给出:

= >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}

根的乘积由下式给出:

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}

从以上两个等式可知，如果 a 的值为 1 ，那么 b 的值为 (-1)*S ，而 c 为 P 。因此，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 <bits/stdc++.h>
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;
}

Java 语言(一种计算机语言，尤用于创建网站)

// Java program for the above approach
import java.io.*;

class GFG{

// Function to find the quadratic
// equation from the given sum and
// products of roots
public static void findEquation(int S, int M)
{

    // Print the coefficients
    System.out.println("1 " + ((-1) * S) + " " + M);   
}

// Driver code
public static void main(String[] args)
{
      int S = 5, M = 6;

    findEquation(S, M);
}
}

// This code is contributed by user_qa7r

Python 3

# Python3 program for the above approach

# Function to find the quadratic
# equation from the given sum and
# products of roots
def findEquation(S, M):

    # Print the coefficients
    print("1 ", ((-1) * S), " " , M)

# Driver Code
S = 5
M = 6

findEquation(S, M)

# This code is contributed by Ankita saini

C

// C# program for the above approach
using System;

class GFG{

// Function to find the quadratic
// equation from the given sum and
// products of roots
public static void findEquation(int S, int M)
{

    // Print the coefficients
    Console.Write("1 " + ((-1) * S) + " " + M);  
}

// Driver code
static void Main()
{
    int S = 5, M = 6;

    findEquation(S, M);
}
}

// This code is contributed by code_hunt

java 描述语言

<script>

// Javascript program for the above approach

// Function to find the quadratic
// equation from the given sum and
// products of roots
function findEquation(S, M)
{

    // Print the coefficients
    document.write("1 " + ((-1) * S) + " " + M);   
}

// Driver Code
var S = 5, M = 6;

findEquation(S, M);

// This code is contributed by Ankita saini

</script>

Output: 

1 -5 6

时间复杂度:O(1) T5辅助空间:** O(1)

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