Given a sum S and product P, the task is to print any pair of integers that has the sum S and the product P. If there exists no such pair, then print -1.
Examples:
Input: S = 5, P = 6
Output: 2, 3
Explanation:
Sum = 2 + 3 = 5, and
Product = 2 * 3 = 6
Input: S = 5, P = 9
Output: -1
Explanation:
No such pair exists
Approach: Let the pair be (x, y). Therefore, according to the problem, the given sum (S) will be (x + y) and the given product (P) will be (x * y)
If the pair is (x, y)
Given that product
P = x * y
y = P / x; (eq.. 1)
Given that sum
S = x + y
S = x + (P / x) from (eq..1)
x2 - Sx + P = 0
which is a quadratic equation in x.
Since this is a quadratic equation, we just need to find it’s roots, using the below equation. 
Here: a = 1 b = -S c = P
Therefore the above equation will be changed as: 
Below is the implementation of the above approach:
CPP
// CPP program to find any pair// which has sum S and product P.#include <bits/stdc++.h>using namespace std;// Prints roots of quadratic equation// ax*2 + bx + c = 0void findRoots(int b, int c){ int a = 1; int d = b * b - 4 * a * c; // calculating the sq root value // for b * b - 4 * a * c double sqrt_val = sqrt(abs(d)); if (d > 0) { double x = -b + sqrt_val; double y = -b - sqrt_val; // Finding the roots int root1 = (x) / (2 * a); int root2 = (y) / (2 * a); // Check if the roots // are valid or not if (root1 + root2 == -1 * b && root1 * root2 == c) cout << root1 << ", " << root2; else cout << -1; } else if (d == 0) { // Finding the roots int root = -b / (2 * a); // Check if the roots // are valid or not if (root + root == -1 * b && root * root == c) cout << root << ", " << root; else cout << -1; } // when d < 0 else { // No such pair exists in this case cout << -1; } cout << endl;}// Driver codeint main(){ int S = 5, P = 6; findRoots(-S, P); S = 5, P = 9; findRoots(-S, P); return 0;} |
Java
// Java program to find any pair// which has sum S and product P.import java.util.*; class GFG {// Prints roots of quadratic equation// ax*2 + bx + c = 0static void findRoots(int b, int c){ int a = 1; int d = b * b - 4 * a * c; // calculating the sq root value // for b * b - 4 * a * c double sqrt_val = Math.sqrt(Math.abs(d)); if (d > 0) { double x = -b + sqrt_val; double y = -b - sqrt_val; // Finding the roots int root1 = (int)(x) / (2 * a); int root2 = (int) (y) / (2 * a); // Check if the roots // are valid or not if (root1 + root2 == -1 * b && root1 * root2 == c) System.out.print( root1 + ", " + root2); else System.out.print( -1); } else if (d == 0) { // Finding the roots int root = -b / (2 * a); // Check if the roots // are valid or not if (root + root == -1 * b && root * root == c) System.out.print(root+ ", "+root); else System.out.print(-1); } // when d < 0 else { // No such pair exists in this case System.out.print( -1); } System.out.println();} // Driver codepublic static void main (String []args){ int S = 5, P = 6; findRoots(-S, P); S = 5; P = 9; findRoots(-S, P);} }// This code is contributed by chitranayal |
Python3
# Python3 program to find any pair# which has sum S and product P.from math import sqrt# Prints roots of quadratic equation# ax*2 + bx + c = 0def findRoots(b, c): a = 1 d = b * b - 4 * a * c # calculating the sq root value # for b * b - 4 * a * c sqrt_val = sqrt(abs(d)) if (d > 0): x = -b + sqrt_val y = -b - sqrt_val # Finding the roots root1 = (x) // (2 * a) root2 = (y) // (2 * a) # Check if the roots # are valid or not if (root1 + root2 == -1 * b and root1 * root2 == c): print(int(root1),",",int(root2)) else: print(-1) elif (d == 0): # Finding the roots root = -b // (2 * a) # Check if the roots # are valid or not if (root + root == -1 * b and root * root == c): print(root,",",root) else: print(-1) # when d < 0 else: # No such pair exists in this case print(-1)# Driver codeif __name__ == '__main__': S = 5 P = 6 findRoots(-S, P) S = 5 P = 9 findRoots(-S, P)# This code is contributed by mohit kumar 29 |
C#
// C# program to find any pair// which has sum S and product P.using System;public class GFG { // Prints roots of quadratic equation // ax*2 + bx + c = 0 static void findRoots(int b, int c) { int a = 1; int d = b * b - 4 * a * c; // calculating the sq root value // for b * b - 4 * a * c double sqrt_val = Math.Sqrt(Math.Abs(d)); if (d > 0) { double x = -b + sqrt_val; double y = -b - sqrt_val; // Finding the roots int root1 = (int)(x) / (2 * a); int root2 = (int) (y) / (2 * a); // Check if the roots // are valid or not if (root1 + root2 == -1 * b && root1 * root2 == c) Console.Write( root1 + ", " + root2); else Console.Write( -1); } else if (d == 0) { // Finding the roots int root = -b / (2 * a); // Check if the roots // are valid or not if (root + root == -1 * b && root * root == c) Console.Write(root+ ", "+root); else Console.Write(-1); } // when d < 0 else { // No such pair exists in this case Console.Write( -1); } Console.WriteLine(); } // Driver code public static void Main (string []args) { int S = 5, P = 6; findRoots(-S, P); S = 5; P = 9; findRoots(-S, P); } }// This code is contributed by Yash_R |
Javascript
<script>// Javascript program to find any pair// which has sum S and product P.// Prints roots of quadratic equation// ax*2 + bx + c = 0function findRoots(b, c){ var a = 1; var d = b * b - 4 * a * c; // calculating the sq root value // for b * b - 4 * a * c var sqrt_val = Math.sqrt(Math.abs(d)); if (d > 0) { var x = -b + sqrt_val; var y = -b - sqrt_val; // Finding the roots var root1 = (x) / (2 * a); var root2 = (y) / (2 * a); // Check if the roots // are valid or not if (root1 + root2 == -1 * b && root1 * root2 == c) document.write(root1 + ", " + root2); else document.write(-1); } else if (d == 0) { // Finding the roots var root = -b / (2 * a); // Check if the roots // are valid or not if (root + root == -1 * b && root * root == c) document.write(root + ", " + root); else document.write(-1); } // when d < 0 else { // No such pair exists in this case document.write(-1); } document.write("<br>");}// Driver codevar S = 5, P = 6;findRoots(-S, P);S = 5, P = 9;findRoots(-S, P);// This code is contributed by rutvik_56.</script> |
Output:
3, 2 -1
Time Complexity: O(1)
Auxiliary Space: O(1)
As constant extra space is used.
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