Find N integers with given difference between product and sum

Given two integers N and D, Find a set of N integers such that difference between their product and sum is equal to D.

Examples :

Input : N = 2, D = 1
Output : 2 3
Explanation: 
product = 2*3 = 6,
Sum = 2 + 3 = 5.
Hence, 6 - 5 = 1(D).

Input : N = 3, D = 5.
Output : 1 2 8
Explanation :
Product = 1*2*8 = 16
Sum = 1+2+8 = 11.
Hence, 16-11 = 5(D).



A tricky solution is to keep the difference D to choose N numbers as N-2 ‘1’s, one ‘2’ and one remaining number as ‘N+D’.
Sum = (N-2)*(1) + 2 + (N+D) = 2*N + D.
Product = 1*2*(N+D) = 2*N+2*D
Difference = (2*N+2*D) – (2*N+D) = D.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP code to generate numbers
// with difference between
// product and sum is D
#include <iostream>
using namespace std;
  
// Function to implement calculation
void findNumbers(int n, int d)
{
    for (int i = 0; i < n - 2; i++)
        cout << "1"  << " ";
  
    cout << "2" << " ";
    cout << n + d << endl;
}
  
// Driver code
int main()
{
    int N = 3, D = 5;
    findNumbers(N, D);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java code to generate numbers
// with difference between
// product and sum is D
import java.io.*;
  
class GFG {
      
    // Function to implement calculation
    static void findNumbers(int n, int d)
    {
        for (int i = 0; i < n - 2; i++)
            System.out.print("1" + " ");
      
        System.out.print("2" + " ");
        System.out.println(n + d);
    }
      
    // Driver code
    public static void main(String args[])
    {
        int N = 3, D = 5;
        findNumbers(N, D);
    }
}
  
/* This code is contributed by Nikita Tiwari.*/

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 code to generate numbers with
# difference between product and sum is D
  
# Function to implement calculation
def pattern(n, d) :
      
    for i in range(0, n - 2) :
        print("1", end=" ")
          
    print("2", end=" ")
    print(n + d)
  
# Driver code
N = 3
D = 5
pattern(N, D)
  
  
# This code is contributed by 'Akanshgupta'

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# code to generate numbers
// with difference between
// product and sum is D
using System;
  
class GFG {
      
    // Function to implement calculation
    static void findNumbers(int n, int d)
    {
        for (int i = 0; i < n - 2; i++)
        Console.Write("1" + " ");
      
        Console.Write("2" + " ");
        Console.Write(n + d);
    }
      
    // Driver code
    public static void Main()
    {
        int N = 3, D = 5;
        findNumbers(N, D);
    }
}
  
/* This code is contributed by vt_m.*/

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP code to generate numbers
// with difference between
// product and sum is D
  
// Function to implement
// calculation
function findNumbers($n, $d)
{
    for ($i = 0; $i < $n - 2; $i++)
        echo "1" ," ";
  
    echo "2" , " ";
    echo $n + $d ,"\n";
}
  
    // Driver Code
    $N = 3; 
    $D = 5;
    findNumbers($N, $D);
  
// This code is contributed by ajit
?>

chevron_right



Output :

 1 2 8


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : jit_t