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++

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// 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;
}

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Java














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// 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.*/



















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Python3














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# 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' 



















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C#














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// 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.*/



















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PHP














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<?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 


?> 



















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Output :


 1 2 8





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