Print n numbers such that their sum is a perfect square

Given an integer n, the task is to print n numbers such that their sum is a perfect square.

Examples:

Input: n = 3
Output: 1 3 5
1 + 3 + 5 = 9 = 32

Input: n = 4
Output: 1 3 5 7
1 + 3 + 5 + 7 = 16 = 42

Approach: The sum of first n odd numbers is always a perfect square. So, we will print the first n odd numbers as the output.



Below is the implementation of the above approach:

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to print n numbers such that
// their sum is a perfect square
void findNumbers(int n)
{
    int i = 1;
    while (i <= n) {
  
        // Print ith odd number
        cout << ((2 * i) - 1) << " ";
        i++;
    }
}
  
// Driver code
int main()
{
    int n = 3;
    findNumbers(n);
}
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// Java implementation of the approach
class GFG {
  
    // Function to print n numbers such that
    // their sum is a perfect square
    static void findNumbers(int n)
    {
        int i = 1;
        while (i <= n) {
  
            // Print ith odd number
            System.out.print(((2 * i) - 1) + " ");
            i++;
        }
    }
  
    // Driver code
    public static void main(String args[])
    {
        int n = 3;
        findNumbers(n);
    }
}
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# Python3 implementation of the approach 
  
# Function to print n numbers such that 
# their sum is a perfect square 
def findNumber(n):
    i = 1
    while i <= n:
  
        # Print ith odd number 
        print((2 * i) - 1, end = " ")
        i += 1
  
# Driver code     
n = 3
findNumber(n)
  
# This code is contributed by Shrikant13
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// C# implementation of the approach
using System;
public class GFG {
  
    // Function to print n numbers such that
    // their sum is a perfect square
    public static void findNumbers(int n)
    {
        int i = 1;
        while (i <= n) {
  
            // Print ith odd number
            Console.Write(((2 * i) - 1) + " ");
            i++;
        }
    }
  
    // Driver code
    public static void Main(string[] args)
    {
        int n = 3;
        findNumbers(n);
    }
}
  
// This code is contributed by Shrikant13
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<?php
// PHP implementation of the approach
  
// Function to prn numbers such that
// their sum is a perfect square
function findNumbers($n)
{
    $i = 1;
    while ($i <= $n
    {
  
        // Print ith odd number
        echo ((2 * $i) - 1) . " ";
        $i++;
    }
}
  
// Driver code
$n = 3;
findNumbers($n);
  
// This code contributed by PrinciRaj1992
?>
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Output:
1 3 5

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