Print squares of first n natural numbers without using *, / and - GeeksforGeeks

Given a natural number ‘n’, print squares of first n natural numbers without using *, / and -.

Input:  n = 5
Output: 0 1 4 9 16

Method 1: The idea is to calculate next square using previous square value. Consider the following relation between square of x and (x-1). We know square of (x-1) is (x-1)² – 2*x + 1. We can write x² as

x2 = (x-1)2 + 2*x - 1 
x2 = (x-1)2 + x + (x - 1)

When writing an iterative program, we can keep track of previous value of x and add the current and previous values of x to current value of square. This way we don’t even use the ‘-’ operator.

void printSquares(int n)

{

    // Initialize 'square' and previous value of 'x'

    int square = 0, prev_x = 0;

    // Calculate and print squares

    for (int x = 0; x < n; x++)

    {

        // Update value of square using previous value

        square = (square + x + prev_x);

        // Print square and update prev for next iteration

        cout << square << " ";

        prev_x = x;

    }

}

Method 2: Sum of first n odd numbers are squares of natural numbers from 1 to n. For example 1, 1+3, 1+3+5, 1+3+5+7, 1+3+5+7+9, ….

void printSquares(int n)

{

    // Initialize 'square' and first odd number

    int square = 0, odd = 1;

 

    // Calculate and print squares

    for (int x = 0; x < n; x++)

    {

        // Print square

        cout << square << " ";

 

        // Update 'square' and 'odd'

        square = square + odd;

        odd = odd + 2;

    }

}

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