Count squares with odd side length in Chessboard in C++

Given a number size as input as dimension of size*size Chessboard. The goal is to find the number of squares that can be formed inside that board having odd lengths.

For Example

Input

size=3

Output

Count of squares with odd side length in Chessboard are: 10

Explanation

All squares will be as shown : and 1 whole square of size 3x3.

Input

size=4

Output

Count of squares with odd side length in Chessboard are: 20

Explanation

there will be 16, 1X1 squares. And 4, 3X3 squares inside it.

Approach used in the below program is as follows −

In this approach we will traverse from length of square as 1 to length as size. For each odd length we will add ( size−i−1)2 to the count.

• Take an integer size as input for Chessboard’s side.

• Function square_odd_length(int size) takes size and returns count of squares with odd side length in Chessboard.

• Take the initial count as 0.

• Traverse from i=1 to i=size increment by 2 for odd values of i.

• For each i take temp=size−i+1.

• Add temp*temp to count.

• At the end of the for loop return count as result.

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
int square_odd_length(int size){
   int count = 0;
   for (int i = 1; i <= size; i = i + 2){
      int temp = size − i + 1;
      count = count + (temp * temp);
   }
   return count;
}
int main(){
   int size = 6;
   cout<<"Count squares with odd side length in Chessboard are: "<<square_odd_length(size);
   return 0;
}

Output

If we run the above code it will generate the following output −

Count squares with odd side length in Chessboard are: 56