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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
#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
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