Minimum Cuts can be made in the Chessboard such that it is not divided into 2 parts in C++

Concept

Given A x B Chessboard, the task is to calculate the Maximum numbers of cuts that we can build in the Chessboard so that the Chessboard is not divided into 2 parts.

Examples 

The examples are given below −

Input 

A = 2, B = 4

Output 

Number of maximum cuts = 3

Input 

A = 2, B = 2

Output 

Number of maximum cuts = 1

Method

• For A = 2, B = 2 ,we can only build 1 cut (mark in red). If we build 1 more cut then the chessboard will divide into 2 pieces

• For A = 2, B = 4 ,we can makes 3 cuts (marks in red). If we build 1 more cut then the chessboard will divide into 2 pieces.

As a result of this, it can be observed that no. of cuts = (A-1) * (B-1).

Example 

 Live Demo

//This is C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
// function that calculates the maximum no. of cuts
int numberOfCuts1(int A, int B){
   int result1 = 0;
   result1 = (A - 1) * (B - 1);
   return result1;
}
// Driver Code
int main(){
   int A = 4, B = 4;
   // Calling function.
   int Cuts = numberOfCuts1(A, B);
   cout << "Maximum cuts = " << Cuts;
   return 0;
}

Output

Maximum cuts = 9