Codeforces Theatre Square Solution 1A in C/C++

Theatre Square is a problem that tests both your mathematical thinking and coding abilities. It presents a scenario where you need to calculate the number of square tiles needed to pave a rectangular theatre square. The challenge lies in efficiently determining this number given the dimensions of the theatre square and the size of the tiles.

Here is a detailed description

https://codeforces.com/problemset/problem/1/A

In this solution article, we will delve into the intricacies of solving the Theatre Square problem on Codeforces. We will walk you through the thought process, algorithm design, and implementation details to provide a clear and concise solution. Whether you're a beginner looking to understand basic algorithms or an experienced coder aiming to sharpen your skills, this article aims to offer valuable insights into tackling Theatre Square and similar problems on Codeforces.

Sequential Description Of Solution 

1. Include Libraries:
• The code includes the bits/stdc++.h> header for standard C++ libraries.
2. Namespace Declaration:
• The using namespace std; line allows for using standard C++ functions and objects without prefixing them with std::.
3. Input Dimensions:
• Read the integer values of n, m, and a from the standard input.
• n: Length of the theatre square.
• m: Width of the theatre square.
• a: Size of each square tile.
4. Check Tile Alignment Logic:
• Check if both n and m are divisible by a.
• If both are divisible:
• Calculate and output the number of tiles required using (n / a) * (m / a).
• If only n is not divisible:
• Calculate the number of rows needed as (n / a) + 1.
• Calculate and output the number of tiles required as ((n / a) + 1) * (m / a).
• If only m is not divisible:
• Calculate the number of columns needed as (m / a) + 1.
• Calculate and output the number of tiles required as (n / a) * ((m / a) + 1).
• If both n and m are not divisible:
• Calculate the number of rows and columns needed as ((n / a) + 1) * ((m / a) + 1).
• Calculate and output the number of tiles required as ((n / a) + 1) * ((m / a) + 1).
5. Output Result:
• Print the calculated number of tiles to the standard output, which represents the minimum number of square tiles needed to pave the theatre square without cutting or breaking any tile.

C++

#include<bits/stdc++.h>
using namespace std;

int main() {
	long long int n, m, a;
	cin >> n >> m >> a;
	if (n % a == 0 && m % a == 0)
	{
		cout << (n / a)*(m / a);
	}
	else if (n % a != 0 && m % a == 0)
	{
		cout << ((n / a) + 1)*(m / a);
	}
	else if (n % a == 0 && m % a != 0)
	{
		cout << (n / a)*((m / a) + 1);
	}
	else
	{
		cout << ((n / a) + 1)*((m / a) + 1);
	}
	return 0;
}
    

C

#include <stdio.h>

int main() {
    long long int n, m, a;
    scanf("%lld %lld %lld", &n, &m, &a);
    
    long long int rows, cols;
    
    if (n % a == 0 && m % a == 0) 
    {
        printf("%lld\n", (n / a) * (m / a));
    } 
    else if (n % a != 0 && m % a == 0) 
    {
        printf("%lld\n", ((n / a) + 1) * (m / a));
    } 
    else if (n % a == 0 && m % a != 0) 
    {
        printf("%lld\n", (n / a) * ((m / a) + 1));
    } 
    else
    {
        printf("%lld\n", ((n / a) + 1) * ((m / a) + 1));
    }    
    return 0;
}

The calculations and conditions are maintained as they were in the C++ version.

Note

We highly recommend you to try hard yourself at first & then if you fail, you can only check the logic from our code. Don't Copy-Paste our code exactly .If you do this, your logic will not be improved/developed. So, try to realize the logic following our description.