HackerRank Geometry – Rectangular Game

Challenge: Rectangular Game

Scor cont: 20.0 / 20

Submission status: Accepted

Submission score: 1.0

Submission ID: 464722129

Limbaj: cpp14

Link challenge: https://www.hackerrank.com/challenges/rectangular-game/problem

Cerinta

You are given an infinite 2-d grid with the bottom left cell referenced as (1,1). All the cells contain a value of zero initially. Let's play a game?  

The game consists of **N** steps wherein each step you are given two integers **a** and **b**. The value of each of the cells in the co-ordinate (u, v) satisfying 1 ≤ u ≤ a and 1 ≤ v ≤ b, is increased by 1. After **N** such steps, if **X** is the largest number amongst all the cells in the rectangular board, can you print the number of **X**'s in the board?

**Input Format**  
The first line of input contains a single integer N. N lines follow.  
Each line contains two integers a and b separated by a single space.  

**Output Format**  
Output a single integer - the number of X's. 

**Constraints**    
1 ≤ N ≤ 100  
1 ≤ a ≤ 10<sup>6</sup>  
1 ≤ b ≤ 10<sup>6</sup>

**Sample Input**  

    3
    2 3
    3 7
    4 1
    
**Sample Output**  

    2
    
**Explanation**

Assume that the following board corresponds to cells (i, j) where 1 ≤ i ≤ 4 and 1 ≤ j ≤ 7.

At the beginning board is in the following state:

    0 0 0 0 0 0 0
    0 0 0 0 0 0 0
    0 0 0 0 0 0 0
    0 0 0 0 0 0 0 

After the first step we will obtain:

    0 0 0 0 0 0 0
    0 0 0 0 0 0 0
    1 1 1 0 0 0 0
    1 1 1 0 0 0 0 

After the second step we will obtain:

    0 0 0 0 0 0 0
    1 1 1 1 1 1 1
    2 2 2 1 1 1 1
    2 2 2 1 1 1 1 
    
Finally, after the last step we will obtain:

    1 0 0 0 0 0 0
    2 1 1 1 1 1 1
    3 2 2 1 1 1 1
    3 2 2 1 1 1 1 

So, the maximum number is 3 and there are exactly two cells which correspond to 3. Hence 2.

Cod sursa

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

typedef long long int ll;

int main(){
    cin.sync_with_stdio(false);
    cin.tie(0);
    ll t;
    cin >> t;
    ll mx=0, yx=0;
    while(t--){
        ll x, y;
        cin >> x >> y;
        if(mx == 0 and yx == 0)
            mx = x, yx = y;
        else if(yx > y)
            yx = y;
        else if(mx > x)
            mx = x;
    }
    cout << mx*yx;
    return 0;
}