Soluție HackerRank pentru Rectangular Game, subdomeniul Geometry, în C++14. Include cerința formatată, exemple, explicația pașilor și cod sursă.
- Problemă: Rectangular Game
- Domeniu: Geometry
- Limbaj: C++14
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
Cerință
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 sursă
#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;
}