PIZZA_BURGER - Editorial

editorial
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PROBLEM LINK:

Practice
Contest: Division 3
Contest: Division 2
Contest: Division 1

Author: Jeevan Jyot Singh
Tester : Takuki Kurokawa
Editorialist: Aman Dwivedi

DIFFICULTY:

Cakewalk

PREREQUISITES:

None

PROBLEM:

Ashish is very hungry and wants to eat something. He has X rupees in his pocket. Since Ashish is very picky, he only likes to eat either PIZZA or BURGER. In addition, he prefers eating PIZZA over eating BURGER. The cost of a PIZZA is Y rupees while the cost of a BURGER is Z rupees.

Ashish can eat at most one thing. Find out what will Ashish eat for his dinner.

EXPLANATION:

  • If X \ge Y, then he will eat PIZZA.

  • if X \ge Z and X < Y, then he is gonna eat BURGER.

  • If X < Z, then he will eat Nothing.

TIME COMPLEXITY:

O(1) per test case

SOLUTIONS:

Author's Solution 
#include <bits/stdc++.h>
using namespace std;

#define IOS ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)

void solve()
{
    int x, y, z; cin >> x >> y >> z;
    if(x >= y)
        cout << "Pizza\n";
    else if(x >= z)
        cout << "Burger\n";
    else
        cout << "Nothing\n";
}

int32_t main()
{
    IOS;
    int T; cin >> T;
    for(int tc = 1; tc <= T; tc++)
    {
        solve();
    }
    return 0;
}
Tester's Solution 
#include <bits/stdc++.h>
using namespace std;

int main() {
    int tt;
    cin >> tt;
    while (tt--) {
        int x, y, z;
        cin >> x >> y >> z;
        if (x >= y) {
            cout << "PIZZA" << '\n';
        } else if (x >= z) {
            cout << "BURGER" << '\n';
        } else {
            cout << "NOTHING" << '\n';
        }
    }
    return 0;
}
Editorialist Solution 
#include<bits/stdc++.h>
using namespace std;

void solve()
{
    int x,y,z;
    cin>>x>>y>>z;

    if(x>=y)
      cout<<"PIZZA"<<"\n";
    else if(x>=z)
      cout<<"BURGER"<<"\n";
    else
      cout<<"NOTHING"<<"\n";
}

int main()
{
  ios_base::sync_with_stdio(0);
  cin.tie(0);

  int t;
  cin>>t;

  while(t--)
    solve();

return 0;
}