 # PIZZA_BURGER - Editorial

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

Cakewalk

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;
}

``````