# PROBLEM LINK:

Practice

Contest: Division 1

Contest: Division 2

Contest: Division 3

Contest: Division 4

* Author:* notsoloud

*raysh07*

**Tester:***iceknight1093*

**Editorialist:**# DIFFICULTY:

TBD

# PREREQUISITES:

None

# PROBLEM:

Chef has 20 neighbors, each of whom want Y plates of food.

Chef cooked X plates - how many of them can he feed completely?

# EXPLANATION:

Suppose Chef wants to feed k neighbors.

Each neighbor wants Y plates of food, so k neighbors will require k\cdot Y plates in total.

This means we must have k\cdot Y \leq X, since there are only X plates of food in total.

Moving Y to the other side, this becomes k \leq \left\lfloor \frac{X}{Y} \right\rfloor as an upper bound on k.

Here, \left\lfloor \ \right\rfloor denotes the floor function.

Further, we must also have k \leq 20, since there are only 20 neighbors.

Putting both constraints together, we see that the answer is just

# TIME COMPLEXITY

\mathcal{O}(1) per testcase.

# CODE:

## Editorialist's code (C++)

```
#include <iostream>
using namespace std;
int main() {
int t; cin >> t;
while (t--) {
int x, y; cin >> x >> y;
cout << min(20, x/y) << '\n';
}
return 0;
}
```

## Editorialist's code (Python)

```
for _ in range(int(input())):
x, y = map(int, input().split())
print(min(20, x//y))
```