# PROBLEM LINK:

Contest Division 1

Contest Division 2

Contest Division 3

Contest Division 4

Setter: Lavish Gupta

Tester: Istvan Nagy, Aryan

Editorialist: Vijay

# DIFFICULTY:

Cakewalk

# PREREQUISITES:

None

# PROBLEM:

The summer is at its peak in Chefland. Chef is planning to purchase a water cooler to keep his room cool. He has two options available:

- Rent a cooler at the cost of X coins per month.
- Purchase a cooler for Y coins.

Chef wonders what is the **maximum** number of months for which he can rent the cooler such that the cost of renting is **strictly less** than the cost of purchasing it.

# EXPLANATION:

## What will be the cost of renting a water cooler for K months ?

The cost of renting a water cooler for one month equals X. Then the cost of renting a water cooler for K months will be equal to X+X+...... K times =K \cdot X.

## Up to when will Chef prefer renting a water cooler instead of buying it?

Chef will rent when the cost of renting is strictly less than the cost of buying a water cooler.

The cost of renting the cooler for K months equals K \cdot X. And, also the cost of buying a new water cooler equals Y. So, Chef will continue to rent up until K months such that K\cdot X<Y=>K= { \frac{Y-1}{X}} which is *rounded down.*

# TIME COMPLEXITY:

O(1) for each test case.

# SOLUTION:

## Editorialist's Solution

```
#include <bits/stdc++.h>
using namespace std;
#define nline '\n'
int main()
{
int t;
cin >> t;
while (t--)
{
int x, y;
cin>>x>>y;
int m=(y-1)/x;
cout<<m<<nline;
}
}
```