### Problem Link:

**Author:** Misha Chorniy

**Tester:** Pushkar Mishra

**Editorialist:** Praveen Dhinwa

### Difficulty:

Cakewalk

### Pre-Requisites:

Implementation, Basic maths.

### Problem Statement

Chef has *money* rupees with him. He spends *jacketCost* rupees for buying a jacket. He buys as many socks as he can by the money left after purchasing the jacket. Cost of a sock is given by *sockCost*. Each day, he uses two socks. He never cleans his socks after using them. You have to find whether there will be a day in which Chef will have one sock remaining.

### Quick Explanation

We just have to check whether \frac{(money - jacketCost)}{sockCost} is even or odd.

### Explanation

Initial money that Chef have = money

We are guaranteed in the problem that jacketCost \leq money.

So the money left after buying a jacket (i.e. money - jacketCost) will be always non-negative.

Using the remaining money, Chef buys as many socks as we can, so he will buy S socks where S = \frac{(money - jacketCost)}{sockCost} where / denotes an integer division. By integer division, we mean floor function (e.g. \lfloor \frac{3}{2} \rfloor = 1 and \lfloor \frac{4}{2} \rfloor = 2).

Now, we need to check if Chef wear two socks daily, will there be a day when he will have to wear a single sock? This is equivalent to checking S is even or not.

For solving subtask 1, we can simply simulate the process of wearing two socks as follows:

```
int remS = S // remS denotes initial number of socks.
while (remS >= 2) {
// wear two socks
remS -= 2;
}
if (remS == 1) return "Unlucky Chef"
else return "Lucky Chef";
```

### Time Complexity

O(S) or O(1) depending on the subtask.

### Solution:

Setter’s solution can be found here

Tester’s solution can be found here

**Please feel free to post comments if anything is not clear to you.**