# PROBLEM LINK:

**Author:** Shreyansh Narayan

**Tester:** Nikhil Sharma

**Editorialist:** Shreyansh Narayan

# DIFFICULTY:

EASY

# PREREQUISITES:

Basic knowledge of motion

# PROBLEM:

At a given instant, you are at the p^{th} position. You will finish the race, once you go beyond the k^{th} position in t second(s). You have a certain velocity v (points/second). Find out if you will qualify for next round or not.

# QUICK EXPLANATION:

Calculate if you can finish the race in the stipulated time, i.e. if you can reach beyond the final position on the X-axis.

# EXPLANATION:

It is important to note that you are given your **present position** on the positive X-axis, **not the initial position** of the race.

So, based on your present position, the problem can be divided into three different cases.

**Case 1**

In the 1^{st} case, your present position is before the final position, i.e. p < k.

Hence, you still need to finish the race.

Again, depending on the velocity, we can have 2 cases.

- If v <= 0, then it is impossible to reach the finish line.
- If v > 0, then we need to check whether it is possible to reach in given time or not.

So, we can use the following formula - displacement = velocity x time, to calculate the time taken and match it with the given time.

**Case 2**

There are again 2 sub-cases.

- In the 1^{st} case, your present position is equal to the final position, i.e. p = p.
- In the 2^{nd} case, your present position is after the final position, i.e. p > k.

Hence, you have already finished the race.

In these situations, no matter what the velocity v and time t are, they donâ€™t matter as you have already finished the race at the given instant of time.

# TIME COMPLEXITY:

O(1)

# SOLUTION:

## Setter's & Editorialist's Solution

```
#include <bits/stdc++.h>
using namespace std;
int main() {
long long k,p;
cin >> k >> p;
int v,t;
cin >> v >> t;
if(p<k) {
if(v>0) {
if(k-p<=v*t)
cout << "Qualified\n";
else
cout << "Sorry\n";
}
else
cout << "Sorry\n";
}
else
cout << "Qualified\n";
return 0;
}
```