1. Abridged Problem Statement  
Given the scheduled start time S and the actual arrival time P (both in seconds), compute how many cups of tea Andrew owes based on how late he is:  
- 0 seconds late → 0 cups  
- 1 to 299 seconds late → 1 cup  
- 300 to 899 seconds late → 2 cups  
- 900 to 1799 seconds late → 3 cups  
- ≥1800 seconds late → 4 cups  

2. Detailed Editorial  
- **Read input**: two integers S (start time) and P (arrival time).  
- **Compute delay**: `delay = P − S`.  
- **Non-late case**: if `delay ≤ 0`, Andrew is on time or early → owes 0 cups.  
- **Convert to minutes (optional)**: dividing seconds by 60 can make threshold checks more intuitive, but you can compare in seconds directly.  
- **Threshold checks**:  
  - If `1 ≤ delay < 5*60`, it’s less than 5 minutes → 1 cup.  
  - If `5*60 ≤ delay < 15*60`, it’s between 5 and 15 minutes → 2 cups.  
  - If `15*60 ≤ delay < 30*60`, it’s between 15 and 30 minutes → 3 cups.  
  - If `delay ≥ 30*60`, it’s 30 minutes or more → 4 cups.  
- **Output** the computed number of cups.  

Time complexity is O(1), memory O(1). Edge cases include P == S (exactly on time) and P < S (early arrival).  

3. C++ Solution with Detailed Comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

// Global variables to store start time (s) and arrival time (t)
double s, t;

// Read input and pre-process
void read() {
    cin >> s >> t;       // read start time S and arrival time P (both in seconds)
    t -= s;              // compute the delay in seconds
    t /= 60.0;           // convert delay from seconds to minutes
}

// Determine how many cups Andrew owes
void solve() {
    int ans = 4;         // default: 4 cups if delay ≥ 30 minutes
    if (t < 30) {        // delay less than 30 minutes
        ans = 3;         // 15 ≤ delay < 30 → 3 cups
    }
    if (t < 15) {        // delay less than 15 minutes
        ans = 2;         // 5 ≤ delay < 15 → 2 cups
    }
    if (t < 5) {         // delay less than 5 minutes
        ans = 1;         // 1 ≤ delay < 5 → 1 cup
    }
    if (t <= 0) {        // on time or early (delay ≤ 0)
        ans = 0;         // owes no cups
    }
    cout << ans << '\n'; // output the answer
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);

    read();              // input and convert
    solve();             // compute and print result
    return 0;
}
```

4. Python Solution with Detailed Comments  
```python
# Read two integers: S = scheduled start time, P = actual arrival time
S, P = map(int, input().split())

# Compute delay in seconds
delay = P - S

# If arrival is on time or early, no cups owed
if delay <= 0:
    print(0)
elif delay < 5 * 60:
    # less than 5 minutes late → 1 cup
    print(1)
elif delay < 15 * 60:
    # between 5 and 15 minutes late → 2 cups
    print(2)
elif delay < 30 * 60:
    # between 15 and 30 minutes late → 3 cups
    print(3)
else:
    # 30 minutes or more late → 4 cups
    print(4)
```

5. Compressed Editorial  
Compute `delay = P − S`. If `delay ≤ 0` → 0 cups. Else compare `delay` against 5 min (300 s), 15 min (900 s), 30 min (1800 s) thresholds in increasing order to assign 1, 2, 3, or 4 cups.