<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

453. Meetings
Time limit per test: 0.25 second(s)
Memory limit: 65536 kilobytes
input: standard
output: standard



Two cities A and B are connected by a straight road that is exactly l meters long. At the initial moment of time a cyclist starts moving from city A to city B at a speed v1 meters/second, and a pedestrian starts moving from city B to city A at a speed v2 meters/second. When one of them reaches a city, the road ends, so the person has to turn around and start moving in the opposite direction by the same road, keeping the original speed. As a result, the cyclist and the pedestrian are traveling between cities A and B indefinitely.

Your task is to calculate the number of times they will meet during the first t seconds. If they meet in exactly t seconds after the initial moment of time, this meeting should also be counted.

Input
The only line of input contains four integer numbers: l, v1, v2 and t. All numbers are between 1 and 109, inclusively.

Output
Print a single integer  — the number of times the cyclist and the pedestrian will meet during the first t seconds.

Example(s)
sample input
sample output
1000 10 1 200
2

sample input
sample output
4 4 3 4
4

<|response|>
## 1) Abridged problem statement (concise)

Two travelers move back and forth on a segment of length `l` between cities **A** and **B**:

- Cyclist starts at **A** toward **B** with speed `v1`.
- Pedestrian starts at **B** toward **A** with speed `v2`.
- On reaching an endpoint, they instantly turn around and continue at the same speed.

Count how many times they meet during the time interval **(0, t]** (a meeting exactly at time `t` counts).  
All inputs are integers up to \(10^9\).

---

## 2) Key observations

1. **Reflections can be “unfolded”**: instead of bouncing on `[0, l]`, imagine reflecting the segment repeatedly to an infinite line. Then each person moves **uniformly** without turning, and the real position is obtained by folding back (triangle wave with period `2l`).

2. Meeting conditions reduce to **two congruences** (two “families” of meetings), both involving **odd multiples** of `l`:
   - **Approaching each other** (relative speed `v1 + v2`):
     \[
     (v_1+v_2)\,t \equiv l \pmod{2l}
     \Rightarrow t=\frac{l(2k+1)}{v_1+v_2}
     \]
   - **One catches up to the other** (relative speed `|v1 - v2|`, only if `v1 != v2`):
     \[
     |v_1-v_2|\,t \equiv l \pmod{2l}
     \Rightarrow t=\frac{l(2k+1)}{|v_1-v_2|}
     \]

3. Counting meetings becomes counting how many **odd multiples** fit into a bound:
   If we need \(l(2k+1) \le X\), then the count is:
   \[
   \left\lfloor \frac{X + l}{2l}\right\rfloor
   \]

4. The two families can overlap (same meeting time appears in both), so use **inclusion–exclusion**:
   \[
   \text{answer} = \text{count}_1 + \text{count}_2 - \text{overlap}
   \]

5. Overlap exists only under a parity condition. Let:
   - \(s=v_1+v_2\), \(d=|v_1-v_2|\), \(g=\gcd(s,d)\)
   - If both \((s/g)\) and \((d/g)\) are **odd**, then overlap times are:
     \[
     t = \frac{l}{g}\cdot(1,3,5,\dots)
     \]
     so overlap count is:
     \[
     \left\lfloor \frac{Tg + l}{2l}\right\rfloor
     \]
   - Otherwise overlap is 0.

All computations fit in signed 64-bit: products are up to about \(2\cdot 10^{18}\).

---

## 3) Full solution approach

Let `T = t` from input.

### Step A: Compute the two base counts

Define:
- `s = v1 + v2`
- `d = abs(v1 - v2)`

**Type 1 (approaching) meetings** occur at \(t=\frac{l(2k+1)}{s}\).  
Count those with \(t \le T\):
\[
l(2k+1) \le Ts
\Rightarrow \text{count}_1 = \left\lfloor \frac{Ts + l}{2l} \right\rfloor
\]

**Type 2 (catch-up) meetings** occur at \(t=\frac{l(2k+1)}{d}\) if `d > 0`.  
Count:
\[
\text{count}_2 =
\begin{cases}
\left\lfloor \frac{Td + l}{2l} \right\rfloor & d>0 \\
0 & d=0
\end{cases}
\]

### Step B: Compute overlap to avoid double counting

If `d == 0`, overlap is 0.

Otherwise:
- `g = gcd(s, d)`
- If `(s/g)` and `(d/g)` are both odd, overlap times are odd multiples of `l/g`, so:
  \[
  \text{overlap}=\left\lfloor \frac{Tg + l}{2l}\right\rfloor
  \]
- Else overlap = 0.

### Step C: Final answer
\[
\boxed{\text{answer} = \text{count}_1 + \text{count}_2 - \text{overlap}}
\]

This is \(O(1)\) time and memory.

---

## 4) C++ implementation (detailed comments)

```cpp
#include <bits/stdc++.h>
using namespace std;

// We use int64_t to safely hold values up to ~2e18.
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int64_t l, v1, v2, T;
    cin >> l >> v1 >> v2 >> T;

    // s: relative speed when they move towards each other
    int64_t s = v1 + v2;

    // d: relative speed when they move in the same direction (catch-up)
    int64_t d = llabs(v1 - v2);

    // Count meetings of type 1:
    // times are t = l*(odd)/s
    // count odd multiples of l <= T*s  => floor((T*s + l) / (2*l))
    int64_t count1 = (T * s + l) / (2 * l);

    // Count meetings of type 2 (only if v1 != v2):
    int64_t count2 = 0;
    if (d > 0) {
        count2 = (T * d + l) / (2 * l);
    }

    // Compute overlap between the two families to avoid double counting.
    int64_t overlap = 0;
    if (d > 0) {
        int64_t g = std::gcd(s, d);

        // Overlap exists only if s/g and d/g are both odd.
        // (Because the meeting formulas use odd multipliers (2k+1).)
        if ( ((s / g) % 2 == 1) && ((d / g) % 2 == 1) ) {
            // Overlap times become t = (l/g) * odd
            // Count odd multiples of (l/g) <= T:
            // Equivalent to count odd multiples of l <= T*g*l? -> simplifies to:
            overlap = (T * g + l) / (2 * l);
        }
    }

    // Inclusion-exclusion:
    cout << (count1 + count2 - overlap) << "\n";
    return 0;
}
```

---

## 5) Python implementation (detailed comments)

```python
import sys
import math

def solve() -> None:
    l, v1, v2, T = map(int, sys.stdin.readline().split())

    # Relative speeds for the two meeting types
    s = v1 + v2
    d = abs(v1 - v2)

    # Type 1: (v1+v2)*t = l (mod 2l)  -> t = l*(odd)/s
    # Count odd multiples of l that are <= T*s:
    count1 = (T * s + l) // (2 * l)

    # Type 2: |v1-v2|*t = l (mod 2l) -> t = l*(odd)/d, only if d>0
    count2 = (T * d + l) // (2 * l) if d > 0 else 0

    # Overlap of both sequences
    overlap = 0
    if d > 0:
        g = math.gcd(s, d)
        # Overlap exists only if reduced values are both odd
        if (s // g) % 2 == 1 and (d // g) % 2 == 1:
            # Overlap times: t = (l/g) * odd
            overlap = (T * g + l) // (2 * l)

    ans = count1 + count2 - overlap
    print(ans)

if __name__ == "__main__":
    solve()
```

