<|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

A straight segment of length **l** connects cities **A** and **B**.
At time 0, a cyclist starts at **A** toward **B** with speed **v1**, and a pedestrian starts at **B** toward **A** with speed **v2**. Whenever someone reaches an endpoint, they immediately turn around and continue with the same speed.

Given integers **l, v1, v2, t** (up to \(10^9\)), compute how many times they **meet** during the time interval \((0, t]\).

---

## 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** producing two families of meeting times:
   - **Approaching each other** (relative speed `v1 + v2`): \((v_1+v_2)\,t \equiv l \pmod{2l}\), giving \(t=\frac{l(2k+1)}{v_1+v_2}\).
   - **One catches up to the other** (relative speed `|v1 - v2|`, only if `v1 != v2`): \(|v_1-v_2|\,t \equiv l \pmod{2l}\), giving \(t=\frac{l(2k+1)}{|v_1-v_2|}\).

3. Counting odd multiples fitting into a bound \(X\): if we need \(l(2k+1) \le X\), the count is \(\left\lfloor \frac{X + l}{2l}\right\rfloor\).

4. The two families can overlap, so use **inclusion-exclusion**: \(\text{answer} = \text{count}_1 + \text{count}_2 - \text{overlap}\).

5. With \(s=v_1+v_2\), \(d=|v_1-v_2|\), \(g=\gcd(s,d)\): overlap exists only if both \((s/g)\) and \((d/g)\) are **odd**. If so, overlap times are \(t = \frac{l}{g}\cdot(1,3,5,\dots)\) and overlap count is \(\left\lfloor\frac{Tg+l}{2l}\right\rfloor\).

---

## 3) Full solution approach

The unfolding trick converts bouncing motion into uniform motion on a line. Two people meet when their triangular-wave positions agree. After case analysis of who is in which half-period:

- **Type 1** meetings (approaching): times \(t = l(2k+1)/(v_1+v_2)\), count = \(\lfloor(T(v_1+v_2)+l)/(2l)\rfloor\).
- **Type 2** meetings (catch-up, only if \(v_1 \ne v_2\)): times \(t = l(2k+1)/|v_1-v_2|\), count = \(\lfloor(T|v_1-v_2|+l)/(2l)\rfloor\).
- **Overlap**: reduce by \(g = \gcd(s,d)\). If both \(s/g\) and \(d/g\) are odd, overlap count = \(\lfloor(Tg+l)/(2l)\rfloor\). Otherwise 0.

Answer: \(\text{count}_1 + \text{count}_2 - \text{overlap}\).

All products fit in 64-bit integers (up to \(2 \times 10^{18}\)).

---

## 4) C++ implementation

```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int64_t l, v1, v2, t;

void read() { cin >> l >> v1 >> v2 >> t; }

void solve() {
    // One way to make this problem easier to argue about is to consider the
    // position of each person as moving along an "unfolded" line, where instead
    // of turning around at walls, we imagine the road being mirrored
    // infinitely. Each person's position can be described by a phase:
    //
    //     phase_c = v1*t for the cyclist,
    //     phase_p = v2*t for the pedestrian.
    //
    // The real position is recovered by "folding" the phase back into [0, l]:
    //
    //   - Cyclist starts at 0 moving right:
    //         real_pos = phase_c mod 2l if <= l,
    //     else:
    //         2l - (phase_c mod 2l).
    //
    //   - Pedestrian starts at l moving left:
    //         real_pos = l - (phase_p mod 2l) if phase_p mod 2l <= l
    //     else
    //         (phase_p mod 2l) - l.
    //
    // They meet when their real positions coincide. Analyzing all cases of
    // which "segment" each is in (moving left or right), we get two types
    // of meetings:
    //
    // 1. Approaching each other: (v1 + v2) * t = l (mod 2l)
    //    Meeting times: t = l*(2k+1)/(v1+v2) for k = 0, 1, 2, ...
    //
    // 2. One catching up to the other: |v1 - v2| * t = l (mod 2l)
    //    Meeting times: t = l*(2k+1)/|v1-v2| for k = 0, 1, 2, ...
    //    (Only exists if v1 != v2)
    //
    // Count of type 1 in (0, T]:
    //     number of k >= 0 with l*(2k+1) <= T*(v1+v2),
    // which equals (T*(v1+v2) + l) / (2*l) using integer division.
    // Similarly for type 2 with |v1-v2|.
    //
    // Overlap: a time t is counted in both if (2k+1)*(v1-v2) =
    // (2m+1)*(v1+v2) for some k, m >= 0. We can use observations from extended
    // Euclid. Let g = gcd(v1+v2, |v1-v2|), s' = (v1+v2)/g, d' = |v1-v2|/g.
    // Since gcd(s', d') = 1, we need s' | (2k+1) and d' | (2m+1). For these to
    // be odd (as required), both s' and d' must be odd. If so, overlap times
    // are t = l*(2k+1)*s'/(v1+v2) = l*(2k+1)/g for odd (2k+1) that are
    // multiples of s', i.e., t = l*n/g for odd n = 1, 3, 5, ... The count is
    // (T*g + l) / (2*l).
    //
    // Then the final answer is count1 + count2 - overlap.

    int64_t s = v1 + v2;
    int64_t d = abs(v1 - v2);

    int64_t count1 = (t * s + l) / (2 * l);
    int64_t count2 = (d > 0) ? (t * d + l) / (2 * l) : 0;

    int64_t overlap = 0;
    if(d > 0) {
        int64_t g = gcd(s, d);
        if((s / g) % 2 == 1 && (d / g) % 2 == 1) {
            overlap = (t * g + l) / (2 * l);
        }
    }

    cout << count1 + count2 - overlap << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

---

## 5) Python implementation

```python
import math
import sys

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

    # Relative speed when they move toward each other
    s = v1 + v2

    # Relative speed when they move in the same direction (catch-up)
    d = abs(v1 - v2)

    # Count type-1 meetings:
    # t = l*(2k+1)/s <= T  <=>  l*(2k+1) <= T*s
    # number of odd multiples of l up to T*s is floor((T*s + l)/(2*l))
    count1 = (T * s + l) // (2 * l)

    # Count type-2 meetings similarly, only if d>0
    count2 = (T * d + l) // (2 * l) if d > 0 else 0

    # Overlap of type-1 and type-2 sequences (avoid double counting)
    overlap = 0
    if d > 0:
        g = math.gcd(s, d)

        # Overlap exists only if (s/g) and (d/g) are both odd
        if ((s // g) % 2 == 1) and ((d // g) % 2 == 1):
            # Overlap times: t = (l/g) * odd
            # Count odd multiples of l/g within T:
            # floor((T*g + l)/(2*l))
            overlap = (T * g + l) // (2 * l)

    ans = count1 + count2 - overlap
    print(ans)

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