## 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 \((0, t]\) (a meeting exactly at time **t** counts).

---

## 2) Detailed editorial (how the solution works)

### Key idea: "unfold" reflections into motion on an infinite line

Bouncing on endpoints is annoying. A standard trick: reflect the segment \([0,l]\) repeatedly to make an infinite line, so that a back-and-forth motion becomes uniform motion without turning.

Let A be position 0 and B be position l.

- Cyclist: starts at 0 moving right with speed \(v_1\). In the unfolded line, its "phase" is \(x_c(t)=v_1 t\).
- Pedestrian: starts at l moving left with speed \(v_2\). In the unfolded line, think of its phase as \(x_p(t)=l - v_2 t\) in the base segment, but after unfolding it also becomes periodic mod \(2l\).

Folding back from unfolded coordinate to real position in \([0,l]\) is done by the "triangle wave": If \(r = x \bmod (2l)\), real position \(= r\) if \(r \le l\), else \(= 2l - r\). Same for both, with different starting offsets/directions. Two people meet when their folded positions are equal.

### Deriving meeting times: two families

When you enumerate cases of who is moving which way (depending on which half of the \(2l\) period they are in), the equality of folded positions reduces to **two congruences**, producing two families of meeting times:

#### Type 1: they approach each other (relative speed \(v_1+v_2\))

Meetings occur when \((v_1+v_2)\,t = l \pmod{2l}\). So:
\[
t = \frac{l(2k+1)}{v_1+v_2}, \quad k=0,1,2,\dots
\]
Count those within \((0, T]\):
\[
\text{count}_1 = \left\lfloor \frac{T(v_1+v_2)+l}{2l}\right\rfloor.
\]

#### Type 2: one catches up to the other (relative speed \(|v_1-v_2|\))

This exists only if \(v_1 \ne v_2\). Meetings occur when \(|v_1-v_2|\,t = l \pmod{2l}\), so:
\[
t = \frac{l(2k+1)}{|v_1-v_2|}.
\]
Count:
\[
\text{count}_2 =
\left\lfloor \frac{T|v_1-v_2|+l}{2l}\right\rfloor
\quad (\text{or }0\text{ if }v_1=v_2).
\]

### Avoid double counting: overlap between the two families

Some times can satisfy both type-1 and type-2 formulas, so we subtract overlaps.

We need:
\[
\frac{l(2k+1)}{v_1+v_2} = \frac{l(2m+1)}{|v_1-v_2|}.
\]
Cancel \(l\):
\[
(2k+1)|v_1-v_2| = (2m+1)(v_1+v_2).
\]

Let \(s = v_1+v_2\), \(d = |v_1-v_2|\), \(g = \gcd(s,d)\), \(s' = s/g\), \(d' = d/g\) with \(\gcd(s',d')=1\).

From \((2k+1)d' = (2m+1)s'\), coprimality implies \(s' \mid (2k+1)\) and \(d' \mid (2m+1)\). But \((2k+1)\) and \((2m+1)\) are **odd**, so for such multiples to exist, both \(s'\) and \(d'\) must be **odd**. If either is even, overlap is 0.

If both are odd, overlap times are:
\[
t = \frac{l}{g} \cdot (1,3,5,\dots)
\]
Count overlaps:
\[
\text{overlap} = \left\lfloor \frac{Tg + l}{2l}\right\rfloor.
\]

### Final answer

\[
\boxed{\text{answer} = \text{count}_1 + \text{count}_2 - \text{overlap}}
\]

All computations fit in 64-bit signed integers because products are up to \(10^9\cdot 2\cdot 10^9 = 2\cdot 10^{18}\), within int64 range.

---

## 3) C++ Solution

```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;
}
```

---

## 4) Python solution (same logic)

```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()
```

---

## 5) Compressed editorial

Unfold reflections: motion on \([0,l]\) with bouncing equals uniform motion mod \(2l\) with a triangular fold. Meeting times split into two congruences:

1. Approaching: \((v_1+v_2)t \equiv l \pmod{2l}\)
   \(\Rightarrow t = \frac{l(2k+1)}{v_1+v_2}\),
   \(\text{count}_1=\left\lfloor\frac{T(v_1+v_2)+l}{2l}\right\rfloor\).

2. Catch-up (if \(v_1\ne v_2\)): \(|v_1-v_2|t \equiv l \pmod{2l}\)
   \(\Rightarrow t = \frac{l(2k+1)}{|v_1-v_2|}\),
   \(\text{count}_2=\left\lfloor\frac{T|v_1-v_2|+l}{2l}\right\rfloor\).

Overlap exists only if with \(s=v_1+v_2\), \(d=|v_1-v_2|\), \(g=\gcd(s,d)\), both \(s/g\) and \(d/g\) are odd; then overlap times are \(t=(l/g)\cdot\text{odd}\) and
\(\text{overlap}=\left\lfloor\frac{Tg+l}{2l}\right\rfloor\).

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