## 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 \(2l\). The code’s reasoning boils down to analyzing the phase within periods of length \(2l\).

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 \(r \le l\),
- real position = \(2l - r\) otherwise.

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 \((0, T]\) (here \(T\) is input \(t\)):
\[
l(2k+1) \le T(v_1+v_2).
\]
Number of odd multiples of \(l\) up to \(T(v_1+v_2)\) is:
\[
\text{count}_1 = \left\lfloor \frac{T(v_1+v_2)+l}{2l}\right\rfloor.
\]
(That “+ l” is a neat way to count how many odd numbers fit using integer division.)

#### 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:
- \(s' \mid (2k+1)\),
- \(d' \mid (2m+1)\).

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 meeting times turn into:
\[
t = \frac{l(2k+1)}{s} = \frac{l \cdot s' \cdot q}{s} = \frac{l q}{g}
\]
where \(q\) must be odd. Thus 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) Provided C++ solution with detailed line-by-line comments

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

// Pretty-print a pair: "first second"
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

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

// Read an entire vector (assumes vector already has correct size)
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for (auto& x: a) {
        in >> x;
    }
    return in;
};

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

// Problem inputs (use int64_t to avoid overflow)
int64_t l, v1, v2, t;

// Read four integers: length, two speeds, and time horizon
void read() { cin >> l >> v1 >> v2 >> t; }

void solve() {
    // s = v1 + v2 is the relative speed when moving towards each other
    int64_t s = v1 + v2;

    // d = |v1 - v2| is the relative speed when moving in same direction
    // (catch-up scenario). If v1==v2 then d=0 and this family doesn't exist.
    int64_t d = abs(v1 - v2);

    // Count type-1 meetings: times t = l*(odd)/(v1+v2) within (0, input_t]
    // count1 = floor((T*s + l) / (2*l))
    int64_t count1 = (t * s + l) / (2 * l);

    // Count type-2 meetings (catch-up): times t = l*(odd)/|v1-v2|
    // Only if d>0.
    int64_t count2 = (d > 0) ? (t * d + l) / (2 * l) : 0;

    // Compute overlap between the two sequences of meeting times to avoid
    // double counting.
    int64_t overlap = 0;
    if (d > 0) {
        // g = gcd(s, d)
        int64_t g = gcd(s, d);

        // Overlap exists only if both reduced numbers s/g and d/g are odd
        // (so that they can divide an odd number 2k+1).
        if ((s / g) % 2 == 1 && (d / g) % 2 == 1) {
            // Overlap times become t = (l/g) * odd.
            // Count odd multiples of (l/g) within T:
            // floor((T*g + l) / (2*l)).
            overlap = (t * g + l) / (2 * l);
        }
    }

    // Inclusion-exclusion:
    // total = type1 + type2 - overlap
    cout << count1 + count2 - overlap << '\n';
}

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

    int T = 1;           // single test case
    // cin >> T;         // (disabled) multiple test cases not used here

    for (int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": "; // (disabled) no case formatting
        solve();
    }

    return 0;
}
```

---

## 4) Python solution (same logic) with detailed comments

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