1. Abridged Problem Statement
Two people arrive independently and uniformly at a meeting point between X and Y hours. The first to arrive waits at most Z minutes for the other; if the other does not come within Z minutes, they leave. Compute the probability that they actually meet. Output with 7 decimal digits.

2. Detailed Editorial

We model the arrival times T₁ and T₂ (in hours) as independent uniform random variables on the interval [X, Y]. They successfully meet if and only if the absolute difference in their arrival times |T₁ − T₂| does not exceed Z minutes (Z/60 hours).

Let D = Y − X be the total interval length in hours, and δ = D·60 the same length in minutes. We convert arrival times to minutes after X—that is, U₁ = (T₁ − X)·60 and U₂ = (T₂ − X)·60—so that U₁, U₂ are independent and uniform on [0, δ]. The condition for meeting becomes |U₁ − U₂| ≤ Z.

Geometrically, the pair (U₁, U₂) is uniformly distributed over the δ×δ square [0,δ]×[0,δ]. The subset where |U₁ − U₂| ≤ Z is the region between the lines U₂ = U₁ + Z and U₂ = U₁ − Z, clipped to the square. Its area equals the total square area minus the areas of the two congruent right triangles in the corners where |U₁ − U₂| > Z:

- Total square area = δ².
- “Failure” region is two right triangles each of legs (δ − Z), so area = 2 · ((δ − Z)² / 2) = (δ − Z)².
- “Success” area = δ² − (δ − Z)² = 2δZ − Z².

Therefore the probability is
  P = (success area) / (total area)
    = [2δZ − Z²] / δ²
    = 2·(Z/δ) − (Z/δ)²
    = 1 − ((δ − Z)/δ)².

The reference code writes the same quantity as Z²/δ² + 2·(δ − Z)·Z/δ², which expands to (2δZ − Z²)/δ². Finally, δ = (Y − X)·60, so compute and print P to 7 decimal places.

Time and memory are trivial; this is an O(1) computation.

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;
};

double X, Y, Z;

void read() { cin >> X >> Y >> Z; }

void solve() {
    // Both arrival times are uniform over a window of delta = (Y - X) * 60
    // minutes, so the sample space is a delta x delta square. They meet when
    // their arrival times differ by at most Z, the diagonal band
    // |t1 - t2| <= Z. Its area is delta^2 minus the two corner triangles of
    // total area (delta - Z)^2, and we report that area divided by delta^2,
    // written here as Z^2/delta^2 + 2*(delta - Z)*Z/delta^2.

    double delta = (Y - X) * 60;
    double ans = (Z * Z) / (delta * delta);
    ans += 2 * (delta - Z) * Z / (delta * delta);
    cout << fixed << setprecision(7) << ans << '\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 with Detailed Comments

```python
import sys

def main():
    # Read input: X and Y are integers (hours), Z is a float (minutes)
    data = sys.stdin.read().strip().split()
    X, Y = map(int, data[:2])
    Z = float(data[2])

    # Convert the total meeting window (Y-X) from hours to minutes
    delta = (Y - X) * 60.0

    # If Z >= delta, they will always meet (the waiting window covers the whole interval)
    # but the formula below handles that case naturally (it gives P = 1).
    # Compute probability using P = (2*delta*Z - Z^2) / delta^2
    numerator = 2.0 * delta * Z - Z * Z
    denominator = delta * delta
    probability = numerator / denominator

    # Print the result with 7 decimal places
    sys.stdout.write(f"{probability:.7f}\n")

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

5. Compressed Editorial

Map arrival times to a δ×δ square (δ = (Y−X)·60). The meeting region is |t₁−t₂|≤Z, whose area is δ²−(δ−Z)²=2δZ−Z². Divide by δ² to get probability P=(2δZ−Z²)/δ²=2(Z/δ)−(Z/δ)², and print it with seven decimal places.
