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.

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. Provided C++ Solution with Detailed Comments  

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

// Main variables: start time X, end time Y (in hours), wait limit Z (in minutes)
double X, Y, Z;

// Read input values
void read() {
    cin >> X >> Y >> Z;
}

// Compute and output the meeting probability
void solve() {
    // Convert interval length (Y - X) from hours to minutes
    double delta = (Y - X) * 60.0;
    // Using the derived formula: P = (2 * delta * Z - Z^2) / (delta^2)
    // We split it as two terms for numerical clarity:
    double term1 = (Z * Z) / (delta * delta);                 // contributes - (Z/delta)^2
    double term2 = 2.0 * (delta - Z) * Z / (delta * delta);   // contributes 2 * Z/delta * (1 - Z/delta)
    double ans = term1 + term2;
    // Output with fixed format and 7 digits after the decimal point
    cout << fixed << setprecision(7) << ans << '\n';
}

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

    // Single test case
    read();
    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.