Certainly! Here's a test case generator for the provided problem statement:

```python
import random
import math

def tcgen():
    A = random.randint(1, 1000)
    B = random.randint(1, 1000)
    H = random.randint(0, 11)
    M = random.randint(0, 59)
    
    return A, B, H, M

# Example usage
for _ in range(5):
    A, B, H, M = tcgen()
    print(A, B, H, M)
```

This generator produces random values for the lengths of the hour and minute hands (`A` and `B`), and the time (`H` hours and `M` minutes). 

### Explanation:
1. `A` and `B` are randomly chosen from the range [1, 1000] to represent the lengths of the hour and minute hands.
2. `H` is randomly chosen from the range [0, 11] to represent the hour.
3. `M` is randomly chosen from the range [0, 59] to represent the minutes.

### Note:
In a competitive programming environment, you would typically run multiple test cases to validate the correctness of your solution. The example usage demonstrates generating 5 test cases, but you can adjust the loop to produce as many as needed.

### Calculating the Output:
To calculate the distance between the ends of the hour and minute hands, you would use the given constraints and the properties of the clock:

1. Calculate the angles of the hour and minute hands:
   - Minute hand angle: \( \theta_m = \frac{M}{60} \times 360 \) degrees
   - Hour hand angle: \( \theta_h = \left(\frac{H}{12} + \frac{M}{720}\right) \times 360 \) degrees

2. Convert these angles to radians.
3. Use the law of cosines to find the distance `d` between the points:
   \[
   d = \sqrt{A^2 + B^2 - 2 \times A \times B \times \cos(\theta_h - \theta_m)}
   \]

The precision requirement is very high (absolute or relative error at most \(10^{-9}\)), so ensure your calculations are precise enough to meet this criterion.