------
## 最终结论
Here's the complete Python code that solves the problem as described:

```python
def min_subscriptions(t, test_cases):
    results = []
    for case in test_cases:
        n, k, d, a = case
        show_count = {}
        unique_shows = 0
        
        # Count unique shows in the first 'd' days
        for i in range(d):
            if a[i] not in show_count:
                show_count[a[i]] = 0
                unique_shows += 1
            show_count[a[i]] += 1
        
        min_unique_shows = unique_shows
        
        # Slide the window through the remaining days
        for i in range(d, n):
            # Remove the show that is sliding out of the window
            if show_count[a[i - d]] == 1:
                unique_shows -= 1
            show_count[a[i - d]] -= 1

            # Add the new show that is entering the window
            if a[i] not in show_count or show_count[a[i]] == 0:
                unique_shows += 1
                show_count[a[i]] = 0
            show_count[a[i]] += 1

            # Update the minimum unique shows encountered
            min_unique_shows = min(min_unique_shows, unique_shows)
        
        # Store the minimum unique shows for each test case in the results list
        results.append(min_unique_shows)
    
    return results

# Input reading and output printing
import sys

input = sys.stdin.read
data = input().split()

t = int(data[0])
index = 1

test_cases = []

for _ in range(t):
    n, k, d = map(int, data[index:index+3])
    index += 3
    a = list(map(int, data[index:index+n]))
    index += n
    
    test_cases.append((n, k, d, a))

results = min_subscriptions(t, test_cases)

for result in results:
    print(result)
```

This code will read from standard input and output the minimum number of subscriptions needed for each test case. It uses a sliding window technique to efficiently calculate the number of unique TV shows within any segment of `d` consecutive days.