1. Abridged Problem Statement  
You have a perfect square chessboard of side length L (without loss of generality, take L=1 so the board’s area is 1). A laser shot burns a perfectly round hole whose diameter equals the side of the board (i.e. radius ½), removing that circular area from the board. Given an integer P (0≤P≤100), find the minimum number of shots needed so that at least P percent of the board’s area is destroyed. Process each P in the input, and print “Case #k: x” for the k-th query.

2. Detailed Editorial  
— Observation 1: One shot makes a circle of radius ½, so it removes area π·(½)²=π/4≈0.7854, i.e. about 78.54% of the board. Thus if P≤78, one shot suffices; if P>78, one shot is not enough.  
— Observation 2: With two circles of radius ½ each, arranged optimally (for instance, by centering them on two opposite corners of the square), you can cover about 95% of the unit square. Hence if 78<P≤95, the answer is 2.  
— Observation 3: Three such circles, placed symmetrically, can cover over 99% of the square. So if 95<P≤99, the answer is 3.  
— Observation 4: To destroy exactly 100% you cannot do it with three circles of that radius—you need four (one in each quadrant)—so for P=100 the answer is 4.  
— Edge case: P=0 → 0 shots.  
Since P is an integer between 0 and 100 and there are at most 100 queries, we precompute these thresholds and answer each query in O(1).  

3. Provided C++ Solution with Line-by-Line Comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

// Overload printing for pair<T1,T2>
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

// Overload reading for pair<T1,T2>
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

// Overload reading for vector<T>
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
}

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

int p;  // percentage required

// Solve one test case: read p, print minimum shots
void solve() {
    // If the required percentage is 0, no shots needed
    if(p == 0) {
        cout << 0 << '\n';
    }
    // One shot covers about 78.54% → if p ≤ 78, answer = 1
    else if(p <= 78) {
        cout << 1 << '\n';
    }
    // Two shots can cover up to about 95% → if p ≤ 95, answer = 2
    else if(p <= 95) {
        cout << 2 << '\n';
    }
    // Three shots can cover up to about 99% → if p ≤ 99, answer = 3
    else if(p <= 99) {
        cout << 3 << '\n';
    }
    // To reach 100%, we need four shots
    else {
        cout << 4 << '\n';
    }
}

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

    // Read percentages until EOF; index tests from 1
    for(int test = 1; cin >> p; test++) {
        cout << "Case #" << test << ": ";
        solve();
    }
    return 0;
}
```

4. Python Solution with Detailed Comments  
```python
import sys

def min_shots(p):
    """
    Given p in [0..100], return the minimum number of radius-0.5 circles
    needed to cover at least p% of the unit square.
    """
    if p == 0:
        return 0
    # One circle covers ~78.54%
    if p <= 78:
        return 1
    # Two circles cover ~95%
    if p <= 95:
        return 2
    # Three circles cover ~99%
    if p <= 99:
        return 3
    # Only four cover 100%
    return 4

def main():
    case_num = 1
    for line in sys.stdin:
        line = line.strip()
        if not line:
            continue
        p = int(line)
        result = min_shots(p)
        print(f"Case #{case_num}: {result}")
        case_num += 1

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

5. Compressed Editorial  
- A single radius‐½ circle burns π/4≈78.5% → if P≤78 ⇒ 1 shot.  
- Two such circles optimally placed cover ≈95% → if 78<P≤95 ⇒ 2 shots.  
- Three cover ≈99% → if 95<P≤99 ⇒ 3 shots.  
- Exactly 100% requires four → P=100 ⇒ 4 shots.  
- P=0 ⇒ 0 shots. Precompute thresholds, answer in O(1).