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.

The approach precomputes these thresholds: for cnt=2 and cnt=3, a brute-force search over laser positions (up to some precision) confirms that 2 circles suffice for 95% and 3 circles suffice for 99%. For 100%, one can show that 3 circles of radius ½ cannot cover the entire unit square (the minimum radius needed is sqrt(65)/16 > 1/2), so 4 shots are required. Since P is an integer between 0 and 100, we answer each query in O(1).

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;
};

int p;

void solve() {
    // The trick in this problem is that we can compute the answer in advance and
    // then answer everything in O(1). This should be obvious by realizing that there
    // are only 100 possible values for p, and we can compute the answer for each one.
    //
    // The other observation is that the answer is small, and certainly less than 4. 
    // 1) For cnt = 1, we can cover the whole board with 1 laser. Assuming that the board 
    //    is 1x1, the diameter is sqrt(2) and the radius of the circle is 1/2. Then the
    //    area is pi * (1/2)^2 = pi/4, or roughly 78.5%.
    // 2) For cnt = 2 and cnt = 3, we can simply brute force all positions of the laser,
    //    up to some precision. We can notice that for 3 lasers, it's quite easy to find
    //    a solution with 99% coverage, and for 2 lasers, we can simply use precision that
    //    is roughly 500 and find a solution with 95% coverage.
    // 4) We can show that to get 100% coverage, we need at 3 circles of radius at least
    //    sqrt(65)/16, which is a bit over the radius of 1/2 we have. A relevant page for 
    //    this is: https://www.quora.com/A-unit-square-is-completely-covered-by-three-
    //             identical-circles-Find-the-smallest-possible-diameter-of-the-circles
    //    Hence, the answer of 99% is satisfactory for 3 lasers too, and we don't have to
    //    use too much compute.

    if(p == 0) {
        cout << 0 << '\n';
    } else if(p <= 78) {
        cout << 1 << '\n';
    } else if(p <= 95) {
        cout << 2 << '\n';
    } else if(p <= 99) {
        cout << 3 << '\n';
    } else {
        cout << 4 << '\n';
    }
}

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

    for(int test = 1; cin >> p; test++) {
        cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

4. Python Solution
```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).
