1. Abridged Problem Statement  
Given \(n\) judges and a uniformly random derangement (permutation without fixed points) of size \(n\), each judge belongs to a cycle of length ≥ 2. In a “Highlander” chase game, each cycle of the permutation collapses to a single survivor who collects all cards in that cycle, and the winners are the survivors of the largest initial cycle(s). A judge has a theoretical chance to win if and only if he is in one of the largest cycles. Compute the expected total number of judges (i.e. sum of sizes of all maximum-length cycles) that have a chance to win.

2. Detailed Editorial  
– **Reformulation**  
  We choose a random derangement \(p\) of \(\{1,\dots,n\}\).  Let the cycle‐decomposition of \(p\) have cycle lengths \(c_1,c_2,\dots\), all \(\ge2\).  Let  
    • \(L=\max_i c_i\) be the largest cycle length.  
    • \(m\) be the number of cycles of length \(L\).  
  Every judge in those \(m\) largest cycles could end up as the survivor of that cycle and thus as an overall winner.  So the quantity we seek is  
  \[
    \mathbb{E}\bigl[m\times L\bigr],
  \]
  where the expectation is over all derangements of size \(n\).

– **DP State**  
  Define
  \[
    \mathrm{dp}[\,\ell\,][\,M\,][\,k\,]
    = \Pr\bigl(\text{a (partial) derangement of size }\ell\text{ has no cycle }>{M},\text{ and exactly }k\text{ cycles of size }M\bigr).
  \]
  We build this up for all \(\ell=0..n\) and \(M=1..n\).  (Cycles of size 1 are forbidden.)

– **Initialization**  
  For every \(M\ge1\), a derangement of size 0 trivially has zero cycles:  
  \(\mathrm{dp}[0][M][0]=1.\)

– **Transition**  
  To form a derangement of size \(\ell>0\), look at which cycle contains element 1.  Suppose that cycle has size \(s\) (with \(2\le s\le\min(M,\ell)\)).  Then we choose its other \(s-1\) members among \(\ell-1\) in \(\binom{\ell-1}{s-1}\) ways, arrange them in a cycle in \((s-1)!\) ways, and interleave with a derangement of the remaining \(\ell-s\) elements.  One can show that the total weight of “adding a cycle of size \(s\)” contributes exactly \(\tfrac1\ell\) to the transition probability.  Thus:
  \[
    \mathrm{dp}[\ell][M][\,k\,]
    += \sum_{s=2}^{\min(M,\ell)}
       \mathrm{dp}[\ell-s][M][\,k - [s=M]\,]\;\times\;\frac1\ell.
  \]
  Here \([s=M]\) is 1 if \(s=M\), else 0.

– **Normalization**  
  We have disallowed 1‐cycles, so the total of \(\mathrm{dp}[n][M][k]\) over all \(M\ge2\), \(k\ge1\) is
  \[
    Z = \sum_{M=2}^{n}\sum_{k=1}^{\lfloor n/M\rfloor}\mathrm{dp}[n][M][k].
  \]
  This \(Z\) is the probability mass of *all* derangements (which is less than 1 if you allowed 1‐cycles in an ordinary permutation).  Divide by \(Z\) to renormalize to the uniform derangement distribution.

– **Answer**  
  Finally
  \[
    \mathbb{E}[mL]
    = \frac1Z\;\sum_{M=2}^n\sum_{k=1}^{\lfloor n/M\rfloor}
      \mathrm{dp}[n][M][k]\;\times\;(k\times M).
  \]
  Compute with double precision, output with 9 decimal places.

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

// Overload << and >> for convenient pair and vector I/O
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& p) {
    return out << p.first << ' ' << p.second;
}
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& p) {
    return in >> p.first >> p.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(const auto& x: a) out << x << ' ';
    return out;
}

int n;

// Read the single integer n
void read() {
    cin >> n;
}

void solve() {
    // dp[ length ][ max_cycle_length ][ count_of_such_cycles ]
    // stores the probability of a permutation of 'length' with no cycle
    // larger than max_cycle_length, and exactly 'count' cycles of size
    // equal to max_cycle_length. We never allow cycles of size 1.
    vector<vector<vector<double>>> dp(n+1, vector<vector<double>>(n+1));
    
    // Allocate correct inner sizes
    for(int len = 0; len <= n; ++len) {
        for(int M = 1; M <= n; ++M) {
            int maxCnt = len / M;
            dp[len][M].assign(maxCnt+1, 0.0);
        }
    }

    // Base case: length=0, zero cycles of any max size => probability=1
    for(int M = 1; M <= n; ++M) {
        dp[0][M][0] = 1.0;
    }

    // Build up for lengths 1..n
    for(int len = 1; len <= n; ++len) {
        for(int M = 1; M <= n; ++M) {
            int maxCnt = len / M;
            for(int cnt = 0; cnt <= maxCnt; ++cnt) {
                // We decide the cycle containing element '1' has size 's'
                // s runs from 2 to min(M, len); each choice contributes 1/len
                // to the probability mass.
                for(int s = 2; s <= min(M, len); ++s) {
                    // If s == M then we are adding one more cycle of size M
                    int prevCnt = cnt - (s == M ? 1 : 0);
                    if(prevCnt < 0) continue;
                    // Ensure dp[len-s][M][prevCnt] is valid
                    if(prevCnt < (int)dp[len-s][M].size()) {
                        dp[len][M][cnt] += dp[len-s][M][prevCnt] / double(len);
                    }
                }
            }
        }
    }

    // Compute normalization Z = total probability of getting a valid derangement
    double Z = 0.0;
    for(int M = 2; M <= n; ++M) {
        int maxCnt = n / M;
        for(int cnt = 1; cnt <= maxCnt; ++cnt) {
            Z += dp[n][M][cnt];
        }
    }

    // Compute the expected sum of sizes of the largest cycles: E[m * M]
    double ans = 0.0;
    for(int M = 2; M <= n; ++M) {
        int maxCnt = n / M;
        for(int cnt = 1; cnt <= maxCnt; ++cnt) {
            double prob = dp[n][M][cnt] / Z;  // renormalized
            ans += prob * (cnt * M);
        }
    }

    // Print with 9 decimal places
    cout << fixed << setprecision(9) << ans << "\n";
}

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

    read();
    solve();
    return 0;
}
```

4. Python Solution with Detailed Comments  
```python
import sys
def main():
    data = sys.stdin.read().strip().split()
    n = int(data[0])

    # dp[l][M][k] = probability of a derangement of size l
    # with no cycle > M, and exactly k cycles of size M
    dp = [ [ [] for _ in range(n+1) ] for __ in range(n+1) ]
    # Pre-allocate inner lists by max possible k = floor(l/M)
    for l in range(n+1):
        for M in range(1, n+1):
            maxCnt = l // M
            dp[l][M] = [0.0] * (maxCnt+1)

    # Base case: empty set has 1 way, zero cycles
    for M in range(1, n+1):
        dp[0][M][0] = 1.0

    # Build DP for lengths 1..n
    for l in range(1, n+1):
        for M in range(1, n+1):
            maxCnt = l // M
            for k in range(maxCnt+1):
                # Choose the cycle size s containing '1'
                # must be in [2..min(M,l)]
                for s in range(2, min(M, l)+1):
                    prev_k = k - (1 if s == M else 0)
                    if prev_k < 0: 
                        continue
                    # We add probability mass 1/l
                    dp[l][M][k] += dp[l-s][M][prev_k] / l

    # Sum up total mass of valid derangements Z
    Z = 0.0
    for M in range(2, n+1):
        for k in range(1, n//M + 1):
            Z += dp[n][M][k]

    # Compute expected total size of largest cycles: E[k * M]
    ans = 0.0
    for M in range(2, n+1):
        for k in range(1, n//M + 1):
            prob = dp[n][M][k] / Z
            ans += prob * (k * M)

    # Print with 9 decimal places
    print(f"{ans:.9f}")

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

5. Compressed Editorial  
- We need the expected sum of sizes of all maximum-length cycles in a random derangement of size \(n\).  
- Define \(\mathrm{dp}[l][M][k]\) = prob. that a derangement of \(l\) elements has no cycle > \(M\) and exactly \(k\) cycles of size \(M\).  Disallow 1-cycles.  
- Base: \(\mathrm{dp}[0][M][0]=1\).  Transition by choosing the cycle through element 1 of size \(s\in[2,\min(M,l)]\), which contributes \(\tfrac1l\) to the transition.  
- After filling dp up to \(l=n\), compute normalization \(Z=\sum_{M=2..n}\sum_{k\ge1}\mathrm{dp}[n][M][k]\).  
- The answer is \(\frac1Z\sum_{M=2..n}\sum_{k\ge1}\mathrm{dp}[n][M][k]\times(k\times M)\), printed with 9 decimals.