1. Abridged Problem Statement

Given three integers n, m, k (1 ≤ n ≤ 60, 1 ≤ m ≤ 15, 0 ≤ k ≤ m ≤ n), count the number of binary strings of length n such that every contiguous substring of length m contains at least k ones.

2. Detailed Editorial

We must count all binary strings S[1..n] over {0,1} such that for every i with 1 ≤ i ≤ n−m+1, the substring S[i..i+m−1] has at least k ones.

A classic way is to use dynamic programming with a bitmask of size m that records the last m bits of the prefix built so far. Let dp[i][mask] be the number of valid length-(i+m) prefixes whose last m bits form the bitmask 'mask'. Here i ranges from 0 up to (n−m), so that i+m runs from m to n.

  – State definition:
    dp[i][mask]: number of valid prefixes of length i+m, ending in the m-bit pattern 'mask'.

  – Base case (i=0):
    We choose the first m bits arbitrarily as 'mask', but only if popcount(mask) ≥ k (to satisfy the constraint on the very first window). Thus:
      dp[0][mask] = 1  if popcount(mask) ≥ k;
                 = 0  otherwise.

  – Transition:
    To extend a valid prefix of length i+m by one more bit b∈{0,1}, we compute a new mask:
      new_mask = ((mask << 1) | b) & ((1<<m)−1)
    which discards the oldest bit and appends b at the least significant position. We only allow this transition if popcount(new_mask) ≥ k, ensuring the newly formed window (positions i+1 through i+m) still has ≥k ones. Hence:
      dp[i+1][new_mask] += dp[i][mask]

  – Final answer:
    After filling dp up to i = n−m, the total number of valid strings of length n is
      sum_{mask=0..2^m−1} dp[n−m][mask]

Time complexity: O((n−m+1) · 2^m · 2) ≲ O(n·2^m). With n≤60, m≤15, this is about 60·32768·2 ≃ 4×10^6 operations, which fits comfortably.

Memory: O((n−m+1)·2^m). One can reduce to O(2^m) by reusing two rolling arrays for dp[i] and dp[i+1].

3. Provided C++ Solution with Detailed Comments

```cpp
#include <bits/stdc++.h>
using namespace std;

// Overload stream operators for convenience (not essential to core logic)
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 n, m, k;

// Read input values n, m, k
void read() {
    cin >> n >> m >> k;
}

void solve() {
    // Number of DP rows is (n-m+1), each row has 2^m masks
    int rows = n - m + 1;
    int totalMasks = 1 << m;

    // dp[i][mask] = number of valid prefixes of length i+m whose last m bits = mask
    vector<vector<int64_t>> dp(rows, vector<int64_t>(totalMasks, 0));

    // Base case: i = 0, first window is exactly the first m bits = mask
    for(int mask = 0; mask < totalMasks; mask++) {
        int cnt = __builtin_popcount(mask);
        if(cnt >= k) {
            dp[0][mask] = 1; // valid initial mask
        }
    }

    // Build dp for i = 1 .. rows-1
    for(int i = 1; i < rows; i++) {
        for(int mask = 0; mask < totalMasks; mask++) {
            int64_t ways = dp[i-1][mask];
            if(ways == 0) continue; // no sequences to extend

            // Try appending bit = 0 or 1
            for(int bit = 0; bit < 2; bit++) {
                // Shift mask left, add new bit, and keep only m LSBs
                int new_mask = ((mask << 1) | bit) & (totalMasks - 1);

                // Check popcount constraint on the new window
                if(__builtin_popcount(new_mask) < k) continue;

                // Accumulate ways
                dp[i][new_mask] += ways;
            }
        }
    }

    // Sum all dp at the final row i = rows-1
    int64_t answer = 0;
    for(int mask = 0; mask < totalMasks; mask++) {
        answer += dp[rows-1][mask];
    }
    cout << answer << "\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():
    # Read input
    data = sys.stdin.read().strip().split()
    n, m, k = map(int, data)

    rows = n - m + 1
    total_masks = 1 << m

    # Initialize DP table: rows x total_masks
    dp = [ [0]*total_masks for _ in range(rows) ]

    # Base case: first m bits form a mask with at least k ones
    for mask in range(total_masks):
        if mask.bit_count() >= k:
            dp[0][mask] = 1

    # Fill DP for each subsequent position
    for i in range(1, rows):
        prev = dp[i-1]
        curr = dp[i]
        for mask in range(total_masks):
            ways = prev[mask]
            if ways == 0:
                continue
            # Try appending 0 or 1
            for bit in (0, 1):
                new_mask = ((mask << 1) | bit) & (total_masks - 1)
                # Enforce at least k ones in this window
                if new_mask.bit_count() < k:
                    continue
                curr[new_mask] += ways

    # The answer is the sum of ways in the last row
    result = sum(dp[rows-1])
    print(result)

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

5. Compressed Editorial

We use a DP over bitmasks of window size m. Let dp[i][mask] count valid prefixes of length i+m ending in 'mask'. Initialize dp[0][mask]=1 for masks with ≥k ones. Transition by appending one bit and shifting the mask, only if new mask has ≥k ones. Finally, sum dp[n−m][*]. Time O(n·2^m), memory O((n−m+1)·2^m).