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. 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 n, m, k;

void read() { cin >> n >> m >> k; }

void solve() {
    // Iron boards are 1 bits. The constraint is that every window of m
    // consecutive boards holds at least k iron boards, which only constrains
    // complete windows; there are n - m + 1 of them.
    //
    // dp[i][mask] = number of ways to fill the first m + i boards such that the
    // i-th window (boards i..i+m-1) equals mask and all windows up to it are
    // valid. The base layer dp[0][mask] enumerates the first window directly,
    // keeping only masks with popcount >= k. To extend, we slide the window by
    // one board: appending bit b turns mask into ((mask << 1) | b) truncated to
    // m bits, again required to have popcount >= k. The answer sums the final
    // layer over all masks.

    vector<vector<int64_t>> dp(n - m + 1, vector<int64_t>(1 << m, 0));

    for(int mask = 0; mask < (1 << m); mask++) {
        int cnt = __builtin_popcount(mask);
        if(cnt < k) {
            continue;
        }

        dp[0][mask] = 1;
    }

    for(int i = 1; i < n - m + 1; i++) {
        for(int mask = 0; mask < (1 << m); mask++) {
            if(dp[i - 1][mask] == 0) {
                continue;
            }

            for(int bit = 0; bit < 2; bit++) {
                int new_mask = (mask << 1) | bit;
                new_mask &= (1 << m) - 1;

                int cnt = __builtin_popcount(new_mask);
                if(cnt < k) {
                    continue;
                }

                dp[i][new_mask] += dp[i - 1][mask];
            }
        }
    }

    int64_t ans = 0;
    for(int mask = 0; mask < (1 << m); mask++) {
        ans += dp[n - m][mask];
    }

    cout << ans << '\n';
}

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

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

    return 0;
}
```

4. Python Solution

```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).
