1. Abridged Problem Statement  
Given integers N and K, and an array D of N decimal digits D[0..N−1]. For each starting index i (0≤i<N), construct an infinite repeating decimal A[i] = 0.(D[i], D[(i+K)%N], D[(i+2K)%N], …). Among all A[i], find the maximum value and print the first N digits of its fractional part.

2. Detailed Editorial  

Overview  
We have N “infinite” decimals, each generated by stepping through the digit-array D in jumps of K mod N and repeating cyclically. Since they’re all between 0 and 1, comparing two such infinite decimals reduces to lexicographically comparing their digit-sequences (extended infinitely).

Key observations  
- The stepping by K mod N partitions the indices {0,…,N−1} into disjoint cycles, each of length L = cycle_size = the smallest positive t with t·K ≡ 0 (mod N). Equivalently, L = N / gcd(N,K).  
- Within a cycle, the infinite decimal starting at different cycle positions are just cyclic rotations of the same finite block of length L. Repeating decimals compare by lexicographically comparing their infinite expansions; for a given cycle, the maximum such decimal is the cyclic rotation of the block that is lexicographically largest.  
- We need the maximum over *all* cycles.

Algorithm steps  
1. Read N, K, and D[0..N−1].  
2. Initialize a boolean used[0..N−1] = {false}.  
3. For each i from 0 to N−1:  
   - If used[i], skip.  
   - Otherwise, follow the cycle starting at i by repeatedly jumping j ← (j+K)%N until you return to an already used index. Record the digits in order into vector `cycle`. Mark each visited j as used.  
   - Compute the lexicographically maximal rotation of `cycle`. We can do this in O(L) using Booth’s algorithm for least rotation, applied to the *negated* sequence to turn “max rotation” into “min rotation”.  
   - Compare that maximal-rotation vector with our global best; if larger lexicographically, replace it.  
4. Denote the chosen cycle’s maximal-rotation block by `ans`. Its length is L. The actual infinite decimal is the infinite repetition of `ans`. We only need the first N digits: output ans[0], ans[1], …, ans[N−1] (taking indices mod L).

Complexities  
- We visit each index exactly once → O(N) overall.  
- Each cycle of length L is processed via Booth in O(L). Sum of all L is N.  
- Total time O(N). N≤150 000 so it’s fast.

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, k;
vector<int> d;

void read() {
    cin >> n >> k;
    string s;
    cin >> s;

    d.resize(n);
    for(int i = 0; i < n; i++) {
        d[i] = s[i] - '0';
    }
}

template<typename T>
int least_rotation(const vector<T>& s) {
    int n = s.size();
    if(n == 0) {
        return 0;
    }

    vector<int> f(2 * n, -1);
    int k_ = 0;
    for(int j = 1; j < 2 * n; ++j) {
        int i = f[j - k_ - 1];
        while(i != -1 && s[j % n] != s[(k_ + i + 1) % n]) {
            if(s[j % n] < s[(k_ + i + 1) % n]) {
                k_ = j - i - 1;
            }
            i = f[i];
        }

        if(i == -1 && s[j % n] != s[(k_ + i + 1) % n]) {
            if(s[j % n] < s[(k_ + i + 1) % n]) {
                k_ = j;
            }
            f[j - k_] = -1;
        } else {
            f[j - k_] = i + 1;
        }
    }

    return k_;
}

template<typename T>
vector<T> max_cyclic_shift(const vector<T>& v) {
    if(v.empty()) {
        return {};
    }

    int m = v.size();
    vector<T> t(m);
    for(int i = 0; i < m; i++) {
        t[i] = -v[i];
    }

    int kk = least_rotation(t);
    vector<T> ans(m);
    for(int i = 0; i < m; ++i) {
        ans[i] = v[(kk + i) % m];
    }

    return ans;
}

void solve() {
    // Following index i by repeated +K mod N visits a cycle of the permutation;
    // the infinite fraction starting at any index of that cycle is periodic with
    // that cycle as its period. Two starts on the same cycle differ only by a
    // rotation of the period, so the largest fraction from a cycle corresponds
    // to its lexicographically maximum cyclic shift (found via Booth's least-
    // rotation algorithm on the negated digits). We take the best shift over all
    // cycles, then print the first N digits by repeating the winning period.

    vector<bool> used(n, false);
    vector<int8_t> ans;
    for(int i = 0; i < n; i++) {
        if(used[i]) {
            continue;
        }

        vector<int8_t> cycle;
        for(int j = i; !used[j]; j = (j + k) % n) {
            used[j] = true;
            cycle.push_back(d[j]);
        }

        cycle = max_cyclic_shift(cycle);
        if(ans.empty() || cycle > ans) {
            ans = cycle;
        }
    }

    for(int i = 0; i < n; i++) {
        cout << (int)ans[i % ans.size()];
    }

    cout << '\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 with Detailed Comments  
```python
import sys
sys.setrecursionlimit(10**7)
def read_input():
    data = sys.stdin.read().split()
    n, k = map(int, data[:2])
    s = data[2].strip()
    D = list(map(int, s))
    return n, k, D

def least_rotation(s):
    # Booth’s algorithm for minimum rotation in O(len(s))
    n = len(s)
    if n == 0:
        return 0
    # 'fail' array is like the prefix-function used in Booth’s
    fail = [-1] * (2*n)
    k = 0  # candidate start
    for j in range(1, 2*n):
        i = fail[j - k - 1]
        while i != -1 and s[j % n] != s[(k + i + 1) % n]:
            if s[j % n] < s[(k + i + 1) % n]:
                k = j - i - 1
            i = fail[i]
        if i == -1 and s[j % n] != s[(k + i + 1) % n]:
            if s[j % n] < s[(k + i + 1) % n]:
                k = j
            fail[j - k] = -1
        else:
            fail[j - k] = i + 1
    return k

def max_cyclic_shift(v):
    # To get maximum rotation, negate and find least rotation
    # Python ints are unbounded so negation works fine
    neg = [-x for x in v]
    start = least_rotation(neg)
    # build the rotated result
    return [v[(start + i) % len(v)] for i in range(len(v))]

def main():
    n, k, D = read_input()
    used = [False]*n
    best = None

    for i in range(n):
        if used[i]:
            continue
        # build one cycle
        cycle = []
        j = i
        while not used[j]:
            used[j] = True
            cycle.append(D[j])
            j = (j + k) % n
        # find its lexicographically maximum cyclic shift
        mx = max_cyclic_shift(cycle)
        if best is None or mx > best:
            best = mx

    # print first n digits of infinite repetition of 'best'
    L = len(best)
    out = ''.join(str(best[i % L]) for i in range(n))
    print(out)

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

5. Compressed Editorial  
Partition the indices mod N by jumps of size K into disjoint cycles. Each cycle yields a repeating block of digits; the value of its infinite decimal is maximized by choosing the lexicographically largest cyclic rotation of that block. Do this for each cycle (using Booth’s O(L) algorithm), track the globally best block, and finally output its first N digits by repeating it. Total time O(N).