rolling hash

也是需要查看，然后修改，rolling hash, recursive hash, polynomial hash, double hash.如果一次不够，那就2次。需要在准备一个线段树，基本的线段树容易些，带lazy标记的区间修改的线段树不是很好写。hash seed key根据需要选择， 我看别人写的，可以写成一个随机数，每次随机选择一个素数作为种子，这样好像好一些。

#include<bits/stdc++.h>
#define pb push_back
#define FOR(i, n) for (int i = 0; i < (int)n; ++i)
#define dbg(x) cout << #x << " at line " << __LINE__ << " is: " << x << endl
typedef long long ll;
using namespace std;
typedef pair<int, int> pii;
const int maxn = 1e6 + 10;
const int mod1 = 1e9 + 7;
const int mod2 = 1e9 + 9;
const int arg1 = 841;
const int arg2 = 151;
int n, k;
string str;
pii H[maxn];
map<pii, int> has[maxn];
map<pii, int> g;
int pw1, pw2;
bool check(map<pii, int> & m) {
    for (auto it = m.begin(); it != m.end(); it++) {
        if(it->second > 1) return 0;
        else if(g.find(it->first) == g.end()) return false;
    }
    return 1;
}
void solve() {
    cin >> n >> k;
    cin >> str;
    pw1 = pw2 = 1;
    for (int i = 0; i < k - 1; i++) {
        pw1 = ll(pw1) * arg1 % mod1;
        pw2 = ll(pw2) * arg2 % mod2;
    }
    pii h = pii(0, 0);
    for (int i = 0; i < k; i++) {
        h.first = (ll(h.first) * arg1 + ll(str[i] - 'a' + 1)) % mod1;
        h.second = (ll(h.second) * arg2 + ll(str[i] - 'a' + 1)) % mod2;
    }
    H[0] = h;
    has[0][h]++;
    for (int i = 1; i < n * k; i++) {
        h.first = (ll(h.first) - ll(str[i - 1] - 'a' + 1) * pw1 % mod1 + mod1) % mod1;
        h.second = (ll(h.second) - ll(str[i - 1] - 'a' + 1) * pw2 % mod2 + mod2) % mod2;
        int ni = (i + k - 1) % (n * k);
        h.first = (ll(h.first) * arg1 + ll(str[ni] - 'a' + 1)) % mod1;
        h.second = (ll(h.second) * arg2 + ll(str[ni] - 'a' + 1)) % mod2;
        has[i % k][h]++;
        H[i] = h;
    }
    int m; cin >> m;
    string cur;
    for (int j = 1; j <= m; j++) {
        cin >> cur;
        pii h = pii(0, 0);
        for (int i = 0; i < k; i++) {
            h.first =  (ll(h.first) * arg1 + ll(cur[i] - 'a' + 1)) % mod1;
            h.second = (ll(h.second) * arg2 + ll(cur[i] - 'a' + 1)) % mod2;
            g[h] = j;
        }
    }
    for (int j = 0; j < k; j++) if(check(has[j])) {
        cout << "YES" << endl;
        bool f = 0;
        for (int i = j; i < n * k; i += k) {
            if(f) cout << " ";
            cout << g[H[i]]; f = 1;
        }
        cout << endl;
        return;
    }
    cout << "NO" << endl;
}
int main() {
    //freopen("test.in", "r", stdin);
    //freopen("test.out", "w", stdout);
    solve();
    return 0;
}