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 + ;
 const int mod1 = 1e9 + ;
 const int mod2 = 1e9 + ;
 const int arg1 = ;
 const int arg2 = ;
 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 > ) return ;
         else if(g.find(it->first) == g.end()) return false;
     }
     return ;
 }
 void solve() {
     cin >> n >> k;
     cin >> str;
     pw1 = pw2 = ;
     for (int i = ; i < k - ; i++) {
         pw1 = ll(pw1) * arg1 % mod1;
         pw2 = ll(pw2) * arg2 % mod2;
     }
     pii h = pii(, );
     for (int i = ; i < k; i++) {
         h.first = (ll(h.first) * arg1 + ll(str[i] - 'a' + )) % mod1;
         h.second = (ll(h.second) * arg2 + ll(str[i] - 'a' + )) % mod2;
     }
     H[] = h;
     has[][h]++;
     for (int i = ; i < n * k; i++) {
         h.first = (ll(h.first) - ll(str[i - ] - 'a' + ) * pw1 % mod1 + mod1) % mod1;
         h.second = (ll(h.second) - ll(str[i - ] - 'a' + ) * pw2 % mod2 + mod2) % mod2;
         int ni = (i + k - ) % (n * k);
         h.first = (ll(h.first) * arg1 + ll(str[ni] - 'a' + )) % mod1;
         h.second = (ll(h.second) * arg2 + ll(str[ni] - 'a' + )) % mod2;
         has[i % k][h]++;
         H[i] = h;
     }
     int m; cin >> m;
     string cur;
     for (int j = ; j <= m; j++) {
         cin >> cur;
         pii h = pii(, );
         for (int i = ; i < k; i++) {
             h.first =  (ll(h.first) * arg1 + ll(cur[i] - 'a' + )) % mod1;
             h.second = (ll(h.second) * arg2 + ll(cur[i] - 'a' + )) % mod2;
             g[h] = j;
         }
     }
     for (int j = ; j < k; j++) if(check(has[j])) {
         cout << "YES" << endl;
         bool f = ;
         for (int i = j; i < n * k; i += k) {
             if(f) cout << " ";
             cout << g[H[i]]; f = ;
         }
         cout << endl;
         return;
     }
     cout << "NO" << endl;
 }
 int main() {
     //freopen("test.in", "r", stdin);
     //freopen("test.out", "w", stdout);
     solve();
     return ;
 }