2022湖北省赛 J. Palindrome Reversion （字符串哈希）

https://codeforces.com/gym/103729/problem/J
题意： 选择一个子串翻转一次，问是否能得到一个回文串

#include<bits/stdc++.h>
//#include <bits/extc++.h>
using namespace std;
// using namespace __gnu_cxx;
// using namespace __gnu_pbds;
#define IOS ios::sync_with_stdio(false) ,cin.tie(0), cout.tie(0);  
//#pragma GCC optimize(3,"Ofast","inline")
#define ll long long
#define ull unsigned long long
#define li __int128_t
//#define int long long
const int N = 1e5 + 5;
const int M = 1e6 + 5;
const int mod = 998244353;
const ll LNF = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1.0);

struct str_hash { //严格鸽板子
   ull base = 131;
   ull p[N]; char s[N];
   unordered_map<ull, ull> mp;
   // 上面可以换成数组：ull mp[N];
   int n;
   void init( string &str ) {
       n = str.length();
       mp.clear();
       for ( int i = 1; i <= n; ++ i ) s[i] = str[i - 1];
       p[0] = 1;
       for ( int i = 1; i <= n; ++ i ) {
           mp[i] = mp[i - 1] * base + s[i];
           p[i] = p[i - 1] * base;
       }
   }
   ull hash( ull l , ull r ) {
       return mp[r] - mp[l - 1] * p[r - l + 1];
   }
}nor, rev;

int main() {
  string s; cin >> s;
  string ss; int len = s.length();
  int st = 0;
  for ( int i = 0; i < len; ++ i ) {
    if(s[i] == s[len - i - 1]) {
      ++ st;
    } else break;
  }
  for ( int i = st; i < len - st; ++ i ) {
    ss += s[i];
  }
  if(ss.size() == 0 ) {
    cout << 1 << " " << 1 << endl; return 0;
  }
  nor.init(ss); reverse(ss.begin(), ss.end());  rev.init(ss);
  int n = ss.length();
  //AAP
  for ( int l = 1; l <= n / 2; ++ l ) {
    if(nor.hash(1, l) == nor.hash( 1 + l, l * 2 ) ) {
      int pst = 2*l + 1;
      if(pst == n + 1) {
        cout << 1 + st << " " << l + st << endl; return 0;
      }
      if( nor.hash( pst, n ) == rev.hash( 1, n - pst + 1) ) {
        cout << l + 1 + st << " " << n + st << endl; return 0;
      }
    }
  }
  //PAA
  for ( int l = 1; l <= n / 2; ++ l ) {
    if(rev.hash(1, l) == rev.hash( 1 + l, l * 2 ) ) {
      int pst = 2*l + 1;
      if(pst == n + 1) {
        cout << 1 + st << " " << l + st << endl; return 0;
      }
      if( rev.hash( pst, n ) == nor.hash( 1, n - pst + 1 ) ) {
        cout << 1 + st << " " << n - l + st << endl; return 0;
      }
    }
  }
  //APA
  for ( int l = 1; l <= n / 2; ++ l ) {
    if(nor.hash(1, l) == nor.hash( n - l + 1, n ) ) {
      int pst = l + 1;

      if( nor.hash( pst, n - l ) == rev.hash( pst, n - l ) ) {
        cout << 1 + st << " " << l + st << endl; return 0;
      }
    }
  }
  cout << -1 << " " << -1 << endl;
  return 0;
}