Submission #990815

#	Time	Username	Problem	Language	Result	Execution time	Memory
990815		ChottuF	Palindromic Partitions (CEOI17_palindromic)	C++17	100 / 100	152 ms	30436 KiB

palindromic

  #include <bits/stdc++.h>
  using namespace std;

  #define int long long

  const int P = 31;
  const int M = 1e9+9;

  vector<int> binpow;
  vector<int> pref;

  int gethash(int l, int r){
      int h = ((pref[r+1] - (binpow[r-l+1] * pref[l]))%M + M)%M;
      return h;
  }

  signed main(){
      ios::sync_with_stdio(0);
      cin.tie(0);
      int t;
      cin >> t;
      while (t--){
          string s;
          cin >> s;
          int n = s.size();
          binpow.clear();
          binpow.resize(n);
          binpow[0] = 1;
          for (int i = 1; i<n; i++){
              binpow[i] = (binpow[i-1]*P) % M;
          }
          pref.clear();
          pref.resize(n+1);
          pref[0] = 0;
          for (int i = 1; i<=n; i++){
              pref[i] = ((pref[i-1]*P)+(s[i-1]-'a'+1)) % M;
          }
          int r = 0;
          int ln = 0;
          int ans = 0;
          vector<bool> vis(n);
          int lst = n-1;
          while (r+ln < n/2){
              int st = r; int stt = r+ln;
              int nd = lst-ln; int ndd = lst;
              int one = gethash(st,stt);
              int two = gethash(nd,ndd);
              //cout << st << " " << stt << "  " << nd << " " << ndd << "\n";
              //cout << one << " " << two << "\n\n";
              if (one == two){
                  for (int i = st; i<=stt; i++){
                      vis[i] = true;
                  }
                  for (int i = nd; i<=ndd; i++){
                      vis[i] = true;
                  }
                  ans += 2;
                  r += ln+1;
                  lst -= ln+1;
                  ln = 0;
              }
              else{
                  ln++;
              }
          }
          int bk = 0;
          for (int i = 0; i<n; i++){
              bk += vis[i];
          }
          if (bk != n) ans++;
          cout << ans << '\n';
      }
      return 0;
  }

• Subtask 1: 15 / 15
• Subtask 2: 20 / 20
• Subtask 3: 25 / 25
• Subtask 4: 40 / 40