Submission #988592

#	Time	Username	Problem	Language	Result	Execution time	Memory
988592		parlimoos	Palindromic Partitions (CEOI17_palindromic)	C++14	100 / 100	5611 ms	16384 KiB

palindromic

//Be Name KHODA
#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
#define pb push_back
#define pp pop_back
#define lb lower_bound
#define ub upper_bound
#define cl clear
#define bg begin
#define arr(x) array<int , x>
#define endl '\n'

vector<int> mods , s;
int inf[2][10] , t , n;

int Sum(int a , int b , int mod){
    int res = a + b;
    if(res > mod) return res - mod;
    if(res < 0) return res + mod;
    return res;
}
int Mul(int a , int b , int mod){
    return (1ll * a * 1ll * b) % mod;
}
int Pow(int a , int b , int mod){
    int res = 1;
    while(b){
        if(b & 1) res = Mul(res , a , mod);
        a = Mul(a , a , mod) , b >>= 1;
    }
    return res;
}
void initMods(){
    for(int i = int(1e9) ; i < int(1e9) + 10 ; i++) mods.pb(i);
}
arr(2) f(int l , int r){
    fill(&inf[0][0] , &inf[1][10] , 0);
    for(int i1 = l , i2 = r ;; i1++ , i2--){
        if(i1 >= i2) return {l , r};
        for(int i = 0 ; i < 10 ; i++){
            inf[0][i] = Mul(inf[0][i] , 26 , mods[i]);
            inf[0][i] = Sum(inf[0][i] , s[i1] % mods[i] , mods[i]);
            inf[1][i] = Sum(inf[1][i] , Mul(s[i2] , Pow(26 , r - i2 , mods[i]) , mods[i]) , mods[i]);
        }
        bool flg = 1;
        for(int i = 0 ; i < 10 ; i++) if(inf[0][i] != inf[1][i]) flg = 0;
        if(flg) return {i1 + 1 , i2 - 1};
    }
}

int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);

    initMods();
    cin >> t;
    for(int ii = 0 ; ii < t ; ii++){
        s.cl();
        string d;
        cin >> d;
        for(char c : d) s.pb(int(c - 'a'));
        int o = 0 , l = 0 , r = (int)s.size() - 1;
        while(true){
            if(l > r) break;
            arr(2) d = f(l , r);
            if(d[0] == l and d[1] == r){
                o++;
                break;
            }
            l = d[0] , r = d[1] , o += 2;
        }
        cout << o << endl;
    }
}

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