Submission #268131

#	Time	Username	Problem	Language	Result	Execution time	Memory
268131		kartel	Palinilap (COI16_palinilap)	C++14	17 / 100	1096 ms	1816 KiB

palinilap

#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define in(x) freopen(x, "r", stdin)
#define out(x) freopen(x, "w", stdout)
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("fast-math")
#pragma GCC optimize("no-stack-protector")
#define F first
#define S second
#define pb push_back
#define N +300500
#define M ll(1e9 + 7)
#define sz(x) (int)x.size()
#define re return
#define oo ll(1e9)
#define el '\n'
#define Max_A int(1e9)
//#define el endl
#define pii pair <int, int>
#define piii pair <int, pair <int, int> >
#define psi pair <string, int>
#define err ld(1e-9)
#define Max_S int(3e6)
#define last(x) x.back()
#define all(x) (x).begin(), (x).end()
#define allr(x) (x).rbegin(), (x).rend()
#define arr_all(x, n) (x + 1), (x + 1 + n)
#define vi vector<int>
using namespace std;
//using namespace __gnu_pbds;
//typedef tree <int, null_type, less_equal <int> , rb_tree_tag, tree_order_statistics_node_update> ordered_set;
typedef long long ll;
typedef long double ld;

ll ans, cur;
string s;
int i, n;
char c;

int main()
{
//    srand(time(0));
    cout.precision(3);
    cout << fixed;
    ios_base::sync_with_stdio(0);
    iostream::sync_with_stdio(0);
    ios::sync_with_stdio(0);
    cin.tie(NULL);
    cout.tie(NULL);

//    in("input.txt");
//    out("output.txt");

    cin >> s;

    n = sz(s);

    for (i = 0; i < n; i++)
    {
        char old_c = s[i];

        for (c = 'a'; c <= 'z'; c++)
        {
            cur = 0;
            s[i] = c;

            vi d1(n);

            int l = 0, r = -1;

            for (int j = 0; j < n; j++)
            {
               int k = ((j > r) ? 1 : min(d1[l + r - j],
                                           r - j + 1));

               while (j+k < n && j-k >= 0 && s[j+k] == s[j-k]) k++;

               d1[j] = k;

               if (j + k - 1 > r)
                 {
                     l = j - k + 1;
                     r = j + k - 1;
                 }
            }

            vi d2(n);

            l = 0;
            r = -1;

            for (int j = 0; j < n; j++)
            {
               int k = ((j > r) ? 0 : min (d2[l + r - j + 1],
                                           r - j + 1));

               while (j+k < n && j-k-1 >= 0 && s[j+k] == s[j-k-1])  ++k;

               d2[j] = k;

               if (j + k - 1 > r)
                 {
                     l = j - k;
                     r = j + k - 1;
                 }
            }

            for (int j = 0; j < n; j++)
                cur += d1[j] + d2[j];

            ans = max(ans, cur);

            s[i] = old_c;
        }
    }

    cout << ans;
}

• Subtask 1: 17 / 17
• Subtask 2: 0 / 37
• Subtask 3: 0 / 46