Submission #43235

#	Time	Username	Problem	Language	Result	Execution time	Memory
43235		krauch	Palindromic Partitions (CEOI17_palindromic)	C++14	0 / 100	1 ms	376 KiB

palindromic

/*
 _    _    _______   _    _
| |  / /  |  _____| | |  / /
| | / /   | |       | | / /
| |/ /    | |_____  | |/ /
| |\ \    |  _____| | |\ \
| | \ \   | |       | | \ \
| |  \ \  | |_____  | |  \ \
|_|   \_\ |_______| |_|   \_\

*/
#include <bits/stdc++.h>

using namespace std;

typedef unsigned long long ull;
typedef long long ll;
typedef double ld;
typedef pair <int, int> PII;
typedef pair <ll, ll> PLL;
typedef pair < ll, int > PLI;


#define F first
#define S second
#define pb push_back
#define eb emplace_back
#define right(x) x << 1 | 1
#define left(x) x << 1
#define forn(x, a, b) for (int x = a; x <= b; ++x)
#define for1(x, a, b) for (int x = a; x >= b; --x)
#define mkp make_pair
#define sz(a) (int)a.size()
#define all(a) a.begin(), a.end()
#define y1 kekekek

#define fname ""

const ll ool = 1e18 + 9;
const int oo = 1e9 + 9, base = 1e9 + 7;
const ld eps = 1e-7;
const int N = 2e6 + 6;

int d[N], mn[N];
char a[N];

bool check(int l1, int r1, int l2, int r2) {
    forn(i, l1, r1) {
        if (a[i] != a[l2 + i - l1]) return 0;
    }
    return 1;
}

int main() {
	ios_base :: sync_with_stdio(0), cin.tie(0), cout.tie(0);

	#ifdef krauch
        freopen("input.txt", "r", stdin);
    #else
        //freopen(fname".in", "r", stdin);
        //freopen(fname".out", "w", stdout);
    #endif

    int T;
    cin >> T;
    while (T--) {
        string s;
        cin >> s;
        int n = sz(s);
        forn(i, 1, n) {
            a[i] = s[i - 1];
        }

        if (n & 1) d[n / 2 + 1] = 1;
        else d[n / 2 + 1] = 0;
        mn[n / 2 + 1] = 0;
        for1(i, n / 2, 1) {
            mn[i] = 0;
            forn(j, 1, (mn[i + 1] ? mn[i + 1] + 1 : n / 2 - i + 1)) {
                if (check(i, i + j - 1, n - i - j + 2, n - i + 1)) {
                    mn[i] = j;
                    break;
                }
            }
            if (mn[i]) d[i] = d[i + mn[i]] + 2;
            else d[i] = 1;
        }
        cout << d[1] << "\n";
    }

	return 0;
}

• Subtask 1: 0 / 15
• Subtask 2: 0 / 20
• Subtask 3: 0 / 25
• Subtask 4: 0 / 40