Submission #1307420

#	Time	Username	Problem	Language	Result	Execution time	Memory
1307420		Nonoze	Palindromic Partitions (CEOI17_palindromic)	C++20	100 / 100	35 ms	3200 KiB

palindromic

/*
*	Author: Nonoze
*	Created: Friday 02/01/2026
*/
#include <bits/stdc++.h>
using namespace std;

#ifndef DEBUG
	#define dbg(...)
#endif

// #define cout cerr << "OUT: "
#define endl '\n'
#define endlfl '\n' << flush
#define quit(x) return (void)(cout << x << endl)

template<typename T> void read(T& x) { cin >> x; }
template<typename T1, typename T2> void read(pair<T1, T2>& p) { read(p.first), read(p.second); }
template<typename T> void read(vector<T>& v) { for (auto& x : v) read(x); }
template<typename T1, typename T2> void read(T1& x, T2& y) { read(x), read(y); }
template<typename T1, typename T2, typename T3> void read(T1& x, T2& y, T3& z) { read(x), read(y), read(z); }
template<typename T1, typename T2, typename T3, typename T4> void read(T1& x, T2& y, T3& z, T4& zz) { read(x), read(y), read(z), read(zz); }
template<typename T> void print(vector<T>& v) { for (auto& x : v) cout << x << ' '; cout << endl; }

#define sz(x) (int)(x.size())
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define make_unique(v) sort(all(v)), v.erase(unique(all(v)), (v).end())
#define pb push_back
#define mp(a, b) make_pair(a, b)
#define fi first
#define se second
#define cmin(a, b) a = min(a, b)
#define cmax(a, b) a = max(a, b)
#define YES cout << "YES" << endl
#define NO cout << "NO" << endl
#define QYES quit("YES")
#define QNO quit("NO")

#define int long long
#define double long double
const int inf = numeric_limits<int>::max() / 4;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int MOD = 1e9+7, LOG=20;



void solve();

signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	int tt=1;
	cin >> tt;
	while(tt--) solve();
	return 0;
}




int n, k, m, q;
vector<int> a;

const int b=31;


void solve() {
	string s; read(s); n=sz(s);
	int cnt=0;
	int hash1=0, hash2=0, p=1;
	for (int i=0; i<n/2; i++) {
		hash1=(hash1*b + s[i]) % MOD;
		hash2=(hash2 + s[n-i-1]*p) % MOD;
		p=(p*b) % MOD;
		if (hash1==hash2) {
			cnt++;
			hash1=0, hash2=0, p=1;
		}
	}
	cnt*=2;
	if (p!=1 || n%2) cnt++;
	cout << cnt << endl;
}

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