Submission #377861

#	Time	Username	Problem	Language	Result	Execution time	Memory
377861		SavicS	Savez (COCI15_savez)	C++14	120 / 120	696 ms	39532 KiB

savez

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>

#define fi first
#define se second
#define pb push_back
#define sz(a) (int)a.size()
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define ff(i,a,b) for(int i=a;i<=b;i++)
#define fb(i,b,a) for(int i=b;i>=a;i--)

using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
const int maxn = 2000005;
const int inf = 1e9 + 5;

template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

// os.order_of_key(k) the number of elements in the os less than k
// *os.find_by_order(k)  print the k-th smallest number in os(0-based)

int n;

ll dp[maxn];

const ll mod = 1e9 + 9;
ll add(ll a, ll b){
    return (a + b) % mod;
}
ll sub(ll a, ll b){
    return (a - b + mod) % mod;
}
ll mul(ll a, ll b){
    return (a * b) % mod;
}
ll power(ll a, ll b){
    if(!b)return 1;
    ll pola = power(a, b / 2);
    pola = mul(pola, pola);
    if(b % 2)pola = mul(pola, a);
    return pola;
}
ll inv(ll a){
    return power(a, mod - 2);
}

const int p = 67;
ll pw[maxn];
ll invz[maxn];
ll hsh[maxn];

ll get(int l, int r){
    if(l > r)return 0;
    return mul(sub(hsh[r], hsh[l - 1]), invz[l - 1]);
}

map<ll,ll> kol;

int main()
{
   	ios::sync_with_stdio(false);
   	cout.tie(nullptr);
  	cin.tie(nullptr);
	cin >> n;
	pw[0] = 1;
	ff(i,1,maxn - 1)pw[i] = mul(pw[i - 1], p);
	invz[0] = 1;
    ff(i,1,maxn - 1)invz[i] = mul(invz[i - 1], inv(p));
	
	ff(i,1,n){
		string s;
		cin >> s;
		// ja bi trebao da hasujem i u nekoj mapi da cuvaj za vrednost hasha max dp do njega
		int len = sz(s);
		s = ' ' + s;
		ff(j,1,len)hsh[j] = add(hsh[j - 1], mul(pw[j - 1], (s[j] - 'A' + 1))); 
		dp[i] = 1;
		ff(j,1,len){
			ll a = hsh[j];
			ll b = get(len - j + 1, len);
			if(a == b)dp[i] = max(dp[i], kol[a] + 1);
		}
		kol[hsh[len]] = max(kol[hsh[len]], dp[i]);
	}
	ll rez = 0;
	ff(i,1,n)rez = max(rez, dp[i]);
	cout << rez << endl;
   	return 0;
}
/**

5
A
B
AA
BBB
AAA


// probati bojenje sahovski ili slicno

**/

• Subtask 1: 12 / 12
• Subtask 2: 12 / 12
• Subtask 3: 12 / 12
• Subtask 4: 12 / 12
• Subtask 5: 12 / 12
• Subtask 6: 12 / 12
• Subtask 7: 12 / 12
• Subtask 8: 12 / 12
• Subtask 9: 12 / 12
• Subtask 10: 12 / 12