oj.uz

답안 #377860

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
377860 2021-03-15T11:08:27 Z SavicS Savez (COCI15_savez) C++14
108 / 120
 711 ms 39532 KB 
#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 = 133;
ll pw[maxn];
ll invz[maxn];
ll hsh[maxn];

ll get(int l, int r){
    if(l == r)return 1;
    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);
		}
		if(kol.count(hsh[len]))kol[hsh[len]] = max(kol[hsh[len]], dp[i]);
		else 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
12.0 / 12.0

# 결과 실행 시간 메모리 Grader output
1 Correct 580 ms 31980 KB Output is correct
2 Correct 582 ms 31668 KB Output is correct
3 Correct 581 ms 31640 KB Output is correct

Subtask #2
0 / 12.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 579 ms 31856 KB Output isn't correct
2 Halted 0 ms 0 KB -

Subtask #3
12.0 / 12.0

# 결과 실행 시간 메모리 Grader output
1 Correct 659 ms 32092 KB Output is correct
2 Correct 665 ms 31980 KB Output is correct
3 Correct 670 ms 31932 KB Output is correct

Subtask #4
12.0 / 12.0

# 결과 실행 시간 메모리 Grader output
1 Correct 584 ms 31852 KB Output is correct
2 Correct 602 ms 31980 KB Output is correct
3 Correct 624 ms 31980 KB Output is correct

Subtask #5
12.0 / 12.0

# 결과 실행 시간 메모리 Grader output
1 Correct 603 ms 32068 KB Output is correct
2 Correct 600 ms 31852 KB Output is correct

Subtask #6
12.0 / 12.0

# 결과 실행 시간 메모리 Grader output
1 Correct 605 ms 31852 KB Output is correct
2 Correct 603 ms 31968 KB Output is correct

Subtask #7
12.0 / 12.0

# 결과 실행 시간 메모리 Grader output
1 Correct 614 ms 32172 KB Output is correct
2 Correct 600 ms 32108 KB Output is correct

Subtask #8
12.0 / 12.0

# 결과 실행 시간 메모리 Grader output
1 Correct 613 ms 32236 KB Output is correct
2 Correct 619 ms 32412 KB Output is correct

Subtask #9
12.0 / 12.0

# 결과 실행 시간 메모리 Grader output
1 Correct 627 ms 33568 KB Output is correct
2 Correct 623 ms 33516 KB Output is correct

Subtask #10
12.0 / 12.0

# 결과 실행 시간 메모리 Grader output
1 Correct 651 ms 34952 KB Output is correct
2 Correct 650 ms 34924 KB Output is correct
3 Correct 711 ms 39532 KB Output is correct