Submission #420450

#	Time	Username	Problem	Language	Result	Execution time	Memory
420450		MetalPower	Cigle (COI21_cigle)	C++14	0 / 100	4 ms	2764 KiB

cigle

#include <bits/stdc++.h>
using namespace std;

const int MX = 5005;

void chmax(int& a, int b){
	if(b > a) a = b;
}

int N, arr[MX], dp[MX][MX];

// Observation #1 : We can easily solve this in O(N^3)
	// dp(i, j) is the largest beauty of the wall if the highest row includes [i, j]
	// the transition is we add another row [j + 1, k] for every j + 1 <= k <= N
	// calculating is easy two pointer

// Observation #2 : if the size of row[j + 1, k] is higher than row[i, j]
	// then for every other k >= current_k 
	// the value of the dp is constant
	// this is also correct for the opposite
	
// By using two pointer we can find the positions where the number changes
	// We update those positions
	// and in the end max each positions with the previous one

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

	cin >> N;

	for(int i = 0; i < N; i++) cin >> arr[i];

	for(int p = 0; p < N; p++){
		int i = p, k = p + 1, cnt = -1, bot = 0, top = 0;

		while(i >= 0 && k < N){
			if(bot == top){
				cnt++; chmax(dp[p + 1][k], dp[i][p] + cnt);
				bot += arr[i--]; top += arr[k++];
			}else if(bot < top){
				bot += arr[i--];
			}else if(top < bot){
				top += arr[k++];
			}
		}

		for(int j = p + 1; j < N; j++) chmax(dp[p + 1][j], dp[p + 1][j - 1]);
	}

	int ans = 0;

	for(int p = 0; p < N; p++) chmax(ans, dp[p][N - 1]);

	cout << ans << '\n';
}

• Subtask 1: 0 / 9
• Subtask 2: 0 / 11
• Subtask 3: 0 / 13
• Subtask 4: 0 / 15
• Subtask 5: 0 / 52