A Twisty Movement_dp

最新推荐文章于 2020-01-29 00:55:08 发布

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本文介绍了一个关于龙舞表演的问题：如何通过反转表演者队列的一部分，使得整个表演序列的最长非递减子序列尽可能长。文章提供了一段C++代码实现，通过动态规划的方法求解最优反转区间。

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A dragon symbolizes wisdom, power and wealth. On Lunar New Year's Day, people model a dragon with bamboo strips and clothes, raise them with rods, and hold the rods high and low to resemble a flying dragon.

A performer holding the rod low is represented by a 1, while one holding it high is represented by a 2. Thus, the line of performers can be represented by a sequence a₁, a₂, ..., a_n.

Little Tommy is among them. He would like to choose an interval [l, r] (1 ≤ l ≤ r ≤ n), then reverse a_l, a_l + 1, ..., a_r so that the length of the longest non-decreasing subsequence of the new sequence is maximum.

A non-decreasing subsequence is a sequence of indices p₁, p₂, ..., p_k, such that p₁ < p₂ < ... < p_k and a_p₁ ≤ a_p₂ ≤ ... ≤ a_{p_k}. The length of the subsequence is k.

Input

The first line contains an integer n (1 ≤ n ≤ 2000), denoting the length of the original sequence.

The second line contains n space-separated integers, describing the original sequence a₁, a₂, ..., a_n (1 ≤ a_i ≤ 2, i = 1, 2, ..., n).

Output

Print a single integer, which means the maximum possible length of the longest non-decreasing subsequence of the new sequence.

Examples

Input

Copy

4
1 2 1 2

Output

Copy

Input

Copy

10
1 1 2 2 2 1 1 2 2 1

Output

Copy

Note

In the first example, after reversing [2, 3], the array will become [1, 1, 2, 2], where the length of the longest non-decreasing subsequence is 4.

In the second example, after reversing [3, 7], the array will become [1, 1, 1, 1, 2, 2, 2, 2, 2, 1], where the length of the longest non-decreasing subsequence is 9.

#include<bits/stdc++.h>
#define L long long
using namespace std;
int n,x[2010],f[2010][5],a[5]={0,1,2,1,2};
int main()
{
//	freopen("in.txt","r",stdin);
	//freopen(".out","w",stdout);
	int i,j;
	scanf("%d",&n);
	for(i=1;i<=n;i++)
	  scanf("%d",&x[i]);
	for(i=1;i<=n+1;i++)
	  for(j=1;j<=4;j++){
        f[i][j]=max(f[i-1][j],f[i-1][j-1])+(x[i]==a[j]);
	  }
	printf("%d\n",f[n+1][4]);
	return 0;
}