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, ..., an.

Little Tommy is among them. He would like to choose an interval [l, r] (1 ≤ l ≤ r ≤ n), then reverse al, al + 1, ..., ar 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, ..., pk, such that p1 < p2 < ... < pk and ap1 ≤ ap2 ≤ ... ≤ apk. The length of the subsequence is k.

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, ..., an (1 ≤ ai ≤ 2, i = 1, 2, ..., n).

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

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.

 1 //C-用dp写

 2 #include<iostream>

 3 #include<cstring>

 4 #include<cstdio>

 5 #include<cmath>

 6 #include<algorithm>

 7 #include<cstdlib>

 8 using namespace std;

 9 const int maxn=2000+10;

10 int dp[maxn][maxn][2];

11 int p1[maxn],p2[maxn],a[maxn];

12 int main(){

13     int n;

14     scanf("%d",&n);

15     p1[0]=0;p2[n+1]=0;

16     for(int i=1;i<=n;i++){

17         scanf("%d",&a[i]);

18         p1[i]=p1[i-1];

19         if(a[i]==1)p1[i]++;

20     }

21     for(int j=n;j>=1;j--){

22         p2[j]=p2[j+1];

23         if(a[j]==2)p2[j]++;

24     }

25     int ans=-1;

26     for(int i=1;i<=n;i++){

27         for(int j=i;j<=n;j++){

28             if(a[j]==2)dp[i][j][1]=dp[i][j-1][1]+1;

29             else dp[i][j][1]=dp[i][j-1][1];

30             if(a[j]==1)dp[i][j][0]=max(dp[i][j-1][0],dp[i][j-1][1])+1;

31             else dp[i][j][0]=max(dp[i][j-1][0],dp[i][j-1][1]);

32             ans=max(p1[i-1]+p2[j+1]+dp[i][j][1],max(ans,p1[i-1]+p2[j+1]+dp[i][j][0]));

33         }

34     }

35     printf("%d\n",ans);

36 }