Each minute, one of the two apple trees drops an apple. Bessie, having much practice, can catch an apple if she is standing under a tree from which one falls. While Bessie can walk between the two trees quickly (in much less than a minute), she can stand under only one tree at any time. Moreover, cows do not get a lot of exercise, so she is not willing to walk back and forth between the trees endlessly (and thus misses some apples).

Apples fall (one each minute) for T (1 <= T <= 1,000) minutes. Bessie is willing to walk back and forth at most W (1 <= W <= 30) times. Given which tree will drop an apple each minute, determine the maximum number of apples which Bessie can catch. Bessie starts at tree 1.

* Lines 2..T+1: 1 or 2: the tree that will drop an apple each minute.

Seven apples fall - one from tree 2, then two in a row from tree 1, then two in a row from tree 2, then two in a row from tree 1. Bessie is willing to walk from one tree to the other twice.

OUTPUT DETAILS:

Bessie can catch six apples by staying under tree 1 until the first two have dropped, then moving to tree 2 for the next two, then returning back to tree 1 for the final two.

　　还要预处理一下。

 #include <cstdio>

 #include <algorithm>

 using namespace std;

 int dp[][],a[];//dp[i][j]：在i时来回j次

 int inv(int m)

 {

     if(m==) return ;

     else return ;

 }

 int main()

 {

     int t,w,a[],maxnum=-;

     scanf("%d%d",&t,&w);

     for(int i=;i<=t;i++) scanf("%d",&a[i]);

     for(int i=;i<=t;i++)

     dp[i][]=dp[i-][]+a[i]%;

     for(int i=;i<=t;i++)

     for(int j=;j<=w;j++)

     {

         dp[i][j]=max(dp[i-][j-],dp[i-][j]);

         if(a[i]==j%+)

         dp[i][j]++;//拿住苹果

     }

     printf("%d",dp[t][w]);

     return ;

 }