D - Apple Catching

Description

It is a little known fact that cows love apples. Farmer John has two apple trees (which are conveniently numbered 1 and 2) in his field, each full of apples. Bessie cannot reach the apples when they are on the tree, so she must wait for them to fall. However, she must catch them in the air since the apples bruise when they hit the ground (and no one wants to eat bruised apples). Bessie is a quick eater, so an apple she does catch is eaten in just a few seconds. 

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.

Input

* Line 1: Two space separated integers: T and W 

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

Output

* Line 1: The maximum number of apples Bessie can catch without walking more than W times.

Sample Input

7 22112211

Sample Output

6

Hint

INPUT DETAILS: 

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. #include <stdio.h>
#include <stdlib.h>
#include <string.h>
int arr[1005];
int dp[1005][35];
int main()
{
int t,w;
while (scanf("%d %d", &t, &w) != EOF)
{
memset(dp, 0, sizeof(dp));
for (int i = 1; i <= t; ++ i )
{
scanf("%d", &arr[i]);
}
//初始化，开始在第一颗树下
if (arr[1] == 1)
{
dp[1][0] = 1;
dp[1][1] = 0;
}
if (arr[1] == 2)
{
dp[1][1] = 1;
dp[1][0] = 0;//  如果第一个苹果是从2位置落的，那么必须动一下才可以，
}


for (int i = 2; i <= t; ++ i)
{
for (int j = 0; j <= w; ++ j)
{
//初始化
if (j == 0)
{
dp[i][j] = dp[i - 1][j] + arr[i] % 2;//上一秒得到的，是上一秒的累加和这一秒是否得到的，j=0则1可以2不可以
continue;
}
//dp状态方程
dp[i][j] = dp[i - 1][j] > dp[i - 1][j - 1] ? dp[i - 1][j] : dp[i - 1][j - 1];
//如果本次是在第i颗树下，就会多收获一个苹果
if (j % 2 + 1 == arr[i])
{
dp[i][j] ++;//若动了基数次，那么只能拿第二个，否则可以拿第一个。
}
}
}
//找最大值
int ans = dp[t][0];
for (int i = 0; i <= w ; ++ i)
{
if (ans < dp[t][i])
{
ans = dp[t][i];
}
}
printf("%d\n", ans);
}
return 0;
}

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.