题解 | #买卖股票的最好时机(四)#

```#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n , k;
    cin >> n >>k;
    int pl =2*k;
    vector<int>prices(n,0);
    for(int i = 0;i<n;i++)
        cin >> prices[i];
    vector<vector<int> >dp(n,vector<int>(pl+1,0));
    dp[0][0] = 0;
    for(int i =1;i<pl;i+=2)
        dp[0][i]=-prices[0];
    for(int i =1;i<n;i++)
        for(int j =0;j<pl;j+=2)
        {
            dp[i][j+1] = max(dp[i-1][j+1],dp[i-1][j]-prices[i]);
            dp[i][j+2] = max(dp[i-1][j+2],dp[i-1][j+1]+prices[i]);
        }
    if(dp[n-1][pl])cout<<dp[n-1][pl];
    else cout<<"0";
    return 0;
}