A Mini locomotive (Dynamic planning 01)

 /*  Test Instructions: Select the number of 3 consecutive numbers   The interval of K to make their and maximum analysis:  dp[j][i]=max (Dp[j-k][i-1]+value[j],dp[j-1][i]);  Dp[j][i]: The maximum and value[j of I continuous intervals   values are selected from the J Number: the number in the J interval is a bit like the backpack, but the  */  #include <cstdio> #include <cstring> #include <iostream>using namespace Std;int dp[50005][4]; int main () {    int t;     scanf ("%d", &t);    while (t--)     {           int n;        scanf ("%d", &n);        int a[n+1],sum[n+1];        Memset (Dp,0,sizeof (DP));        memset (A,0,sizeof (a));        memset (sum , 0,sizeof (sum));         for (int i=1; i<=n; i++)         {    &NB Sp       scanf ("%d", &a[i]);            sum[i]=sum[i-1]+a[i];//prefix and, For the number of consecutive K        }     &NBSp     int k=0;        scanf ("%d", &k);        int value[n+1];  &NB Sp     memset (value,0,sizeof (value));        for (int i=1; i<=k; i++)       &NB Sp     value[i]=value[i-1]+a[i];        for (int i=k+1; i<=n; i++)             value[i]=sum[i]-sum[i-k];//consecutive K numbers and Value[i] represent the interval length k for the interval            for (int j=k; j<=n; j + +)             for (int i=1; i<=3; i++)                 Dp[j][i]=max (dp[j-k][i-1]+value[j],dp[j-1][i]);//Select the I interval from the J number, It is equivalent to the number of previous (J-k) selected (I-1) The basis of the interval (value[j]), if not selected is equivalent to the number of (j-1) Select I range                printf ("%d\n", Dp[n][3]);      }     return 0;}

A Mini locomotive (Dynamic planning)