一个dp。
/*
POJ: 3661 RUNNING
*/
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MaxN = 10005;
const int MaxM = 505;
int dp[MaxN][MaxM];
int n, m;
int dis[MaxN];
void solve()
{
memset(dp, 0, sizeof(dp));
dp[1][1] = dis[1];
for(int i = 2; i <= n; i++) {
for(int j = 1; i - j >= 0 && j <= m; j++)
dp[i][j] = dp[i-1][j-1] + dis[i];
dp[i][0] = dp[i-1][0];
for(int k = 1; i - k >= k && k <= m; k++)
dp[i][0] = max(dp[i][0], dp[i-k][k]);
}
}
int main()
{
//freopen("data.in", "rb", stdin);
while(scanf("%d%d", &n, &m) != EOF) {
for(int i = 1; i <= n; i++) {
scanf("%d", &dis[i]);
}
solve();
printf("%d\n", dp[n][0]);
}
return 0;
}