虽然知道有更加简便的做法，但还是要问为什么以下做法的正确性不成立：

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 2e3 + 10;
ll n, m, p[N][N];
ll a[N], dp[N][N];
int main () {
    cin >> n >> m;
    for (int i = 1; i <= n; ++ i) {
        cin >> a[i];
    }
    memset (p, 0, sizeof p);
    for (int i = 1; i <= n; ++ i) {
        for (int j = m; j >= 0; -- j) {
            if (j >= 1) {
                if (dp[i - 1][j] >= dp[i - 1][j - 1] + a[i] * (p[i - 1][j - 1] + 1)) {
                    dp[i][j] = dp[i - 1][j];
                    p[i][j] = p[i - 1][j];
                }
                else {
                    p[i][j] = p[i - 1][j - 1] + 1;
                    dp[i][j] = max (dp[i - 1][j], dp[i - 1][j - 1] + a[i] * p[i][j]);
                }
            }
            else {
                dp[i][j] = dp[i - 1][j];
                p[i][j] = p[i - 1][j];
            }
        }
    }
    cout << dp[n][m];
    return 0;
}

$\tt p$ 数组为记录当前状态下选了多少个，用来解决原序列和子序列下标不对应的问题。