完全背包

$f_k$ 表示当前花费不超过 $k$ 个金币能获得的最大额外价值。

对于第 $i$ 天的第 $j$ 件物品：

• 不买：$f_k$
• 买了第二天卖出：$f_{k-p_{i,j}} +p_{i+1,j}-p_{i,j}$

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 105, M = 1e4 + 5;
const int INF = 0x3f3f3f3f, MOD = 1E9 + 7;
int t, n, m, p[N][N], f[M];

int main(int argc, char* argv[]) {
    cin >> t >> n >> m;
    for (int i = 1; i <= t; i++)
        for (int j = 1; j <= n; j++)
            cin >> p[i][j];

    for (int i = 1; i < t; i++) {
        memset(f, 0x00, sizeof f);
        for (int j = 1; j <= n; j++)
            for (int k = p[i][j]; k <= m; k++)
                f[k] = max(f[k], f[k - p[i][j]] + p[i + 1][j] - p[i][j]);
        m += f[m];
    }
    cout << m;
    return 0;
}