关键就是每一天的买卖是完全独立的，在每一天尽量赚取最多的钱 因为没有后效性 所以可以使用$\Large\text{DP}$大法！

$$dp[k - a[i][j]] = max(dp[k - a[i][j]], dp[k] + a[i + 1][j] - a[i][j]); $$$$\ \ \ \ \ \ \ max(don't \ buy \ anything,原本的钱+明天赚的钱-今天的成本) $$

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;
int dp[10005], a[105][105];//dp[i]存手中剩i元时，全卖掉时剩下的钱

int main() {
    int t, n, m, awa;
    scanf("%d%d%d", &t, &n, &m);
    
    for (int i = 1; i <= t; ++i) {
        for (int j = 1; j <= n; ++j) {
            scanf("%d", &a[i][j]);
        }
    }
    
    awa = m;
    for (int i = 1; i < t; i++)//每一天都推，赚到最多的钱
      {
        
        memset(dp, ~0x3f, sizeof(dp));
       
        dp[awa] = awa;//初始剩awa块能拿awa块钱（因为你没买没卖啊）
        
        for (int j = 1; j <= n; j++) {
            
            for (int k = awa; k >= a[i][j]; --k) {
              
                dp[k - a[i][j]] = max(dp[k - a[i][j]], dp[k] + a[i + 1][j] - a[i][j]);
            }
        }//背包啊
        int ma = 0;
        for (int j = 0; j <= awa; ++j) {
            ma = max(ma, dp[j]);//所有情况找最大值；
        }
        awa = ma;
    }
    cout << awa << endl;
    return 0;
}