完全背包变形
题目链接

思路

每天手里的钱作为背包容量，商品今天的价格作为体积，明天的价格作为价值，设$dp[i]$为花费达到$i$元时最大的净利润

ACcode

#include<bits/stdc++.h>using namespace std;const int M = 2e4 + 9;
int dp[M], w[M][M];int main()
{ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);int t, n, m;cin >> t >> n >> m;for (int i = 1;i <= t;i++)for (int j = 1;j <= n;j++)cin >> w[j][i];for (int i = 1;i < t;i++) {for (int p = 0;p < M;p++)dp[p] = 0;for (int j = 1;j <= n;j++) {for (int k = w[j][i];k <= m;k++) {dp[k] = max(dp[k], dp[k - w[j][i]] + w[j][i + 1] - w[j][i]);}}m += dp[m];}cout << m;return 0;
}