【题解】LuoGu5662：纪念品

原题传送门

• 枚举当前是第几天
• $d{p}_i$表示今天本金是$i$的情况下，今天买进，明天卖出后会获得最多的利润
• 就是说，转化了题意，把$p_{i+1,j}-p_{i,j}$当做一个物品
• 枚举商品，$d{p}_k=max(d{p}_k,d{p}_{i-p_{i,j}}+p_{i+1,j}-p_{i,j})(i为第几天，j为商品，k为本金)$
• 每天更新金币上限,$m=max(m,m+d{p}_m)$

Code:

#include 
#define maxn 10010
using namespace std;
int t, n, m, p[110][110], dp[maxn];

inline int read(){
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s=  (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}

int main(){
	t = read(), n = read(), m = read(); 
	for (int i = 1; i <= t; ++i)
		for (int j = 1; j <= n; ++j) p[i][j] = read();
	for (int i = 1; i < t; ++i){
		memset(dp, 0, sizeof(dp));
		for (int j = 1; j <= n; ++j)
			for (int k = p[i][j]; k <= m; ++k) dp[k] = max(dp[k], dp[k - p[i][j]] + p[i + 1][j] - p[i][j]);
		m = max(m, dp[m] + m);
	}
	printf("%d\n", m);
	return 0;
}