先上題

此題的關鍵在於要使剩下5元去買到最貴的菜，則使m-=5，之後就是01揹包的裸題了

#include<iostream>
using namespace std;

#include<algorithm>

int main()
{
	int n;
	while (scanf("%d", &n) != EOF, n)
	{
		int price[1010] = { 0 };
		int dp[1010] = { 0 };
		for (int i = 1; i <= n; i++)
		{
			scanf("%d", &price[i]);
		}
		int m;
		scanf("%d", &m);
		if (m < 5)
		{
			cout << m << endl;
			continue;
		}
		sort(price + 1, price + 1 + n);
		int Maxn = price[n];
		m -= 5;
		for (int i = 1; i < n; i++)
		{
			for (int j = m; j >= price[i]; j--)
			{
				dp[j] = max(dp[j], dp[j - price[i]] + price[i]);
			}
		}
		cout << m + 5 - dp[m] - Maxn << endl;
	}
	return 0;
}