ZOJ-2156

仍然是多重背包DP，我的算法基本和上题一模一样，但是本题物品比较多，需要组合一下，然后01背包来搞，注意组合物品和DP过程中物品数的累加不要惯性地用加一来DP了。这个在算法在时间上不是最优的，背包九讲上有O(VN)的算法，但我看不大懂，一知半解，就不想去重写了，只好作罢。下次有碰到的话再说吧，我提交的执行时间是150ms，看到排第一的有0ms的，估计用了O(VN)的算法

#include<stdio.h>
#include<string.h>

static int min(int a, int b)
{
	return a < b ? a : b;
}

static int pack[50], amount[50], tt;

static void fill_pack(int total, int cents)
{
	int sum = 0, k = 1;
	while (1)
	{
		if (sum + k >= total)
			break;
		sum += k;
		pack[tt] = k * cents;
		amount[tt++] = k;
		k *= 2;
	}
	if (total - sum)
	{
		pack[tt] = (total - sum) * cents;
		amount[tt++] = total - sum;
	}
}

int main()
{
	int p, c1, c2, c3, c4, dp[10001], path[10001], res[26];
	while (scanf("%d %d %d %d %d", &p, &c1, &c2, &c3, &c4), p || c1 || c2 || c3
			|| c4)
	{
		c1 = min(p, c1), c2 = min(p / 5, c2), c3 = min(p / 10, c3), c4 = min(
				p / 25, c4);
		tt = 0;
		fill_pack(c1, 1);
		fill_pack(c2, 5);
		fill_pack(c3, 10);
		fill_pack(c4, 25);
		int i, j;
		dp[0] = 0;
		for (i = 1; i <= p; i++)
			dp[i] = -1;
		for (i = 0; i < tt; i++)
			for (j = p; j >= pack[i]; j--)
				if (dp[j - pack[i]] != -1
						&& dp[j - pack[i]] + amount[i] > dp[j])
				{
					dp[j] = dp[j - pack[i]] + amount[i];
					path[j] = i;
				}
		if (dp[p] != -1)
		{
			memset(res, 0, sizeof(res));
			while (p)
			{
				int ii = path[p];
				res[pack[ii] / amount[ii]] += amount[ii];
				p -= pack[ii];
			}
			printf(
					"Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n",
					res[1], res[5], res[10], res[25]);
		}
		else
			puts("Charlie cannot buy coffee.");
	}
	return 0;
}