Problem C.Storage Keepers 

Background

   Randy Company has N (1<=N<=100) storages. Company wants some men to keep them safe. Now there are M (1<=M<=30) men asking for the job. Company will choose several from them. Randy Company employs men following these rules:

1.       Each keeper has a number Pi (1<=Pi<=1000) , which stands for their ability.

2.       All storages are the same as each other.

3.       A storage can only be lookd after by one keeper. But a keeper can look after several storages. If a keeper’s ability number is Pi, and he looks after K storages, each storage that he looks after has a safe number Uj=Pi div K.(Note: Uj, Pi and K are all integers). The storage which is looked after by nobody will get a number 0.

4.       If all the storages is at least given to a man, company will get a safe line L=min Uj

5.       Every month Randy Company will give each employed keeper a wage according to his ability number. That means, if a keeper’s ability number is Pi, he will get Pi dollars every month. The total money company will pay the keepers every month is Y dollars.

  Now Randy Company gives you a list that contains all information about N,M,P, your task is give company a best choice of the keepers to make the company pay the least money under the condition that the safe line L is the highest.

 

Input

The input file contains several scenarios. Each of them consists of 2 lines:

  The first line consists of two numbers (N and M), the second line consists of M numbers, meaning Pi (I=1..M). There is only one space between two border numbers.

  The input file is ended with N=0 and M=0.

 

Output

  For each scenario, print a line containing two numbers L(max) and Y(min). There should be a space between them.

 

Sample Input

2 1

7

1 2

10 9

2 5

10 8 6 4 1

5 4

1 1 1 1

0 0

 

Sample Output

3 7

10 10

8 18

0 0

题意：有m个仓库， n个小伙伴，每个小伙伴有个能力值p，要这些小伙伴去守护仓库，每个小伙伴的雇佣金是p，每个小伙伴看守的仓库安全值为p/k(每个小伙伴看守仓库数)。仓库的安全值为所有仓库中，安全值最小的仓库的安全值。

要求出最大安全值和最大安全值下的最小开销。

思路： 背包， 首先是第一个问题，我们把每个小伙伴看成物品，要看守的仓库数看成背包容量，每个小伙伴看守的仓库数为k，价值为p[i]/k。 状态转移方程为dp[j] = max(dp[j], min(dp[j - k], p[i]/k).。

然后是第二个问题。在第一个问题求出的最大安全值maxx下，求最小价值，依然是背包，k表示每个小伙伴看守的仓库数，状态转移方程为dp[j] = min(dp[j], dp[j - k] + p[i]);

代码：

 

#include 
    
     #include 
     
      const int INF = 1 << 30;int n, m, p[105], i, j, k, dp[1005], maxx, minn;int max(int a, int b) {	return a > b ? a : b;}int min(int a, int b) {	return a < b ? a : b;}int main() {	while (~scanf("%d%d", &m, &n) && m || n) {		memset(dp, 0, sizeof(dp));		dp[0] = INF;		for (i = 0; i < n; i ++)			scanf("%d", &p[i]);		for (i = 0; i < n; i ++) {			for (j = m; j >= 0; j --) {				for (k = 1; k <= p[i] && k <= j; k ++) {					dp[j] = max(dp[j], min(dp[j - k], p[i] / k));				}			}		}		maxx = dp[m];		if (maxx == 0) {			printf("0 0\n");			continue;		}		for (i = 1; i <= m; i ++) 			dp[i] = INF;		dp[0] = 0;		for (i = 0; i < n; i ++)			for (j = m; j >= 0; j --)				for (k = min(j, p[i]/maxx); k > 0; k --) {					dp[j] = min(dp[j], dp[j - k] + p[i]);				}		printf("%d %d\n", maxx, dp[m]);	}	return 0;}