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

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