uva 10163 Storage Keepers （dp）

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

最小值最大，二分L,在L已知的情况下，dp求对应的Y即可。

#include<cstdio>
#include<map>
#include<queue>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<list>
#include<set>
#include<cmath>
using namespace std;
const int maxn = 100 + 5;
const int INF = 1e9;
const double eps = 1e-6;
typedef unsigned long long ULL;
typedef long long LL;
typedef pair<int, int> P;
#define fi first
#define se second

int p[maxn];
int dp[maxn][maxn];
int n, m;

int solve(int ability){
    if(ability==0)
        return 0;
    for(int i = 0;i <= m;i++){
        dp[i][0] = 0;
        for(int j = 1;j <= n;j++)
            dp[i][j] = INF;
    }
    for(int i = 1;i <= m;i++){
        int num = p[i]/ability;
        for(int j = 1;j <= num;j++)
            dp[i][j] = min(dp[i-1][j], p[i]);
        for(int j = num+1;j <= n;j++){
            dp[i][j] = min(dp[i-1][j], p[i]+dp[i-1][j-num]);
        }
    }
    return dp[m]
;
}

int main(){
    while(cin >> n >> m){
        if(n == 0 && m == 0)
            break;
        for(int i = 1;i <= m;i++)
            cin >> p[i];
        int ans = INF;
        int l = 0, r = 1000;
        while(l <= r){
            int mid = (l+r)/2;
            int tem = solve(mid);
            if(tem < INF){
                ans = tem;
                l = mid+1;
            }
            else
                r = mid-1;
        }
        cout << r << ' ' << ans << endl;
    }
    return 0;
}