Multiple optimal service order problems

Description

There are N Customers waiting for a service at the same time. The service time required by customer I is tI, 1 ≤ I ≤ n. A total of S can provide this service. How should we arrange the service order of N customers to minimize the average waiting time? The average waiting time is the total waiting time for N Customers divided by N.

Calculate the optimal service order for the given service time and S value required by N Customers.

Input

The first line of the input data has two positive integers n (n ≤ 10000) and S (S ≤ 1000), indicating that there are N Customers and S can provide services required by customers. There are n positive integers in the next line, indicating the service time required by N Customers.

Output

The output data has only one INTEGER (the calculation result is rounded to the nearest integer), indicating the minimum average wait time.

Sample Input

10 2 56 12 1 99 1000 234 33 55 99 812

Sample output

336

 Greedy policy: the shortest service time is preferred.

 #include <iostream>

 #include <algorithm>

 #include <cstring>

 #include <cstdio>

 using namespace std;

 int main()

 {

     int i,n,j,k,minx;

     int s;

     double t;

     int a[10005],b[1005];

     while(cin>>n>>s)

     {

         for(i=0;i<n;i++)

             cin>>a[i];

         sort(a,a+n);

         memset(b,0,sizeof(b));

         for(i=0;i<n;i++)

         {

             minx=0x7fffffff;

             k=0;

             for(j=0;j<s;j++)

             {

                 if(minx>b[j])

                 {

                     minx=b[j];

                     k=j;

                 }

             }

             b[k]+=a[i];

             a[i]=b[k];

         }

         t=0;

         for(i=0;i<n;i++)

             t+=a[i];

         t/=n;

         printf("%d\n",(int)(t+0.5));

     }

     return 0;

 }