Add All
Input: standard input
Output: standard output

Yup!! The problem name reflects your task; just add a set of numbers. But you may feel yourselves condescended, to write a C/C++ program just to add a set of numbers. Such a problem will simply question your erudition. So, let’s add some flavor of ingenuity to it.

 

Addition operation requires cost now, and the cost is the summation of those two to be added. So, to add 1 and 10, you need a cost of 11. If you want to add 1, 2 and 3. There are several ways –

 


1 + 2 = 3, cost = 3

3 + 3 = 6, cost = 6

Total = 9

1 + 3 = 4, cost = 4

2 + 4 = 6, cost = 6

Total = 10

2 + 3 = 5, cost = 5

1 + 5 = 6, cost = 6

Total = 11


 

I hope you have understood already your mission, to add a set of integers so that the cost is minimal.

 

Input

Each test case will start with a positive number, N (2 ≤ N ≤ 5000) followed by N positive integers (all are less than 100000). Input is terminated by a case where the value of N is zero. This case should not be processed.

 

Output

For each case print the minimum total cost of addition in a single line.

 

Sample Input                           Output for Sample Input

3

1 2 3

4

1 2 3 4

0



9

19






把所有数加起来，每次的花费是两个数的和，所以每次都选最小的花费之和便最少。

由于n不大，暴力也是可以的，不过堆的操作要快的多。

这里用有限队列来写，代码量少了好多。

#include<iostream>  
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<string>
#include<cstring>
#include<cstdlib>
#include<queue>
#include<vector>
#include<functional>
using namespace std;
const int maxn = 5005;
priority_queue<int, vector<int>, greater<int>> q; // >>在较早的编译器里这个被强制认定为是位运算的符号，有必要加个空格在中间。
int n, a, sum, b;

int main(){
  while (cin >> n, n)
  {
    for (int i = 0; i < n; i++)
    {
      scanf("%d", &a);
      q.push(a);
    }
    sum = 0;
    while (q.size()>1)
    {
      a = q.top(); q.pop();
      b = q.top(); q.pop();
      sum += a + b;
      q.push(a + b);
    }
    printf("%d\n", sum);  q.pop();
  }
  return 0;
}