uva 10954 - Add All

Problem F

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

想起了noip的合并果子，那时候还年轻，只会快排+插排每次找最小的2个，然后知道了可以用堆去实现维护一个优先队列，现在终于会用stl偷懒了，优先级比较函数比较难记，老忘记每次都要百度出来；

#include<stdio.h>
#include<queue>
using namespace std;
struct cmp
{
bool operator() (const int &a, const int &b)
{return a>b;}
};
priority_queue<int,vector<int>,cmp> q;
int main()
{
int n,i,x,y;
while (scanf("%d",&n),n)
{
for (i=1;i<=n;i++)
{
scanf("%d",&x);
q.push(x);
}
x=0;
while (q.size()>1)
{
y=q.top();
q.pop();
y+=q.top();
q.pop();
x+=y;
q.push(y);
}
q.pop();
printf("%d\n",x);
}
return 0;
}