Ultraviolet A 10954-Add All

Original question:
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 shoshould not be processed.
 
Output
For each case print the minimum total cost of addition in a single line.
 
Sample input:
3
1 2 3
4
1 2 3 4
0

Sample output:
9
19

Question:
There are some numbers. We need to sum up all these numbers to calculate their sum. When an addition is not executed, the result of this addition is a cost. Addition can be performed in any number order, so the order is different. Calculate the minimum total cost.

Analysis summary:
According to the greedy idea, in order to minimize the total cost, the final result must be the least if each time is spent.
The priority queue can be used, so that the two numbers obtained each time for addition must be the smallest two in the queue, and the cost is also the least.

Code:
[Cpp]
/*
* Ultraviolet A: 10954-Add All
* Result: Accept
* Time: 0.024 s
* Author: D_Double
*/
# Include <cstdio>
# Include <queue>
# Define maxn5005
Using namespace std;
 
Priority_queue <long, vector <long>, greater <long> que;
 
Int main (){
Int N;
Long m;
 
While (scanf ("% d", & N ){
While (! Que. empty () que. pop ();
For (int I = 0; I <N; ++ I ){
Scanf ("% lld", & m );
Que. push (m );
}
Long sum = 0, ans = 0, a, B, t;
While (! Que. empty ()){
A = que. top ();
Que. pop ();
If (que. empty () break;
B = que. top ();
Que. pop ();
Sum = a + B;
Ans + = sum;
Que. push (sum );
}
Printf ("% lld \ n", ans );
}
Return 0;
}

Author: shuangde800