UVA 10954 Add All(优先队列)

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

输入的整数按从小到大的顺序，依次相加，在把它们的和放到优先队列里，把中途产生的所有和加起来即为最后结果

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stdlib.h>
#include <cmath>
#include <queue>
#define  eps 1e-8
#define N 1000010
using namespace std;
int main()
{
     int n;
     while(scanf("%d",&n),n)
     {
         priority_queue<int, vector<int>, greater<int> > q;
         int x;
         for(int i=0;i<n;i++)
         {
             scanf("%d",&x);
             q.push(x);
         }
         int ans=0;
         while(!q.empty())
         {
             int x1=q.top();
             q.pop();
             int x2=q.top();
             q.pop();
             ans+=x1+x2;
             if(!q.empty())
             q.push(x1+x2);
         }
         printf("%d\n",ans);

     }
    return 0;
}