UVA 10954 Add All

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 of11. 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.

#include<iostream>
#include<queue>
using namespace std;
int main()
{
	int n;
	while(cin>>n)
	{
		if(n==0) return 0;
		priority_queue<int, vector<int>, greater<int> > q;//建立越小的数优先级越大的队列 
		int x;
		for(int i=0;i<n;i++)
		{
			cin>>x;
			q.push(x);//将元素x放入队列中 
		}
		long long sum=0;
		for(int i=0;i<n-1;i++)
		{
			int a=q.top();q.pop();//连续拿出两次拿出最小的数 
			int b=q.top();q.pop();
			sum=sum+a+b;
			q.push(a+b);//将a+b的和放入队列中 
		} 
		cout<<sum<<endl;
	}
}