原題：
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.
樣例輸入:
3
1 2 3
4
1 2 3 4
0
樣例輸出：
9
19
題目大意:
有一些數字， 要把所有這些數字加起來求它們的總和。 沒執行一次加法時，這次加法的結果就是一個花費。可以以任意的數序進行加法，所以不同的順序花費不同的。求總花費最小是多少。
分析總結：
按照貪心的思路，為了讓總花費最少，那麼要使得每一次的花費最少，最終結果一定是最少的。
可以用優先隊列，這樣每次取出來用來相加的兩個數一定是隊列中最小的兩個數，花費也是最少的。
代碼：
[cpp]
/*
* UVa: 10954 - Add All
* Result: Accept
* Time: 0.024s
* Author: D_Double
*/
#include<cstdio>
#include<queue>
#define MAXN 5005
using namespace std;
priority_queue<long long, vector<long long>, greater<long long> >que;
int main(){
int N;
long long m;
while(scanf("%d",&N) && N){
while(!que.empty()) que.pop();
for(int i=0; i<N; ++i){
scanf("%lld",&m);
que.push(m);
}
long 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;
}
