Add All

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

Submit Status

Description

Problem F

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 –

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

Problem setter: Md. Kamruzzaman, EPS

题意：有n个数的集合s，每次可以从s中删除两个数，然后把他们的和放回集合，直至剩下一个数。每次的开销等于删除的两个数的和，求最小总开销。

思路：使用从小到大排序的优先队列。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <set>
#include <queue>
#include <stack>
#include <map>
using namespace std;
const int inf=0x3f3f3f3f;
int main()
{
    int n,i,a;
    int sum;
    while(~scanf("%d",&n)){
        if(n==0) break;
        priority_queue<int ,vector<int >,greater<int > > q;
        while(n--){
            scanf("%d",&a);
            q.push(a);
        }
        sum=0;
        while(!q.empty()){
            int x=q.top();
            q.pop();
            int y=q.top();
            q.pop();
            sum+=x+y;
            if(!q.empty())
            q.push(x+y);
        }
        printf("%d\n",sum);
    }
}