程序師世界是廣大編程愛好者互助、分享、學習的平台,程序師世界有你更精彩!
首頁
編程語言
C語言|JAVA編程
Python編程
網頁編程
ASP編程|PHP編程
JSP編程
數據庫知識
MYSQL數據庫|SqlServer數據庫
Oracle數據庫|DB2數據庫
 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> UVA 10954- Add All(優先隊列)

UVA 10954- Add All(優先隊列)

編輯:C++入門知識

UVA 10954- Add All(優先隊列)


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

Description

Download as PDF

Problem F
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


Problem setter: Md. Kamruzzaman, EPS


題意:有n個數的集合s,每次可以從s中刪除兩個數,然後把他們的和放回集合,直至剩下一個數。每次的開銷等於刪除的兩個數的和,求最小總開銷。

思路:使用從小到大排序的優先隊列。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
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,greater > 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);
    }
}


  1. 上一頁:
  2. 下一頁:
Copyright © 程式師世界 All Rights Reserved