Add All

Add All

 Time limit3000 ms

 OSLinux

 

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

3

1 2 3

4

1 2 3 4

0

Sample Output

9

19

 

 

大意：给一串数字，求出符合题意的最大值。

思路：给一个数组，每次排序后把最小的加在后一个，这样最小的加最多次。

 

 

代码：

#include <iostream>
#include <cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
int main(){
　　int n;
　　while(scanf("%d",&n)!=EOF){
　　　　if(n==0) return 0;
　　　　int a[n];
　　　　for(int i=0;i<n;i++){
　　　　　　cin>>a[i];
　　　　}
　　　　int x=0,sum=0,y=0;
　　　　while(x<n-1){
　　　　　　sort(a+x,a+n);
　　　　　　a[x+1]+=a[x];
　　　　　　sum+=a[x+1];
　　　　　　x++;
　　　　}
　　　　cout<<sum<<endl;
　　}
　　return 0;
}

 

转载于:https://www.cnblogs.com/yiyunwang/p/9387806.html