Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needs N (1 ≤ N ≤ 20,000) planks of wood, each having some integer length L_i (1 ≤ L_i ≤ 50,000) units. He then purchases a single long board just long enough to saw into the N planks (i.e., whose length is the sum of the lengths L_i). FJ is ignoring the "kerf", the extra length lost to sawdust when a sawcut is made; you should ignore it, too.

FJ sadly realizes that he doesn‘t own a saw with which to cut the wood, so he mosies over to Farmer Don‘s Farm with this long board and politely asks if he may borrow a saw.

Farmer Don, a closet capitalist, doesn‘t lend FJ a saw but instead offers to charge Farmer John for each of the N-1 cuts in the plank. The charge to cut a piece of wood is exactly equal to its length. Cutting a plank of length 21 costs 21 cents.

Farmer Don then lets Farmer John decide the order and locations to cut the plank. Help Farmer John determine the minimum amount of money he can spend to create the N planks. FJ knows that he can cut the board in various different orders which will result in different charges since the resulting intermediate planks are of different lengths.

Line 1: One integer N, the number of planks 
Lines 2..N+1: Each line contains a single integer describing the length of a needed plank

Line 1: One integer: the minimum amount of money he must spend to make N-1 cuts

3
8
5
8

34

题目大意：FJ需要修补牧场的围栏，他需要 N 块长度为 Li 的木头（N planks of woods）。开始时，FJ只有一块无限长的木板，因此他需要把无限长的木板锯成 N 块长度

为 Li 的木板，Farmer Don提供FJ锯子，但必须要收费的，收费的标准是对应每次据出木块的长度，比如说测试数据中 5 8 8，一开始，
FJ需要在无限长的木板上锯下长度 21 的木板（5+8+8=21），第二次锯下长度为 5 的木板，第三次锯下长度为 8 的木板，至此就可以将长度分别为 5 8 8 的木板找出
题目思路：题意可以转换成赫夫曼树的构造过程，用数组存储所有需要的木板长度，然后排序，将前两段最短的生成新节点，将生成的新节点进行插入排序，最后统计生成赫夫曼树后所有花费。

#include<iostream>
#include<algorithm>
#include<string.h>


using namespace std;


int main(){
    int q[20005];
    memset(q,0,sizeof(q));

    int i,j,n,sum=0;
    long long ans=0;

    cin>>n;
    for(i=0;i<n;i++){
        cin>>q[i];
    }
    sort(q,q+n);//将需要的木板排序

    for(i=0;i<n-1;i++){
        sum = q[i]+q[i+1];
        ans+=sum;//对需要的花费进行统计
        for(j=i+2;j<n;j++)
        {
            if(sum>q[j]){
                q[j-1]=q[j];
            }
            else{
                q[j-1]=sum;
                break;
            }
            
        }//for循环插入排序
        if(j==n)
            q[j-1]=sum;//for循环不起作用跳出，即sum比所有剩余元素还要大的情况
    }
    cout<<ans<<endl;
}