Solution：只需要從板的集合裡取出最短的兩塊，並且把長度為兩塊板長度之和的板加入集合中即可，用優先佇列可以高效的實現。

AC_Code:

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#include<set>
#include<cstdlib>
#define ll long long
#define ull unsigned long long
#define mem(a,x) memset((a),(x),sizeof ((a)))//x只能是0或-1或false或true
#define debug(x) cout<<"X: "<<(x)<<endl
#define de cout<<"************"<<endl
#define lowbit(x) ((x)&(-x))
#define lson rt<<1
#define rson rt<<1|1
#define gcd(a,b) __gcd(a,b)
#define lcm(a,b) a*b/(__gcd(a,b))
#define inf 0x3f3f3f3f//1e9+6e7
#define eps 1e-8
#define mod 1e9+7
#define N 10010
const double pi=acos(-1.0);
using namespace std;
int a[N];
int main()
{
    int n;
    cin>>n;
    for(int i=0; i<n; i++)
        scanf("%d",&a[i]);
    ll ans=0;
    //宣告一個從小到大取出數值的優先佇列
    priority_queue<int,vector<int>,greater<int>> que;
    for(int i=0; i<n; i++)
        que.push(a[i]);
    while(que.size()>1)
    {
        int l1,l2;
        l1=que.top();
        que.pop();
        l2=que.top();
        que.pop();
        //把兩塊木塊合併
        ans+=l1+l2;
        que.push(l1+l2);
    }
    cout<<ans<<endl;
    return 0;
}