Fence Repair

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.

Input

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

Output

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

Sample Input

3
8
5
8

Sample Output

34

Hint本次

He wants to cut a board of length 21 into pieces of lengths 8, 5, and 8.
The original board measures 8+5+8=21. The first cut will cost 21, and should be used to cut the board into pieces measuring 13 and 8. The second cut will cost 13, and should be used to cut the 13 into 8 and 5. This would cost 21+13=34. If the 21 was cut into 16 and 5 instead, the second cut would cost 16 for a total of 37 (which is more than 34).

题解：题意很简单，一刀两断求花费最少。只要将顺序用优先队列从小到大排序，由最小的两个段的和求出来压入队列（和即为本次切割的花费)并且将他们pop()出队列，直到队列元素个数不大于1时结束；

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<vector>
using namespace std;
int main()
{
     int i,j,m,n,sum=0;
     priority_queue<int,vector<int>,greater<int> >s;
     scanf("%d",&n);
     for(i=0;i<n;i++)
     {
        scanf("%d",&j);
        s.push(j);

     }
     long long int ans=0;//由于数据比较大，可能爆long long
     while(s.size()>1)
     {
         m=s.top();
         s.pop();
         n=s.top();
         s.pop();
         ans+=m+n;//统计每次切割的费用
         s.push(m+n);//压入队列
     }
     printf("%lld
",ans);
    return 0;
}