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.

3
8
5
8

34

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

 using namespace std;

 int main()
 {
     int t;
     __int64 sum1,sum;
     __int64 a[20002];
     bool flag=false;
     while(~scanf("%d",&t))
     {
        priority_queue<int,vector<int>,greater<int> > q;
        for(int i=0;i<t;i++)
         {
             cin>>a[i];
             q.push(a[i]);
         }
         sum1=0;
         sum=0;
         while(1)
         {
             if(t==1)
             {
                 cout<<a[0]<<endl;
                 flag=true;
                 break;
             }
             int x=q.top();
             q.pop();
             int y=q.top();
             q.pop();
             sum1=x+y;
             sum+=sum1;
             if(q.empty())
             break;
             q.push(sum1);

         }
         if(!flag)
             cout<<sum<<endl;
     }
     return 0;
 }

注释：

题意：取原序列里两个最小的    加和放到序列里（保证每次取得都为队列里最小的两个）  最后生一个最大的即为不用亲自割的

1：优先队列