1125 Chain the Ropes （25 分）

1125 Chain the Ropes （25 分）

Given some segments of rope, you are supposed to chain them into one rope. Each time you may only fold two segments into loops and chain them into one piece, as shown by the figure. The resulting chain will be treated as another segment of rope and can be folded again. After each chaining, the lengths of the original two segments will be halved.

Your job is to make the longest possible rope out of N given segments.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (2≤N≤10​4​​). Then Npositive integer lengths of the segments are given in the next line, separated by spaces. All the integers are no more than 10​4​​.

Output Specification:

For each case, print in a line the length of the longest possible rope that can be made by the given segments. The result must be rounded to the nearest integer that is no greater than the maximum length.

Sample Input:

8
10 15 12 3 4 13 1 15

Sample Output:

14

分析：贪心题，自己做的时候直接排序然后从小开始两两对折最后输出结果就AC了，但不知道原因。。（做贪心有时候就凭感觉，没有严格证明）。后来看柳神题解才恍然大悟，若从大的开始对折，每次对折损耗会更多，而从小的开始损耗的就越少。比如10000，100,10,1，我从10000开始对折的话总共对折3次，相当于会损耗大的部分3次，如果最后再对折就损耗少了！

 /**
 * Copyright(c)
 * All rights reserved.
 * Author : Mered1th
 * Date : 2019-02-28-13.41.03
 * Description : A1125
 */
 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<string>
 #include<unordered_set>
 #include<map>
 #include<vector>
 #include<set>
 using namespace std;
 ;
 int a[maxn];
 int main(){
 #ifdef ONLINE_JUDGE
 #else
     freopen("1.txt", "r", stdin);
 #endif
     int n;
     scanf("%d",&n);
     ;i<n;i++){
         scanf("%d",&a[i]);
     }
     sort(a,a+n);
     ];
     ;i<n;i++){
         sum=(a[i]+sum)/;
     }
     cout<<int(sum);
     ;
 }