1125 Chain the Ropes -PAT甲级

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本文介绍了一种通过折叠和链接绳段以创建尽可能长的单根绳子的算法。核心思路在于优先处理较短的绳段，以确保最终绳子的最大长度。文章详细解释了输入输出规范，并提供了一个高效的解决方案，利用排序和精确计算避免精度损失。

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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≤104). Then N positive integer lengths of the segments are given in the next line, separated by spaces. All the integers are no more than 104.

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:

解题思路：为了保证最后的长度最长，需要先折短的，越早被折的折的次数越多

在计算的过程中应保证精度不被丢失，最后向下取整即可，若一开始就用整数相除，很有可能会造成精度的丢失，造成结果不正确

满分代码如下：

#include<bits/stdc++.h>
using namespace std;
const int N=10005;
int len[N];
int n;
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>len[i];
	}
	sort(len+1,len+n+1);
	double sum=len[1];
	for(int i=2;i<=n;i++){
		sum=(sum+len[i])/2.0;
	}
	int res=floor(sum);
	cout<<res<<endl;
	return 0;
}