【笨方法学PAT】1125 Chain the Ropes （25 分）_your job is to make the longest possible rope out -优快云博客

本文探讨了如何通过折叠和连接n段不同长度的绳子，以获得尽可能长的绳子的问题。通过将绳子按长度排序并从最短的两段开始，每次将两段绳子折叠并连接，然后继续此过程直到所有绳子都连接成一根。最终输出的绳子长度为求解结果。

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一、题目

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:

二、题目大意

给n段绳子，进行折叠，求最大长度。

三、考点

逻辑

四、注意

1、注意折叠方式。

五、代码

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main() {
	//read
	int n;
	cin >> n;
	vector<int> v(n);
	for (int i = 0  ; i < n; ++i) {
		cin >> v[i];
	}

	//sort
	sort(v.begin(), v.end());

	//solve
	int len = v[0];
	for (int i = 1; i < n; ++i) {
		len = (len + v[i]) / 2;
	}

	//output
	cout << len << endl;

	system("pause");
	return 0;
}