1125 Chain the Ropes （25 分）

最新推荐文章于 2025-07-13 12:44:43 发布

原创最新推荐文章于 2025-07-13 12:44:43 发布 · 169 阅读

0 ·

CC 4.0 BY-SA版权

PAT甲专栏收录该内容

53 篇文章

订阅专栏

本文介绍了一个有趣的编程问题：如何将给定的多个绳段通过特定的折叠和链接方式，制作成尽可能长的单根绳子。文章详细解释了输入输出规格，并提供了一个C++代码示例，展示了如何通过算法实现这一目标。

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:

#include<iostream>
#include<cstdio>
#include<string>
#include<vector>
#include<map>
#include<algorithm>
using namespace std;
int main(){
	int n;
	vector<int> v;
	cin >> n;
	for(int i = 0;i < n;i++){
		int a;
		cin >> a;
		v.push_back(a);
	} 
	sort(v.begin(),v.end());
	int sum = v[0];
	for(int i = 1;i < n;i++){
		sum = (sum+v[i])/2+0.5;
	}
	cout << sum << endl;
	return 0;
}