PAT 甲级 7-1 1125 Chain the Ropes (25分)-优快云博客

本文介绍了一种算法，用于解决如何将多段绳子通过打结连接成一根尽可能长的绳子的问题。通过每次选取最短的两段绳子进行连接，并确保原始绳长在每次连接后减半，以达到最大化总绳长的目的。

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.

在这里插入图片描述

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:

绳子每次打结长度都会减小到原来的一半，那么打结的顺序会影响到最终的长度。
先小的打结，大的最后打。

#include<bits/stdc++.h>
using namespace std;
int n,len;
int main() {
	cin >> n;
	vector<int> v(n);
	for (int i = 0; i < n; i++) scanf("%d", &v[i]);
	sort(v.begin(), v.end());
	len = v[0];
	for (int i = 1; i < n; i++) {
		len = (len + v[i]) / 2;
	}
	printf("%d", len);
	return 0;
}

#include<bits/stdc++.h>
using namespace std;
int v[10010],n.i; 
int main() {
	cin >> n;
	for (i = 0; i < n; i++) scanf("%d", &v[i]);
	sort(&v[0],&v[n]);
	int len = v[0];
	for (int i = 1; i < n; i++) {
		len = (len + v[i]) / 2;
	}
	printf("%d", len);
	return 0;
}