PAT甲级1125解题报告

最新推荐文章于 2022-03-21 22:47:50 发布

Tate_Brown

最新推荐文章于 2022-03-21 22:47:50 发布

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分类专栏： C++ PAT

本文链接：https://blog.youkuaiyun.com/TateBrwonJava/article/details/84258362

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本文介绍了一个绳子链接问题的算法解决方案。给定一系列绳子的长度，目标是通过连续将两段绳子对折并连接的方式，制作出尽可能长的绳子。文章详细解释了解题思路，即从最短的绳子开始，将其与下一根绳子对折连接，直到最后形成一根总绳子。这种方法确保最终绳子的长度不会超过原始绳子中最长的那一根。

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好几天没做了已经。。感觉有点变难了，这几天也比较忙就一天一题吧，挑了半天挑了道会做了。

1125 Chain the Ropes （25 point(s)）

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:

题目大意：给一堆绳子，给出他们的长度，求一个最大的长度不大于他最长绳子的长度，绳子可以对接，但对接的话他们的长度都要对折。

解题思路：常识吧，任意两数相加的平均数小于等于这两者的最大值，所以就排个序从最短的一直对折到最后一根，结果可以证明比最长的那根小，这样得到的也就是最长了。

#include<iostream>
#include<string.h>
#include<vector>
#include<algorithm>
#include<iomanip>
#include<time.h>
#include<math.h>
#include<set>
#include<list>
#include<climits>
#include<queue>
#include<cstring>
#include<map>
#include<stack>
#include<string>
using namespace std;
vector<int> a;
int main()
{
	int N;
	scanf("%d", &N);
	for (int i = 0; i < N; i++) {
		int tmp;
		scanf("%d", &tmp);
		a.push_back(tmp);
	}
	if (a.size() == 1) {
		printf("%d\n", a[0]);
	}
	else {
		sort(a.begin(), a.end());
		double res = 0;
		res = a[0];
		for (int i = 1; i < a.size(); i++) {
			res =(res+ a[i])/2;
		}
		int t = res;
		printf("%d\n", t);
	}
	return 0;
}