1125 Chain the Ropes (25 分)

1125 Chain the Ropes (25 分),第1张

概述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 chai

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). Then N positive integer lengths of the segments are given in the next line,separated by spaces. All the integers are no more than 1.

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:
810 15 12 3 4 13 1 15
Sample Output:
14
#include<iostream>#include<algorithm>using namespace std;const int maxn = 100010;using namespace std;int main(){    int arr[maxn];    int n;    cin >> n;    for(int i = 0; i < n; i++){        cin >> arr[i];    }    sort(arr,arr+n);    int result = arr[0];    for(int i = 1; i < n; i++){        result = (result+arr[i])/2;    }    printf("%d",result);    return 0;}
总结

以上是内存溢出为你收集整理的1125 Chain the Ropes (25 分)全部内容,希望文章能够帮你解决1125 Chain the Ropes (25 分)所遇到的程序开发问题。

如果觉得内存溢出网站内容还不错,欢迎将内存溢出网站推荐给程序员好友。

欢迎分享,转载请注明来源:内存溢出

原文地址: http://outofmemory.cn/yw/1031120.html

(0)
打赏 微信扫一扫 微信扫一扫 支付宝扫一扫 支付宝扫一扫
上一篇 2022-05-23
下一篇 2022-05-23

发表评论

登录后才能评论

评论列表(0条)

保存