标签：space ecif tput div sed 一半 origin color another
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 <= 10^4^). Then N positive integer lengths of the segments are given in the next line, separated by spaces. All the integers are no more than 10^4^.
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:
14
题目大意：把两条绳子结合在一起，两者的长度会减少一半。给出一些绳子，要求找把所有的绳子链接在一起，要求最后得到的绳子长度最长
思路：要得到的长度最长就应该最后连接最长的绳子，减少长度长的绳子的折叠次数1 #include<iostream> 2 #include<vector> 3 #include<algorithm> 4 using namespace std; 5 int main(){ 6 int n, i; 7 cin>>n; 8 vector<int> v(n); 9 for(i=0; i<n; i++) scanf("%d", &v[i]); 10 sort(v.begin(), v.end()); 11 int ans=(v[0]+v[1])/2; 12 for(i=2; i<n; i++) ans = (ans+v[i])/2; 13 cout<<ans<<endl; 14 return 0; 15 }
标签：space ecif tput div sed 一半 origin color another
原文地址：https://www.cnblogs.com/mr-stn/p/9160784.html
