PAT A1125 Chain the Ropes

来源：互联网发布：sql 双竖线用法编辑：程序博客网时间：2024/06/05 14:35
/*---------------------------------------------------------------PAT A1125 Chain the Ropes题意：绳子环环相扣，各段（halved）减半，求可能的最大长度；题解：权为 1/2的赫尔曼树，取最小的两段配对成新的节点；练习一下快速排序。---------------------------------------------------------------*/#include<iostream>#include<fstream>#include<math.h>using namespace std;//快速排序复杂度O(nlog(n))void Quick_Sort(int (&Sequence)[10001], const int& Left, const int& Right){if (Left >= Right)    return;int Low = Left, High = Right;    Sequence[0] = Sequence[Low]; //取0位为枢轴while (Low < High){while (Low < High && Sequence[High] >= Sequence[0])//左边 --High;Sequence[Low] = Sequence[High];while (Low < High && Sequence[Low] <= Sequence[0])//右边++Low;Sequence[High] = Sequence[Low];}Sequence[Low] = Sequence[0];              //枢轴归位int Center = Low;Quick_Sort(Sequence, Left, Center - 1);   //递归快速排序Quick_Sort(Sequence, Center + 1, Right);  //类似于递归建树，二分法递归查询}int main(){#ifdef _DEBUGifstream cin("Input.txt");ofstream cout("Output.txt");#endifint Number;cin >> Number;int Segments[10001];for (int i = 1; i <= Number; ++i)cin >> Segments[i];Quick_Sort(Segments, 1, Number);//for (int i = 1; i <= Number; ++i)     //查看快速排序是否正确//cout << Segments[i] << ' ';double Sum = Segments[1];for (int i = 2; i <= Number; ++i)Sum  = 0.5 * (Sum + Segments[i]);cout << floor(Sum);                   //取整输出#ifdef _DEBUGcin.close();cout.close();#endifreturn 0;}
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