Who's in the Middle - poj 2388 (快速排序寻找中位数)

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Who's in the Middle - poj 2388 (快速排序寻找中位数)

题意; 寻找中位数利用快速排序来寻找中位数. #include <iostream> using namespace std; int N; ]; int Median(int left,int right,int pos){ ,r=left; int pirior=op[right]; for(int i=left;i<right;i++){ if(op[i]<pirior){ int temp=op[i]; op[i]=op[r]; op[r]=temp; r++; l++;…

poj 2388 insert sorting

/** \brief poj 2388 insert sorting 2015 6 12 * * \param * \param * \return * */ #include <iostream> #include <cstdio> #include <cstring> using namespace std; const int N=10000; int Arr[N]; void insertSort(int len) { for(int j=1;j<len;…

POJ 2388&&2299

排序(水题)专题,毕竟如果只排序不进行任何操作都是极其简单的. 事实上,排序算法十分常用,在各类高级的算法中往往扮演着一个辅助的部分. 它看上去很普通,但实际的作用却很大.许多算法在失去排序后将会无法实现. 以上在扯P 看2388,求中位数. 好,sort一遍(与衢州2017市赛PJ T1一样水). CODE #include<cstdio> #include<algorithm> using namespace std; ; int n,a[N]; inline char tc(…

poj 2388 Who's in the Middle（快速排序求中位数）

一.Description FJ is surveying his herd to find the most average cow. He wants to know how much milk this 'median' cow gives: half of the cows give as much or more than the median; half give as much or less. Given an odd number of cows N (1 <= N < 10…

POJ 2388 Who's in the Middle(水~奇数个数排序求中位数)

题目链接:http://poj.org/problem?id=2388 题目大意: 奇数个数排序求中位数解题思路:看代码吧! AC Code: #include<stdio.h> #include<algorithm> using namespace std; int main() { int n; while(scanf("%d",&n)!=EOF) { ]; ; i<n; i++) scanf("%d",&na[i…

poj 2388 Who's in the Middle

点击打开链接 Who's in the Middle Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 28324 Accepted: 16396 Description FJ is surveying his herd to find the most average cow. He wants to know how much milk this 'median' cow gives: half of the cow…

POJ 2388 Who's in the Middle (快速选择算法：O(N)求数列第K大)

[题意]求数列中间项. ---这里可以扩展到数列第K项. 第一次做的时候直接排序水过了= =--这一次回头来学O(N)的快速选择算法. 快速选择算法基于快速排序的过程,每个阶段我们选择一个数为基准,并把区间划分成小于这个数和大于这个数的两个子区间,此时便可以判断这个数是不是第k大项,如果比K大,则去左区间找,否则去右区间找. #include #include #include #include #include using namespace std; template doubleORint…