A1101 Quick Sort (25 分)】的更多相关文章

一.技术总结 这里的一个关键就是理解调换位置排序是时,如果是元主,那么它要确保的条件就只有两个一个是,自己的位置不变,还有就是前面的元素不能有比自己大的. 二.参考代码 #include<iostream> #include<algorithm> #include<vector> using namespace std; int record[100000]; int main(){ int n, max = 0, count = 0; scanf("%d&q…
题意: 输入一个正整数N(<=1e5),接着输入一行N个各不相同的正整数.输出可以作为快速排序枢纽点的个数并升序输出这些点的值. trick: 测试点2格式错误原因:当答案为0时,需要换行两次…
1101. Quick Sort (25) 时间限制 200 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CAO, Peng There is a classical process named partition in the famous quick sort algorithm. In this process we typically choose one element as the pivot. Then the elements…
题目地址 https://pta.patest.cn/pta/test/16/exam/4/question/676 5-14 Insertion or Heap Sort   (25分) According to Wikipedia: Insertion sort iterates, consuming one input element each repetition, and growing a sorted output list. Each iteration, insertion s…
1101 Quick Sort (25 分) There is a classical process named partition in the famous quick sort algorithm. In this process we typically choose one element as the pivot. Then the elements less than the pivot are moved to its left and those larger than th…
There is a classical process named partition in the famous quick sort algorithm. In this process we typically choose one element as the pivot. Then the elements less than the pivot are moved to its left and those larger than the pivot to its right. G…
There is a classical process named partition in the famous quick sort algorithm. In this process we typically choose one element as the pivot. Then the elements less than the pivot are moved to its left and those larger than the pivot to its right. G…
There is a classical process named partition in the famous quick sort algorithm. In this process we typically choose one element as the pivot. Then the elements less than the pivot are moved to its left and those larger than the pivot to its right. G…
树状数组+离散化 #include<cstdio> #include<cstring> #include<cmath> #include<map> #include<algorithm> using namespace std; +; int a[maxn],ans[maxn],c[maxn],b[maxn]; int n; map<int,int>m; int lowbit(int x) { return x&(-x); }…
快速排序有一个特点,就是在排序过程中,我们会从序列找一个pivot,它前面的都小于它,它后面的都大于它.题目给你n个数的序列,让你找出适合这个序列的pivot有多少个并且输出来. 大水题,正循环和倒着循环一次,统计出代码中的minnum和maxnum即可,注意最后一定要输出'\n',不然第三个测试会显示PE,格式错误. #include <iostream> #include <cstdio> #include <algorithm> #include <map&…