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&…

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); }…

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…

博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6789787.html特别不喜欢那些随便转载别人的原创文章又不给出链接的所以不准偷偷复制博主的博客噢~~ 给出n个couple和m个宾客如果宾客没有couple或者couple没来,则被认为lonely问你有多少个lonely的宾客,并且按照id的升序输出 #include <iostream> #include <cstdio> #include…

本文同步发布在CSDN:https://blog.csdn.net/weixin_44385565/article/details/90613846 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 th…

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…

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…

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 the…

如题,大水题...贴个代码完事,就这么任性~~ #include <iostream> #include <cstdio> #include <algorithm> #include <string.h> #define INF 0x3f3f3f3f using namespace std; /* 求满足下面的最大的E: 恰好有E个天的行程都大于E miles 从小到大排序,倒着循环一遍,直到最小值<=天数为止,此时的天数-1即为答案. */ +; i…