dex reverse number

The question is very simple. Find the number of exchanges in the worst case of quicksort. After understanding it, we know that it should be a reverse-order pair problem of the sequence. It is found that the nlog (n) sorting speed of Merge Sorting is very good, while in the sorting process, there is a merge process Merge (), here we will merge two ordered series into an ordered series. In the process of me

A sequence composed of 0 .. n-1. Each time you move the first element of the team to the end of the team, Evaluate the number of minimum reverse pairs of the N sequences Algorithm : Inserts an element into the line segment tree in sequence. The reverse logarithm of each addition is larger than the inserted Number of

Reprint please indicate the source, thank you http://blog.csdn.net/ACM_cxlove? Viewmode = ContentsBy --- cxlove Returns a reverse_prime, which is a 7-digit number and a prime number First, find all such numbers. Two operations Q k: deletes the number K. Http://uva.onlinejudge.org/index.php? Option = com_onlinejudge Itemid = 8 page = show_problem pro

consists of n positive integers separated by a space, and the first I represents the size of the data that is numbered I.The third line is a positive integer q, indicating that Mato will look at the data for a few days.After the Q line of two positive integers per line L, R, indicating Mato this day to see the [l,r] interval of the file.OutputQ lines, one positive integer per line, indicating the number of times the Mato need to be exchanged this day

Returns information such as Network, mask, broadcast, reverse resolution, subnet number, IP type, etc. based on the IP or subnet enteredModule required for IPY module here is the Python3 version#!/usr/bin/envpython#-*-coding:utf-8-*-fromipyimportipip_s= input (' pleaseinputanipornet-range: ') # Parameters are IP or network segment Ips=ip (ip_s) Iflen (IPs) >1: #网络地址个数 print (' net:%s ' %ips.net ()) #输出网络地

Ideas:In the process of merging and sorting, one step is to remove the small element from the left and right two arrays in the tmp[] array.The array on the right is actually the element on the right side of the original array. When you take the element to the right without taking the element to the left, the remaining elements on the left side are larger than the first element on the right, i.e. they can form an inverse pair. Assuming that there are now n elements left on the right, the inverse

ans[i]=ans[i-1]+ (n+1) -2*num[i]Num[i] is the data at inputAns[i] is the number of reverse order when m=iThe reverse order of ans[0 in a tree-like array#include #include #include using namespace Std;const int maxn=5010;int NUM[MAXN];int TREE[MAXN];int lowbit (int i){Return i (-i);}int getsum (int i){int sum=0;while (i>0){Sum+=tree[i];I-=lowbit (i);}return sum;}vo

poj2299 Ultra-quicksort tree-like array for inverse number Ultra-quicksort Time Limit: 7000MS Memory Limit: 65536K Total Submissions: 49587 Accepted: 18153 DescriptionIn this problem, you has to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping, adjacent sequence elements until the sequence is Sorted in ascending order. For the i

Poj 2274 The Race (reverse number + line segment tree)The Race Time Limit:15000 MS Memory Limit:65536 K Total Submissions:3237 Accepted:664 Case Time Limit:3000 MS DescriptionDuring the Annual Interstellar Competition for Tuned Spaceships, N spaceships will be competing. each spaceship I is tuned in such a way that it can accelerate in zero time to its maximu

Topic Links:http://acm.hdu.edu.cn/showproblem.php?pid=2838Main topic:There are n cows lined up in a row. Each cow has a unique "bad temper" value. The range of bad temper is 1~100000. will nowThe cows are reordered so that the cows are ranked in the order in which the bad temper increases. All cows can exchange their positions with each other. But the exchange of spleenThe gas value is x, y two cows, the time required is x+y. Now ask: What is the shortest time it takes to rearrange the cows?Idea

Frosh WeekTime limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 2772 Accepted Submission (s): 923Problem descriptionduring Frosh Week, students play various games-get to know all other and compete against other Teams. In one such-game, all the Frosh-a team stand in a line, and is then asked to arrange themselves according to some crit Erion, such as their height, their birth date, or their student number

InversionTime limit:2000/1000 MS (java/others) Memory limit:131072/131072 K (java/others)Total submission (s): 3171 Accepted Submission (s): 1154Problem Descriptionbobo has a sequence a1,a2,...,. He is allowed to swapadjacentNumbers for no more than k times.Find the minimum number of inversions after his swaps.Note:the number of inversions is the number of pair

exchanges to make the sequence orderlyResolving Merge sort to find reverse orderAC Code#include #includeusing namespacestd;Const intMAXN = 1e5+ -, inf=0x3f3f3f3f; typedefLong Longll;Const intmod=1e9+7;intA[maxn],t[maxn],ans;voidMerge_sort (int*a,intXintYint*t) { if(y-x>1) { intm=x+ (y-x)/2;//Divide intP=x,q=m,i=x; Merge_sort (a,x,m,t); //Recursive SolutionMerge_sort (a,m,y,t);//Recursive Solution while(p//as long as a seq