归并排序求逆序数 Time Limit:7000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Description In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjac
圆桌会议 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3671 Accepted Submission(s): 2575 Problem Description HDU ACM集训队的队员在暑假集训时常常要讨论自己在做题中遇到的问题.每当面临自己解决不了的问题时,他们就会围坐在一张圆形的桌子旁进行交流,经过大家的讨论后一般没有解决不了的问
Problem Description The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj. For a given sequence of numbers a1, a2, ..., an, if we move the first m >= 0 numbers to the end of
Minimum Inversion Number Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 1 Accepted Submission(s) : 1 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description The inversion
DNA Sorting Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 80832 Accepted: 32533 Description One measure of ``unsortedness'' in a sequence is the number of pairs of entries that are out of order with respect to each other. For instanc
题意:给出一序列,你可以循环移动它(就是把后面的一段移动到前面),问可以移动的并产生的最小逆序数. 求逆序可以用并归排序,复杂度为O(nlogn),但是如果每移动一次就求一次的话肯定会超时,网上题解都说可以用并归做,想了好久,最后发现"the next line contains a permutation of the n integers from 0 to n-1",坑爹的家伙,这些数竟然是从0到n-1的. 这样就可以做了,推导一下可以发现每移动一位,数列的逆序数就会又规律的变化
/** \brief poj2299 * * \param date 2014/8/5 * \param state AC * \return memory 4640K time 3250ms * */ #include <iostream> #include <fstream> #include <cstdio> #include <cstring> using namespace std; const int MAXN=500000; int