题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1394 Minimum Inversion Number                        Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)                                            Total Submission(s): 10…

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 the seq…

Minimum Inversion Number Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 15675 Accepted Submission(s): 9569 Problem Description The inversion number of a given number sequence a1, a2, -, an is the…

Minimum Inversion Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 15517    Accepted Submission(s): 9467 Problem Description The inversion number of a given number sequence a1, a2, ..., a…

题目传送门 Minimum Inversion Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 25116    Accepted Submission(s): 14827 Problem Description The inversion number of a given number sequence a1, a2,…

之前写过树状数组的,再用线段树写一下--- #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<cmath> #include<vector> using namespace std; #define lp (p << 1) #define rp (p << 1 | 1) #define getmi…

题意:给定一个序列,求分别将前m个数移到序列最后所得到的序列中,最小的逆序数. 分析:m范围为1~n,可得n个序列,求n个序列中最小的逆序数. 1.将序列从头到尾扫一遍,用query求每个数字之前有多少个大于该数字的数,方法如下. (1)将已经扫过的数字所对应的位置标记,通过query求该数字之后有多少个数被标记过 (2)该数字之后所有被标记的数字,都是在该数字之前出现过的(i<j),而这些数字又大于该数字(ai>aj),因此该数字之后所有的标记和就是该数字之前比该数字大的数的个数. 2.su…

Minimum Inversion Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9342    Accepted Submission(s): 5739 Problem Description The inversion number of a given number sequence a1, a2, ..., an…

Minimum Inversion Number Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 10326 Accepted Submission(s): 6359 Problem Description The inversion number of a given number sequence a1, a2, ..., an is t…

一.题目描述 There is a permutation P with n integers from 1 to n. You have to calculate its inversion number, which is the number of pairs of Pi and Pj satisfying the following conditions: iPj. 二.输入 The input may contain several test cases. In each test c…