ZOJ 1484 Minimum Inversion Number

歸並排序求逆序數。。。。

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 seqence, we will obtain another sequence. There are totally n such sequences as the following:

a1, a2, ..., an-1, an (where m = 0 - the initial seqence)
a2, a3, ..., an, a1 (where m = 1)
a3, a4, ..., an, a1, a2 (where m = 2)
...
an, a1, a2, ..., an-1 (where m = n-1)

You are asked to write a program to find the minimum inversion number out of the above sequences.

Input

The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 5000); the next line contains a permutation of the n integers from 0 to n-1.

Output

For each case, output the minimum inversion number on a single line.

Sample Input

10
1 3 6 9 0 8 5 7 4 2

Sample Output

16

#include 
#include 
#include 
#include 

using namespace std;

int a[5500],b[5500],t[5500],n,nixushu=0;

void merge_sort(int l,int r)
{
    if(l>=r) return;
    int m=(l+r)/2;
    merge_sort(l,m); merge_sort(m+1,r);
    int q=0,left=m-l+1,right=r-m,nl=0,nr=0;
    while(nl