逆序数对 （常规）

http://acm.hdu.edu.cn/showproblem.php?pid=1394

求下列数列中逆序数对最少的一个，并求出逆序数对数量？（n <= 5000）

a1, a2, ..., an-1, an

a2, a3, ..., an, a1
a3, a4, ..., an, a1, a2
...

an, a1, a2, ..., an-1 

#pragma warning (disable:4786)  
#include<iostream>
#include<algorithm>
using namespace std;
#define MAX 5000
int arr[MAX + 10];
int main(){
    int n,i,j;
    while( scanf("%d",&n) != EOF ){
        for( i = 0; i < n; i ++ ){
            scanf( "%d",&arr[i] );
        }
		//常规 n^2 方法求逆序数对
        int count = 0;
        for(i = 0; i < n; i ++){
            for( j = i + 1; j < n; j ++ ){
                if( arr[i] > arr[j] )
                    count ++;
            }
        }

		//求出将每一位数移动到最后时较前一个数组的逆序数对数量改变值
        int c1, c2,min = count;
        for( i =0; i < n; i ++ ){
            c1 = 0, c2 = 0;
            for( j = 0; j < i; j ++ ){
                if( arr[i] > arr[j] )
                    c1 ++;
                else c2 ++;
            }
            for( j = i + 1; j < n; j ++ ){
                if( arr[i] > arr[j] )
                    c1 ++;
                else c2 ++;
            }
            count = count + c2 - c1;
            if( count < min )
                min = count;
        }
        
        printf("%d\n", min );

    }
    return 0;
}