Minimum Inversion Number

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 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<stdio.h>#include<string.h>#define Max 5010int tree[Max];int input[Max];int n;int lowbit(int x){    return x & (-x);}void modify(int x,int add)//一维{      while(x<=n)      {              tree[x]+=add;            x+=lowbit(x);     }}__int64 get_sum(int x)//求1到x之间的和{      __int64 ret=0;     while(x!=0)      {               ret+=tree[x];           x-=lowbit(x);       }      return ret;}int main(){//freopen("b.txt","r",stdin);int i,x;__int64 ans,maxn;while(scanf("%d",&n)==1){memset(tree,0,sizeof(tree));memset(input,0,sizeof(input));ans=0;for(i=0;i<n;i++){scanf("%d",&x);input[i]=x;ans+=get_sum(n)-get_sum(x+1);//如果在x+1放入树状数组之前，在x+1至n有数放入，则该区间数的个数，就是逆序数的对数modify(x+1,1);}maxn=ans;for(i=0;i<n;i++){ans=ans-input[i]-1+n-input[i];if(ans<maxn) maxn=ans;}printf("%I64d\n",maxn);}return 0;}