HDU_1394_Minimum Inversion Number

给定一组数，然后依次的挪动该组数的元素共得到n种序列。求这n中序列中逆序数最少的个数。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
#include<queue>
using namespace std;
#define N 5005
int n,c[N]; 
int sum(int x)
{
    int ret=0;
    while(x)
    {
        ret+=c[x];
        x-=x&-x;
    }
    return ret;
}
void add(int x)
{
    while(x<=n)
    {
        c[x]++;
        x+=x&-x;
    }
}
int main()
{
    int a[N],i,j,k;
    while(~scanf("%d",&n))
    {
        memset(c,0,sizeof(c));
        int ans=0;
        for(i=1;i<=n;++i)
        {
            scanf("%d",&a[i]);
            a[i]++;//a[i]从0开始，而树状数组下标从1开始，所以自增1
            add(a[i]);
            ans+=(i-sum(a[i])); //求得的逆序对
        }
        int Min=ans; 
        /*从后往前看，把a[n]移到第一位，逆序对增加a[i]-1，逆序对减少n-a[i]*/
        for(i=n;i>1;--i)
        {
            ans+=2*a[i]-1-n;
            Min=min(ans,Min);
        }
        printf("%d\n",Min);
    }
    return 0;
}