HDU - 1394 Minimum Inversion Number 求逆序数对 树状数组 （水题的理解）

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

题意:求一个不断变化的序列（变化规则为第一数字到最后）的最小逆序对数是多少 (i < j and ai > aj)

思路：树状数组，可以将原本序列的逆序数求得之后，把第一个数移到最后的逆序数是可以直接得到的。
比如原来的逆序数是ans,把a[i]移到最后后，减少逆序数a[i]，同时增加逆序数n-a[i]-1个,就是ans-a[i]+n-a[i]-1;

因为数字为0----n-1所以数字整体要+1，达到数字本身和原数组的索引相对应的目的

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
#include <set>
#include <cmath>
using namespace std;
const int MAXN=5050;
int c[MAXN];
int a[MAXN];
int n;
int lowbit(int x)
{
    return x&(-x);
}
void add(int i,int val)
{
    while(i<=n)
    {
        c[i]+=val;
        i+=lowbit(i);
    }
}
int sum(int i)
{
    int s=0;
    while(i>0)
    {
        s+=c[i];
        i-=lowbit(i);
    }
    return s;
}

int main()
{
    while(scanf("%d",&n)==1)
    {
       int ans=0;
       memset(c,0,sizeof(c));
       for(int i=1;i<=n;i++){    //计算初始状态的逆序对
        scanf("%d",&a[i]);
        a[i]++;                  //+1处理
        ans+=sum(n)-sum(a[i]);  //计算当前数字的逆序对
        add(a[i],1);           //将该数组放入树状数组中
       }
       int re=ans;
       for(int i=1;i<n;i++){   //可以省略最后一次变化
        ans+=n-a[i]-(a[i]-1);
         re=min(re,ans);
       }
       printf("%d\n",re);
    }
    return 0;
}