仍旧在练习线段树中。。这道题一开始没有完全理解搞了一上午，感到了自己的shabi。。
Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15527 Accepted Submission(s): 9471
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

题目大意：
给出一个n，和0-n的一个数列，求这个数列的最小逆序对数，最小逆序对表示，给定数列每次都将第一个数放到最后一个位置，如此变化n次至变换回原状，在这n种不同的数列中，逆序对数的最小值即为所求（N<=5000）每组样例多组数据

明白题意后想到求逆序对的三种方法，第一种暴力法，第二种归并，第三种树状数组和线段树，由于在学习线段树于是果断练习用线段树编写
此处线段树的作用是求出原数列的逆序对
对于每次变换，我们发现，第一个数移到最后一位，只需要在上一步里求出的逆序对减去上一步第一位的数，并加上比上一步第一位要大的数即可

下面是代码：

#include <cstdio>
#include <algorithm>
using namespace std;
#define maxn 5005
int sum[maxn<<2];
int a[maxn];void updata(int now)
{sum[now]=sum[now<<1]+sum[now<<1|1];
}void build(int l,int r,int now)
{sum[now]=0;if(l==r) return;int mid=(l+r)>>1;build(l,mid,now<<1);build(mid+1,r,now<<1|1);
}int query(int L,int R,int l,int r,int now)
{if(L<=l && r<=R)return sum[now];int mid=(l+r)>>1;int total=0;if(L<=mid) total+=query(L,R,l,mid,now<<1);if(R>mid) total+=query(L,R,mid+1,r,now<<1|1);return total;
}void point_change(int loc,int l,int r,int now)
{if(l==r){sum[now]++;return;}int mid=(l+r)>>1;if(loc<=mid) point_change(loc,l,mid,now<<1);else point_change(loc,mid+1,r,now<<1|1);updata(now);
}int main()
{int n;while(~scanf("%d",&n)){build(0,n-1,1);int number=0;for(int i=0;i<n;i++){scanf("%d",&a[i]);number+=query(a[i],n-1,0,n-1,1);point_change(a[i],0,n-1,1);}int ans=number;for(int i=0;i<n;i++){number+=n-a[i]-a[i]-1;ans=min(ans,number);}printf("%d\n",ans);}return 0;
}