Minimum Inversion Number HDU

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：给你一串由0-n-1组成的序列，每次将头元素移动至序列尾部，对每个变换求一次逆序对，问在这些变换的序列中序列逆序对最小是多少。
思路：首先求逆序对，线段树区间 （l，r）的和 ，这个和表示序列到目前为止l，r中出现数字的次数，然后当前位置数x的逆序对就是(x+1,n)的结果，由于根据题意序列是0-n-1，元素不重复，所以避免越界可以query(x,n)。。。我建树从1建到n而且会有越界风险感觉很low2333来神的线段树好像就，神TM高端2333
然后把第一个初始序列的逆序对处理出来之后，由于这个题意这个操作啊，就很神奇 ，毛估估是有规律的对不对。看 序列 2 4 1 3 我们操作要移动2。这个2在队首对这个序列提供的逆序对贡献是比这个数小的所有数的个数，2移到队尾对逆序对的贡献是所有比2大的数字个数之和。那么我们对上一次的总数减去比当前数小的个数 加上比当前数大的个数 取一个最小就可以了
恕我直言 大半夜脑子不大好使。。我printf里面一开始写了个tot在那里狂debug。。。一度绝望。。

#include<iostream>#include<algorithm>#include<vector>#include<queue>#include<vector>#include<cmath>#include<cstdio>#include<cstring>#include<string>#include<stack>#include<map>using namespace std;//thanks to pyf ...#define INF 0x3f3f3f3f#define CLR(x,y) memset(x,y,sizeof(x))#define mp(x,y) make_pair(x,y)typedef pair<int,int> PII;typedef long long ll;const int N = 1e5+5;struct Tree{    int l,r,sum;}t[N*4];int a[N];void push_up(int step){    t[step].sum = t[step*2].sum + t[step*2+1].sum;}void build(int l,int r,int step){    t[step].l = l;    t[step].r = r;    t[step].sum = 0;    if(l==r)    {        t[step].sum = 0;        return ;    }    int mid = (l+r)/2;    build(l,mid,step*2);    build(mid+1,r,step*2+1);}void update(int x,int step){    if(t[step].l==t[step].r)    {        t[step].sum ++;        return;    }    int mid = (t[step].l + t[step].r)/2;    if(x<=mid)        update(x,step*2);    else        update(x,step*2+1);    push_up(step);}int query(int l,int r,int step){//  cout << l << " " << r << " " << t[step].l << " " << t[step].r << endl;    if(t[step].l == l&&t[step].r == r)        return t[step].sum;    int mid = (t[step].l + t[step].r)/2;    if(r<=mid)        return query(l,r,step*2);    else if(l>mid)        return query(l,r,step*2+1);    else        return query(l,mid,step*2) + query(mid+1,r,step*2+1);}int main(){    int n;    while(scanf("%d",&n)==1)    {        for(int i=0;i<n;i++)        {            int x;            scanf("%d",&x);            a[i] = x+1;        }        build(1,n,1);        int tot = 0;        for(int i=0;i<n;i++)        {//          cout << a[i] << endl;            tot += query(a[i],n,1);            update(a[i],1);                 }        int ans = tot;        for(int i=0;i<n;i++)        {            tot += n-a[i];            tot -= a[i] -1;//          cout << ans << " "<< tot << endl;            ans = min(ans,tot);        }        printf("%d\n",ans);    }}