Description

In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence

9 1 0 5 4 ,

Ultra-QuickSort produces the output

0 1 4 5 9 .

Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.

Input

The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.

Output

For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.

Sample Input
5
9
1
0
5
4
3
1
2
3
0

Sample Output
6
0

题目大意，给你一串每个元素不相同的序列，每次只能相邻的两个元素进行交换，求，最小多少次交换可以使该序列成为上升序列。

注意：明白一个规律，一个数x，肯定要和在它左边且比它大的数进行交换，即求逆序数。

struct node{
    int elem;
    int old;
};
node arr[500010];
int a[500010];
int c[500010];
int n;
int cmp(node a, node b)
{
    return a.elem <= b.elem;
}
int main()
{
    int i;
    while(scanf("%d", &n) && n)
    {
        memset(c, 0, sizeof(c));
        arr[0].elem = -1;
        arr[0].old = 0;
        for(i = 1; i <= n; i )
        {
            scanf("%d", &arr[i].elem);
            arr[i].old = i;
        }
        sort(arr, arr n 1, cmp);//排序
        for(i = 1; i <= n; i )//将新排好序的下标一一对应存下来；
        {
            a[arr[i].old] = i;
        }
        double sum_ = 0;
        for(i = 1; i <= n; i )
        {
            add(a[i], 1);//将这个数插入
            sum_ = i - sum(a[i]);//得到比当前这个数大，并且初始时在它的左边的数的个数；即得到它的最优移动次数。
        }
        printf("%.0lf\n", sum_);
    }
    return 0;
}
int lowbit(int x)
{
    return x&(-x);
}
double sum(int end_)
{
    double sum = 0;
    while(end_ > 0)
    {
        sum = c[end_];
        end_ -= lowbit(end_);
    }
    return sum;
}
void add(int i, int elem)
{
    while(i <= n)
    {
        c[i] = elem;
        i = lowbit(i);
    }
    return ;
}