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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 题意就是输入一个n，接着输入n个数，然后求出这n个数的逆序数。

最简单的求逆序数就是用冒泡排序，每次比较，如果交换说明整两个数是逆序的，ans++，也就是说冒泡排序交换次数就是逆序数；但是时间复杂度是O(n^2).

再就是可以用归并排序来求逆序数，隐隐约约记得之前做过一道题是可以的来求的，虽然现在忘了啊。 这里用树状数组来求。

先说离散化

如果数据量少，但是数据范围大，那么开数组的话会浪费很大一部分空间，所以需要离散化处理(通俗点就是把离散较大的数据紧凑起来)

for (int i=1; i<=n; i++){   	scanf("%d", &init[i].val);	init[i].order=i;}sort(init+1, init+n+1, cmp);for (int i=1; i<=n; i++)	a[init[i].order]=i;

这样最后a数组的下标就是原来数据的元素的order位置，值就是被处理后的值。从而做到把离散的数据紧凑起来，节省空间。

这样就是把9 1 0 5 4变成5 2 1 4 3 再说求逆序数 这里用树状数组来求，就是每个元素的前面比它大的数量的和就是最终的逆序数。

在说明一点，i-sum(a[i])是本来一共有i个数，sum(a[i])是所有小于等于a[i]的个数，相减就是大于a[i]的个数，也就是在新的数前面大于它的元素的个数。

#include 
   
    #include 
    
     #include 
     
      #include 
      
       using namespace std;int n;int a[500005];int c[500005];struct node{   	int order;	int val;}init[500005];int cmp(node a, node b){   	return a.val
       
        =1){   		ans+=c[index];		index-=lowbit(index);	}	return ans;}int main(){   	while (scanf("%d", &n) && n){   		for (int i=1; i<=n; i++){   			scanf("%d", &init[i].val);			init[i].order=i;		}		sort(init+1, init+n+1, cmp);		for (int i=1; i<=n; i++)			a[init[i].order]=i;		memset(c, 0, sizeof(c));		long long ans=0;		for (int i=1; i<=n; i++){   			update(a[i], 1);			ans+=(i-sum(a[i]));		}		printf("%lld\n", ans);	}	return 0;}
       
      
     
    
   

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