Ultra-QuickSort(归并排序求逆序对数)

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 34283		Accepted: 12295

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

//归并排序模板，只加了一句 cnt += m-p;

 1 #include<stdio.h>

 2 #include<string.h>

 3 long long  cnt;

 4 int num[500010],tmp[500010];

 5 void merge_sort(int num[],int x, int y, int tmp[])

 6 {

 7     if(y-x > 1)

 8     {

 9         int m = x + (y-x)/2;

10         int p = x, q = m, i = x;

11         merge_sort(num,x,m,tmp);

12         merge_sort(num,m,y,tmp);

13         while(p < m || q < y)

14         {

15             if(q >= y || (p < m && num[p] <= num[q]))

16                 tmp[i++] = num[p++];

17             else

18             {

19                 tmp[i++] = num[q++];

20                 cnt += m-p;

21             }

22         }

23         for(i = x; i < y; i++)

24             num[i] = tmp[i];

25     }

26 }

27 int main()

28 {

29     int n;

30     while(~scanf("%d",&n) && n)

31     {

32         for(int i = 0; i < n; i++)

33             scanf("%d",&num[i]);

34         cnt = 0;

35         merge_sort(num,0,n,tmp);

36         printf("%I64d\n",cnt);

37     }

38     return 0;

39 }

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