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

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
",cnt);
37     }
38     return 0;
39 }

View Code