?Ultra-QuickSort

Ultra-QuickSort

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
题目大意就是让你计算一个冒泡排序中，需要交换的次数。以为数据量较大，所以我们这里用到了Merge Sort :

 1 #include<stdio.h>
 2 #include<string.h>
 3 int temp[500005] ;
 4 int a[500005] ;
 5 __int64 number ;
 6 void MergeSort( int a[] , int fir , int end )
 7 {
 8     int len = end - fir ;
 9     if( len <= 1 )
10         return ;
11     int mid = fir + len/2 ;
12     MergeSort( a , fir , mid ) ;
13     MergeSort( a , mid , end ) ;
14     int p1 = fir , p2 = mid ;
15     for( int i = fir ; i < end ; i++ )
16     {
17         if( p1 == mid )
18         {
19             temp[i] = a[p2++] ;
20         }
21         else if( p2 == end )
22         {
23             temp[i] = a[p1++] ;
24         }
25         else
26         {
27             if( a[p1] >= a[p2] )
28             {
29                 temp[i] = a[p2++] ;
30                 number += mid - p1 ;//在Merge Sort 排序中仅仅多加了这句话
31             }
32             else
33             {
34                 temp[i] = a[p1++] ;
35             }
36         }
37     }
38
39     for(int i = fir ; i < end ; i++ )
40     {
41         a[i] = temp[i] ;
42     }
43 }
44 int main()
45 {
46    // freopen("a.txt" ,"r" , stdin );
47     int n ;
48     while( scanf("%d" , &n ) != EOF )
49     {
50         if( n == 0 ) break;
51         number = 0 ;
52         for(int i = 0 ; i < n ; i++ )
53             scanf("%d" , &a[i] ) ;
54         MergeSort( a , 0 , n ) ;
55
56         printf("%I64d\n" , number );
57     }
58     return 0 ;
59 }

AC