http://poj.org/problem?id=2299
方法：
利用merge sort求逆序数，复杂度nlgn
如果比较每一对儿数（或使用bubble sort），复杂度为n^2，太慢
对于一对儿逆序对儿，有三种情况：两数都在数组左半边，两数都在数组右半边，两数分别在左右两个半边。
由于Merge时两个半边都为sorted，所以只会出现第三种情况。
计算逆序数只需要在Merge中加一句即可，当a[lb]>a[rb]时,a[lb...le]>a[rb]，所以逆序数为(le-lb+1)
由于题目中n<500,000，所以逆序数为O(n^2)，需要使用__int64存放，不然会溢出
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
2: __int64 totalReverse = 0 ;
3:
int re)
5: {
int le = rb-1 ;
int index = lb ;
int cpback = lb ;
while( lb<=le && rb<=re )
10: {
if( a[lb] <= a[rb] )
12: tmpArr[index++] = a[lb++] ;
else
14: {
15: tmpArr[index++] = a[rb++] ;
//*** counting the reverse-pairs
17: }
18: }
19:
//copy the rest to the temp array
while(lb<=le)
22: tmpArr[index++] = a[lb++] ;
while(rb<=re)
24: tmpArr[index++] = a[rb++] ;
25:
//copy back
for( ; cpback<=re ; ++cpback )
28: a[cpback] = tmpArr[cpback] ;
29: }
30:
int e)
32: {
if(b<e)
34: {
int mid = b+(e-b)/2 ;
36: Msort(a,tmpArr,b,mid);
37: Msort(a,tmpArr,mid+1,e);
38: Merge(a,tmpArr,b,mid+1,e);
39: }
40: }
41:
int n)
43: {
44: Msort(a,tmpArr,0,n-1) ;
45: }
46:
void run2299()
48: {
int n ;
int i;
int[500000];
int[500000];
while(cin>>n)
54: {
if(n==0)
break ;
57:
for(i=0;i<n;++i)
59: cin>>a[i] ;
60: MergeSort(a,tmpArr,n) ;
61: cout<<totalReverse<<endl;
62: totalReverse = 0 ;
63: }
delete []a;
delete []tmpArr;
66: }