复习--Ultra-QuickSort（归并排序求逆序数）

Ultra-QuickSort
Time Limit:7000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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

每一次归并时，如果要放右侧的数时，那么逆序数 += （ mid - i + 1 ）

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
__int64 a[500010] , b[500010] , num ;
void f(int x,int y)
{
int i , j , k ;
int m = (x + y) / 2 ;
if( y-x > 1)
{
f(x,m);
f(m+1,y);
}
i = x ;
j = m+1 ;
k = x ;
while( i <= m || j <= y )
{
if( j > y || ( i <= m && a[i] < a[j] ) )
b[k++] = a[i++] ;
else
{
b[k++] = a[j++] ;
num += m-i+1 ;
}
}
for(i = x ; i <= y ; i++)
a[i] = b[i] ;
}
int main()
{
int i , j , n ;
while(scanf("%d", &n) && n)
{
num = 0 ;
for(i = 1 ; i <= n ; i++)
scanf("%I64d", &a[i]);
f(1,n);
printf("%I64d\n", num);
}
return 0;
}