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

59105431230

Sample Output

60

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;}