ACM归并排序

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

解题思路：

题目大意是输入个数字，将他们按从小到大的顺序排列，问需要经过多少次的交换。这个题目我们要用归并排序的方法解决。

介绍一下归并排序：1、把序列分成元素个数尽量相等的两半；2、把两半的元素分别排序；3、把两个有序表合并成一个。归并排序的时间复杂度为O（nlogn）
注意事项：只要有一个序列非空，就要继续合并。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
int  A[500004],T[500004];
long long  s;
using namespace std;
void merge_sort(int *A,int x,int y,int *T)
{
    if(y-x>1)
    {
        int m=x+(y-x)/2;
        int p=x,q=m,i=x;
        merge_sort(A,x,m,T);
        merge_sort(A,m,y,T);
        while(p<m||q<y)
        {
            if(q>=y||(p<m&&A[p]<=A[q]))//只要一个序列为空就要继续合并
                T[i++]=A[p++];
            else
            {
                T[i++]=A[q++];
                s=s+m-p;

            }
        }
        for(i=x;i<y;i++)
            A[i]=T[i];
    }
}
int main()
{
    int n;
    while(scanf("%d",&n)&&n)
    {
        for(int i=0;i<n;i++)
            scanf("%d",&A[i]);
        s=0;
        //memset(T,0,sizeof(T));
        merge_sort(A,0,n,T);//不能把n写成n-1，否则会答案错误
        printf("%I64d
",s);
    }

    return 0;
}