POJ2299--归并排序--Ultra-QuickSort

最新推荐文章于 2014-04-10 16:23:43 发布

SCUT_Pein

最新推荐文章于 2014-04-10 16:23:43 发布

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分类专栏：排序文章标签： POJ2299 归并排序

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本文探讨了一种名为Ultra-QuickSort的特定排序算法，该算法通过交换相邻元素来对整数序列进行升序排列。文章详细介绍了算法的输入输出格式，并提供了一个示例输入输出。此外，还给出了一段C++代码实现，用于计算排序过程中所需的交换次数。

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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

Sample Output

比较坑，输出既然要用longlong才能过

#include <cstdio>
using namespace std;
int A[500008],AA[500008];
long long int sum;
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+1,i=x;
		merge_sort(A,x,m,T);
		merge_sort(A,m+1,y,T);
		while(p<=m||q<=y)
		{
			if(q>y||(p<=m&&A[p]<=A[q]))
			{
				T[i++]=A[p++];
			}
			else
			{
				T[i++]=A[q++];
				sum+=q-i;
			}
		}
		for(int i=x;i<=y;i++)
		{
			A[i]=T[i];
		}
	}
}
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF&&n)
	{
		sum=0;
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&A[i]);
		}
		merge_sort(A,1,n,AA);
		printf("%lld\n",sum);
	}
	return 0;
}