POJ 2299 Ultra-QuickSort

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 33404		Accepted: 11953

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

Source

Waterloo local 2005.02.05

题目要求是找出冒泡排序中交换的次数，就是统计逆序数，但是用冒泡排序会超时。

所以选择归并排序。

下面是代码。

#include <iostream>
#include <cstdio>
#include <string.h>
using namespace std;
long long int sum;
int mergearray(int a[],int start,int mid,int end,int temp[])
{
int m=start,n=mid+1;
int k=start;
while(m<=mid&&n<=end)
{
if(a[m]<=a
)
{
temp[k]=a[m];
m++;
k++;
}
else
{
temp[k]=a
;
n++;
k++;
sum+=mid-m+1;
}
}
if(m<=mid)
while(m<=mid)
{
temp[k]=a[m];
k++;
m++;
}
else
while(n<=end)
{
temp[k]=a
;
k++;
n++;
}
for(int i=start;i<=end;i++)
{
a[i]=temp[i];
}
}
int mergesort(int a[],int start,int end,int temp[])
{
if(start==end)
temp[start]=a[start];
else if(end>start)
{
int mid=(start+end)/2;
mergesort(a,start,mid,temp);
mergesort(a,mid+1,end,temp);
mergearray(a,start,mid,end,temp);
}

}
int main()
{

int limit;int a[500050],b[500050];
while(scanf("%d",&limit)&&limit)
{
for(int i=1;i<=limit;i++)
scanf("%d",&a[i]);
mergesort(a,1,limit,b);
printf("%lld\n",sum);
sum=0;
}
return 0;

}