POJ 2299, Ultra-QuickSort

Time Limit: 7000MS Memory Limit: 65536K
Total Submissions: 12873 Accepted: 4518

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

// POJ2299.cpp : Defines the entry point for the console application.
//

#include <iostream>
using namespace std;

static unsigned long long cnt = 0;
void merge(int* a, int* b, int beg, int mid, int end)
{
int i = beg,k = 0,j = mid + 1;
while (i <= mid && j <= end)
if (a[i] <= a[j])
{
b[k] = a[i];
++i;
++k;
}
else
{
b[k] = a[j];
++j;
++k;
cnt += mid + 1 - i;
};

while (i <= mid)
{
b[k] = a[i];
++i;
++k;
}

while (j <= end)
{
b[k] = a[j];
++j;
++k;
}

copy(&b[0],&b[k],&a[beg]);
};

void mergeSort(int* a, int* b, int beg, int end)
{
if (beg < end)
{
int mid = (beg + end) >> 1;
mergeSort(a, b, beg, mid);
mergeSort(a, b, mid + 1, end);
merge(a, b, beg, mid, end);
}
};

int main(int argc, char* argv[])
{
int N;
int a[500001], b[500001];
while (cin >> N && N != 0)
{
for (int i = 0; i < N; ++i) scanf("%d", &a[i]);
cnt = 0;
mergeSort(a,b,0,N - 1);

cout << cnt <<endl;
}
return 0;
}