Counting the inversions in an Array

This is my implementation for counting the inversions in an Array.

	#include <stdio.h>
	#include <stdlib.h>
	#include <string.h>
	#include <math.h>
	#define MAXSIZ 100010
	long long int sortandcount(long long int a[],long long int low,long long int high);
	long long int mergeandcountsplit(long long int b[],long long int lowi,long long int midi,long long int highi);
	int main(int argc, char const *argv[])
	{
	long long int array[MAXSIZ]/*input array*/, n/*array size*/;
	long long int inversions,i;
	scanf("%lld", &n);
	for (i = 0; i < n; ++i)
	scanf("%lld", &array[i]);/*reading*/
	inversions=sortandcount(array,0,n-1);
	printf("%lld\n", inversions);
	return 0;
	}
	long long int sortandcount(long long int a[],long long int low,long long int high)
	{
	long long int x=0,y=0,z=0,mid=(low+high)/2;
	if (low<high)
	{
	x=sortandcount(a,low, mid);
	y=sortandcount(a,mid+1,high);
	z=mergeandcountsplit(a,low,mid,high);
	}
	return x+y+z;
	}
	long long int mergeandcountsplit(long long int b[],long long int lowi,long long int midi,long long int highi)
	{
	long long int stepmother[MAXSIZ];/*step mother for our sorting array*/
	/*It's very pity that, every time step
	mother grows his child in a sorted way
	from an unsorted and then every time
	the natural mother reclaims the sorted son*/
	/*too pity*/
	long long int i=lowi,j=midi+1,k=0;
	long long int inversion_S=0;
	while(i<=midi&&j<=highi)
	{
	if (b[i]<=b[j])
	{
	stepmother[k++]=b[i++];
	}
	else
	{
	inversion_S+=midi-i+1;/*Birthday Party*/
	stepmother[k++]=b[j++];
	}
	}
	while(i<=midi)
	stepmother[k++]=b[i++];
	while(j<=highi)
	stepmother[k++]=b[j++];
	while(k--)
	b[lowi+k]=stepmother[k];
	return inversion_S;
	}

view raw inversions.c hosted with ❤ by GitHub

I was really amazed to find that counting inversions in an array is used in collaborative filtering i.e for generating recommendations for a new user.

Now I am getting that Codejam problem called Rope Intranet which is about finding the Inversions of the second building values as the first building values will be given in a sorted order.