Countring the number of inversions in an array using divide and conguer algorithm

gistfile1.txt

#include <iostream>
#include <string>
#include <sstream>
#include <queue>
#include <stdio.h>
#include <fstream>
using namespace std;

#define INF 19999999
long long int glob;

long long int merge( int *arr, int beg, int mid, int end ) {
  queue<int> left;
  queue<int> right;

  for( int i=beg; i<=mid; ++i ) {
    left.push(arr[i]);
  }
  for( int i=mid+1; i<=end; ++i ) {
    right.push(arr[i]);
  }

  int index=beg;
  int ret=0;

  while( !left.empty() && !right.empty() ) {
    if( left.front() <= right.front() ) {
      arr[index++] = left.front();
      left.pop();
    } else {
      arr[index++] = right.front();
      right.pop();
      ret+=left.size();
    }
  }

  while( !left.empty() ) { arr[index++]=left.front();left.pop(); }
  while( !right.empty() ) { arr[index++]=right.front();right.pop(); }

  return ret;
}


void mergesortInvCount(int *arr, int beg, int end ) {
  if( beg < end ) {
    int mid = (int)((beg+end)/2);
    mergesortInvCount( arr, beg, mid );
    mergesortInvCount( arr, mid+1, end );
    glob += merge(arr, beg, mid, end );
  }
}

int main() {
  int arr[] =  {2, 4, 1, 3, 5};//{1, 20, 6, 4, 5};
  
  mergesortInvCount(arr,0,4);
  printf(" Number of inversions are %d \n",glob);

  system("pause");
  return 0;
}