inversion count

I am trying to solve this(http://www.spoj.com/problems/INVCNT/) question... I ma able to sort the array correctly using merge sort but am not able to get the correct inversion count...any help? my solution

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
#define getcx getchar_unlocked
#ifndef ONLINE_JUDGE
    #define getcx getchar
#endif
#define l long long
#define rep(i,n) for(l i=0;i<n;++i)
#define pi(n) printf("%d\n",n)
#define pl(n) printf("%lld\n",n)
#define MAX 100000
using namespace std;
inline void string(char *str)
{
    register char c = 0;
    register int i = 0;
    while (c < 33)
        c = getcx();
    while (c != '\n') {
        str[i] = c;
        c = getcx();
        i = i + 1;
    }
    str[i] = '\0';
}
inline int inp()
{
   int n=0;
   int ch=getcx();int sign=1;
   while( ch < '0' || ch > '9' ){if(ch=='-')sign=-1; ch=getcx();}

   while(  ch >= '0' && ch <= '9' )
           n = (n<<3)+(n<<1) + ch-'0', ch=getcx();
   return n*sign;
}
inline long long in()
{
   long long n=0;
   long long ch=getcx();long long sign=1;
   while( ch < '0' || ch > '9' ){if(ch=='-')sign=-1; ch=getcx();}

   while(  ch >= '0' && ch <= '9' )
           n = (n<<3)+(n<<1) + ch-'0', ch=getcx();
   return n*sign;
}
l merge(l a[200001],l p,l q,l r)
{
    l n=q-p+1,m=r-q,count=0;
    l b[n+1],c[m+1];
    for(l i=0;i<n;++i)  //loop to store left sub array in temporary array
    b[i]=a[p+i];
    for(l i=0;i<m;++i)//loop to store right sub array in temporary array
    c[i]=a[q+1+i];
    l k=p,i=0,j=0;
    for(;i<n&&j<m;++k)  //storing the values in original array
    {
        if(b[i]<=c[j])
        {
            a[k]=b[i];
            ++i;    
        }
        else
        {
            a[k]=c[j];
            ++j;
            count+=q-i;//if element in left sub array is greater than element in right sub array then number of inversions = q(mid)-i
        }
    }
    while(i<n)
    {
        a[k]=b[i];
        ++i;++k;
    }
    while(j<m)
    {
        a[k]=c[j];
        ++k;++j;
    }
    return count;
}
l mergesort(l a[200001],l p,l r)
{
    l count=0;
    if(p<r)
    {
        l q=p+(r-p)/2;
        count=mergesort(a,p,q);
        count+=mergesort(a,q+1,r);
        count+=merge(a,p,q,r);
    }       
    return count;
}
main()
{
    l t,n,a[200001],count;
    t=in();
    while(t--)
    {
        n=in();
        for(l i=0;i<n;++i)
        a[i]=in();
        count=mergesort(a,0,n-1);
    //  for(l i=0;i<n;++i)
    //  cout<<a[i]<<" ";
        cout<<count;
        cout<<endl;
    }
}