C++ Counting inversions in array, Fatal Signal 11 (BIT)

I was given this challenge in a programming "class". Eventually I decided to go for the "Binary Indexed Trees" solution, as data structures are a thing I'd like to know more about. Implementing BIT was somewhat straight forward, things after that - not so much. I ran into "Fatal Signal 11" when uploading the solution to the server, which, from what I've read, is somewhat similar to a Null pointer exception. Couldn't figure out the problem, decided to check out other solutions with BIT but stumbled upon the same problem.

#include<iostream>
using namespace std;



/*    <BLACK MAGIC COPIED FROM geeksforgeeks.org>     */
int getSum(int BITree[], int index){
    int sum = 0;
    while (index > 0){
        sum += BITree[index];
        index -= index & (-index);
    }
    return sum;
}
void updateBIT(int BITree[], int n, int index, int val){
    while (index <= n){
       BITree[index] += val;
       index += index & (-index);
    }
}
/*    <BLACK MAGIC COPIED FROM geeksforgeeks.org>     */




int Count(int arr[], int x){
    int sum = 0;
    int biggest = 0;

    for (int i=0; i<x; i++) {
        if (biggest < arr[i]) biggest = arr[i];
    }

    int bit[biggest+1];

    for (int i=1; i<=biggest; i++) bit[i] = 0;

    for (int i=x-1; i>=0; i--)
    {
        sum += getSum(bit, arr[i]-1);
        updateBIT(bit, biggest, arr[i], 1);
    }
    return sum;
}

int main(){
    int x;
    cin >> x;


    int *arr = new int[x];
    for (int temp = 0; temp < x; temp++) cin >> arr[temp];

        /*sizeof(arr) / sizeof(arr[0]); <-- someone suggested this, 
        but it doesn't change anything from what I can tell*/

    cout << Count(arr,x);

    delete [] arr;
    return 0;
}

I am quite stumped on this. It could be just some simple thing I'm missing, but I really don't know. Any help is much appreciated!

Solution

You have condition that every number lies between 1 and 10¹⁸. So, your biggest number can be 10¹⁸. This is too much for the following line:

int bit[biggest+1];