If the sum of all numbers is bounded by $$$C$$$, we can solve the problem in $$$O\left(\frac{nC} {64}\right) $$$ using bitsets.

i don't know but first thing i think about and i think it's wrong or need some optimization if i use sliding window to get the number of even length then make pointer in the first of the window and another in the end of the window then i do ( right — left ) / 2 if it was even then count++

ok i don't have Competitive programmers friends to ask them so i give my idea to ChatGPT and it will take O($$$n^2$$$) too but gpt said use prefix sum i really didn't understand it well but i hope i was useful or i have discussion in this blog and try to find optimal solution and learn more things

GPT code

void solve(vector<int>& arr) {
    int n = arr.size();
    int count = 0;

    for (int left = 0; left < n; left++) {
        for (int right = left + 1; right < n; right += 2) { 
            int mid = (left + right) / 2;
            int sum1 = accumulate(arr.begin() + left, arr.begin() + mid + 1, 0);
            int sum2 = accumulate(arr.begin() + mid + 1, arr.begin() + right + 1, 0);
            if (sum1 == sum2) {
                count++;
            }
        }
    }
    cout << count << "\n";
}

Maybe it can be done using binarysearch+hashing

https://en.wikipedia.org/wiki/3SUM

i think it's related (or i guess check "is answer = 0" is equal hard than 3SUM)