XOR Sum 2

#include <bits/stdc++.h>
// #include <atcoder/all>
using namespace std;
#define int long long
#define all(x) (x).begin(), (x).end()
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pair<int, int>> vpi;
typedef pair<int, int> pi;
#define f first
#define s second
#define pb push_back
#define endl "\n"
#define yes cout << "YES" << endl
#define no cout << "NO" << endl
#define init(x, a) memset(x, a, sizeof(x))
const int mod1 = 1e9 + 7, mod2 = 998244353, INF = 2e18, N = 2e5 + 5, L = 19;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
// -----------------------------------------------------------------------------

void solve()
{
    int n;
    cin >> n;
    vi a(n); // a[i] < 2^20
    vi prefxor(n + 1, 0), prefsum(n + 1, 0);
    for (int i = 0; i < n; i++)
    {
        cin >> a[i];
        prefxor[i + 1] = prefxor[i] ^ a[i];
        prefsum[i + 1] = prefsum[i] + a[i];
    }
    // acc to pigeonhole principle, we can have subarray size of atmost 20
    // if it's more than 20, then 1 bit would he repeated, so carry over would be there so XOR != SUM
    int ans = 0;
    for (int k = 0; k < 25; k++) // subarray length
    {
        for (int i = 0; i < n; i++)
        {
            if (i + k >= n)
            {
                break;
            }
            int x = prefxor[i] ^ prefxor[i + k + 1];
            int y = prefsum[i + k + 1] - prefsum[i];
            if (x == y)
            {
                ans++;
            }
        }
    }
    cout << ans << endl;
    return;
}

signed main()
{
    // __START__;
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t = 1;
    // cin >> t;
    while (t--)
    {
        solve();
    }
    // __END__;
    return 0;
}