HackerEarth Bit operations problem solution

1. L R X: For each element in the range [L, R] such as y, set y = y|X.
2. L R X: For each element in the range [L,R] like y, set y = y&X.
3. L R X: For each element in the range [L,R] like y, set y = y xor x.
4. L R: Print the sum of the elements in range[L,R].
5. L R: Print the XOR of the elements in range [L,R].

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e6 + 17;
int n, q, a[maxn];
int main(){
    ios::sync_with_stdio(0), cin.tie(0);
    cin >> n >> q;
    while(q--){
        int t, l, r, x;
        cin >> t >> l >> r;
        l--;
        if(t < 4){
            cin >> x;
            while(l < r)
                if(t == 1)
                    a[l++] |= x;
                else if(t == 2)
                    a[l++] &= x;
                else
                    a[l++] ^= x;
        }
        else if(t == 4)
            cout << accumulate(a + l, a + r, 0ll) << 'n';
        else
            cout << accumulate(a + l, a + r, 0, [](int a, int b){  return a ^ b;  }) << 'n';
    }
}