Xor sum

// ~/.vim/cpp_template.cpp
#include <bits/stdc++.h>
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>

#define REP(i,a,b) for (int i = a; i < b; i++)
#define ll long long 

// Type Aliases for 1D and 2D vectors with initialization
#define vi(n, val) vector<int>(n, val) // 1D vector of ints with size n, initialized to val

#define vll(n, val) vector<long long>(n, val) // 1D vector of long longs with size n, initialized to val
#define ll long long
#define vvi(n, m, val) vector<vector<int>>(n, vector<int>(m, val)) // 2D vector of ints (n x m), initialized to val
#define vvll(n, m, val) vector<vector<long long>>(n, vector<long long>(m, val)) // 2D vector of long longs (n x m), initialized to val


using namespace std;

template <typename T>
void pV(const std::vector<T>& vec, const std::string& label = "Vector") {
    std::cout << label << ": [ ";
    for (const auto& elem : vec) {
        std::cout << elem << " ";
    }
    std::cout << "]" << std::endl;
}


using namespace std;


void dfs(int s, vector<bool> *visited, vector<int> (*adj)[]) {
    if ((*visited)[s]) return;
    (*visited)[s] = true;
    // process node s


    for (auto u: (*adj)[s]) {
        dfs(u, visited, adj);
    }
}
/*
vector<int> adj[N];                                                          
vector<bool> visited(N, false);
int u, v;
for(int i = 0; i < M;i++){
    cin >> u >> v;
    u--;
    v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
}
*/

int qxor(int a, int b, ll n, vector<int>* tree) {
    a += n; b += n;
    int s = 0;
    while (a <= b) {
        //cout << a << ' ' << b << endl;
        if (a%2 == 1) s = (*tree)[a++] ^ s;
        if (b%2 == 0) s = (*tree)[b--] ^ s;
        a /= 2; b /= 2;
    }
    return s;
}

void change(int k, int u, ll n, vector<int>* tree) {
    k += n;
    (*tree)[k] = u;
    for (k /= 2; k >= 1; k /= 2) {
        (*tree)[k] = min((*tree)[2*k], (*tree)[2*k+1]);
    }
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    // Your code starts here
    int n, q;
    cin >> n >> q;
    ll n_2 = pow(2, ceil(log(n)/log(2))); 

    vector<int> v = vi(2*n_2+1, 1e9);
    for(ll i = n_2; i < n_2+n; i++){
        cin >> v[i];
    }

    for(ll i = n_2-1; i > 0; i--){
        v[i] = v[2*i] ^  v[2*i+1];
    }

    REP(i, 0, q){
        int a, b;
        cin >> a >> b;
        ++a;++b;
        if(a == b){
            cout << v[n_2-2+a] << endl;
            continue;
        }
        cout << qxor(a, b, n_2-2, &v) << endl;
    }



    
    
    return 0;
}

Test 1

Verdict: ACCEPTED

Test 2

Verdict: ACCEPTED