Xor sum

#ifdef ONLINE_JUDGE
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
#endif
#include "bits/stdc++.h"
#define fast ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define fill(arr,val) memset(arr,val,sizeof(arr))
#define FOR(i, a, b) for(__typeof(b) i = a, _b = b; i <= _b; ++i)
#define FORD(i, a, b) for(__typeof(a) i = a, _b = b; i >= _b; --i)
#define ALL(a) (a).begin(), (a).end()
#define YES cout << "YES\n"
#define NO cout << "NO\n"
#define ll long long
#define fi first
#define se second
#define pb push_back
#define pf push_front
#define ii pair<int,int>
#define iii pair<int,pair<int,int>>
#define dq deque<int>
#define nend '\n'
using namespace std;

const ll inf = 1e18;
const int dx[] = {1,0,-1,0};
const int dy[] = {0,1,0,-1};
const int MOD = 1e9 + 7;
const ll hashi = 2e9 + 11;

inline ll add(ll a, ll b) { return (a + b) % MOD; }
inline ll sub(ll a, ll b) { return ((a - b) % MOD + MOD) % MOD; }
inline ll mul(ll a, ll b) { return (a * b) % MOD; }

inline ll binpow(ll a, ll b) {
    ll res = 1;
    a %= MOD;
    while (b > 0) {
        if (b & 1) res = res * a % MOD;
        a = a * a % MOD;
        b >>= 1;
    }
    return res;
}

inline ll modInverse(ll n) {
    return binpow(n, MOD - 2);
}

inline ll modDivide(ll a, ll b) {
    return (a % MOD * modInverse(b)) % MOD;
}

ll extended_gcd(ll a, ll b, ll& x, ll& y) {
    if (b == 0) {
        x = 1; y = 0;
        return a;
    }
    ll x1, y1;
    ll d = extended_gcd(b, a % b, x1, y1);
    x = y1;
    y = x1 - y1 * (a / b);
    return d;
}

void init_code() {
    fast;
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif
}
/// CSES - Range Xor Queries
/// No updates -> prefix xor, O(1) per query
/// p[i] = x[1]^x[2]^...^x[i]
/// xor of [a,b] = p[b]^p[a-1]  (x^x = 0, so the part before a cancels,
///                              same idea as prefix sum: s[b]-s[a-1])
/// If the problem had updates: use segment tree, replace + with ^
const int N=2e5+2;
int n,q,p[N];
void solve() {
    cin >> n >> q;
    FOR(i,1,n) {
        int x;
        cin >> x;
        p[i]=p[i-1]^x;
    }
    while (q--) {
        int a,b;
        cin >> a >> b;
        cout << (p[b]^p[a-1]) << nend;
    }
}
int main() {
    init_code();

    int t = 1;
    // cin >> t;
    while (t--) {
        solve();
    }

}

Test 1

Verdict: ACCEPTED

Test 2

Verdict: ACCEPTED