HackerEarth Xor sequence problem solution

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

#define ms(s, n) memset(s, n, sizeof(s))
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define FORd(i, a, b) for (int i = (a) - 1; i >= (b); --i)
#define FORall(it, a) for (__typeof((a).begin()) it = (a).begin(); it != (a).end(); it++)
#define sz(a) int((a).size())
#define present(t, x) (t.find(x) != t.end())
#define all(a) (a).begin(), (a).end()
#define uni(a) (a).erase(unique(all(a)), (a).end())
#define pb push_back
#define pf push_front
#define mp make_pair
#define fi first
#define se second
#define prec(n) fixed<<setprecision(n)
#define bit(n, i) (((n) >> (i)) & 1)
#define bitcount(n) __builtin_popcountll(n)
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pi;
typedef vector<int> vi;
typedef vector<pi> vii;
const int MOD = (int) 1e9 + 7;
const int FFTMOD = 1007681537;
const int INF = (int) 1e9;
const ll LINF = (ll) 1e18;
const ld PI = acos((ld) -1);
const ld EPS = 1e-9;
inline ll gcd(ll a, ll b) {ll r; while (b) {r = a % b; a = b; b = r;} return a;}
inline ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
inline ll fpow(ll n, ll k, int p = MOD) {ll r = 1; for (; k; k >>= 1) {if (k & 1) r = r * n % p; n = n * n % p;} return r;}
template<class T> inline int chkmin(T& a, const T& val) {return val < a ? a = val, 1 : 0;}
template<class T> inline int chkmax(T& a, const T& val) {return a < val ? a = val, 1 : 0;}
inline ll isqrt(ll k) {ll r = sqrt(k) + 1; while (r * r > k) r--; return r;}
inline ll icbrt(ll k) {ll r = cbrt(k) + 1; while (r * r * r > k) r--; return r;}
inline void addmod(int& a, int val, int p = MOD) {if ((a = (a + val)) >= p) a -= p;}
inline void submod(int& a, int val, int p = MOD) {if ((a = (a - val)) < 0) a += p;}
inline int mult(int a, int b, int p = MOD) {return (ll) a * b % p;}
inline int inv(int a, int p = MOD) {return fpow(a, p - 2, p);}
inline int sign(ld x) {return x < -EPS ? -1 : x > +EPS;}
inline int sign(ld x, ld y) {return sign(x - y);}
#define db(x) cerr << #x << " = " << (x) << " ";
#define endln cerr << "n";

const int maxn = 2e4 + 5;
const int logn = 31;
const int magic = 150;
int n, m, t;
int a[maxn];
int b[maxn];

int ptr;
int nxt[maxn * logn][2];

void clear() {
    FOR(i, 0, ptr + 1) {
        nxt[i][0] = nxt[i][1] = 0;
    }
    ptr = 0;
}

void add(int val) {
    int k = 0;
    FORd(i, logn, 0) {
        int x = bit(val, i);
        if (!nxt[k][x]) nxt[k][x] = ++ptr;
        k = nxt[k][x];
    }
}

int query(int val) {
    int k = 0, res = 0;
    FORd(i, logn, 0) {
        int x = bit(val, i);
        res <<= 1;
        if (nxt[k][x ^ 1]) {
            k = nxt[k][x ^ 1];
            res |= 1;
        }
        else {
            k = nxt[k][x];
        }
    }
    return res;
}

void chemthan() {
    cin >> n >> m >> t;
    assert(1 <= n && n <= 2e4);
    assert(1 <= m && m <= 2e4);
    assert(0 <= t && t <= 1);
    FOR(i, 1, n + 1) {
        cin >> a[i];
        assert(0 <= a[i] && a[i] <= 1e9);
        b[i] = a[i], a[i] ^= a[i - 1];
    }
    FORd(i, n + 1, 1) b[i] ^= b[i + 1];
    static int f[magic][maxn];
    static int g[magic][maxn];
    for (int st = 1; st <= n; st += magic) {
        clear();
        int ix = (st - 1) / magic;
        int mx = 0;
        FOR(i, st, n + 1) {
            add(a[i - 1]);
            chkmax(mx, query(a[i]));
            f[ix][i] = mx;
        }
    }
    for (int st = magic; st <= n; st += magic) {
        clear();
        int ix = st / magic - 1;
        int mx = 0;
        FORd(i, st + 1, 1) {
            add(b[i + 1]);
            chkmax(mx, query(b[i]));
            g[ix][i] = mx;
        }
    }
    int ans = 0;
    while (m--) {
        int x, y; cin >> x >> y;
        assert(0 <= x && x <= 1e9);
        assert(0 <= y && y <= 1e9);
        int l = (x + (long long) ans * t) % n + 1;
        int r = (y + (long long) ans * t) % n + 1;
        if (l > r) swap(l, r);
        if (r - l + 1 <= magic) {
            ans = 0;
            clear();
            FOR(i, l, r + 1) {
                add(a[i - 1]);
                chkmax(ans, query(a[i]));
            }
            cout << ans << "n";
            continue;
        }
        int il = (l - 1 + magic - 1) / magic;
        int ir = r / magic - 1;
        ans = max(f[il][r], g[ir][l]);
        clear();
        FOR(i, l, il * magic + 1) {
            add(a[i - 1]);
        }
        FOR(i, (ir + 1) * magic + 1, r + 1) {
            chkmax(ans, query(a[i]));
        }
        cout << ans << "n";
    }
}

int main(int argc, char* argv[]) {
    ios_base::sync_with_stdio(0), cin.tie(0);
    if (argc > 1) {
        assert(freopen(argv[1], "r", stdin));
    }
    if (argc > 2) {
        assert(freopen(argv[2], "wb", stdout));
    }
    chemthan();
    cerr << "nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "msn";
    return 0;
}