# HackerEarth Big Number Array problem solution

In this HackerEarth Big Number Array problem solution, we have given an array of n integers. Initially, all elements are zero. You are asked to complete q queries of two kinds:
1. x y l r: for each i in range [x, y] flip the l-th bit to r-bit of i-th element.
2. x y: check if x-th element equals y-th element.

## HackerEarth Big Number Array 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 pconent(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 = 1 << 20;long long val[maxn];int n, q;int op[maxn];int x[maxn];int y[maxn];int l[maxn];int r[maxn];int myrand() {    return abs(2311 * rand() * rand() + 1992 * rand());}long long fen[maxn];void upd(int p, long long val) {    p++;    for (; p <= n; p += p & -p) {        fen[p] ^= val;    }}long long query(int p) {    p++;    long long res = 0;    for (; p > 0; p -= p & -p) {        res ^= fen[p];    }    return res;}void chemthan() {    srand(2311);    FOR(i, 0, maxn) val[i] = (long long) myrand() * myrand();    int test; cin >> test;    assert(1 <= test && test <= 100);    int sumn = 0, sumq = 0;    while (test--) {        cin >> n >> q;        assert(1 <= n && n <= 1e5);        assert(1 <= q && q <= 1e5);        sumn += n, sumq += q;        assert(1 <= sumn && sumn <= 2e5);        assert(1 <= sumq && sumq <= 2e5);        vi dc;        FOR(i, 0, q) {            cin >> op[i] >> x[i] >> y[i], x[i]--, y[i]--;            assert(1 <= op[i] && op[i] <= 2);            assert(0 <= x[i] && x[i] <= y[i] && y[i] < n);            if (op[i] == 1) {                cin >> l[i] >> r[i];                assert(0 <= l[i] && l[i] <= r[i] && r[i] <= 1e9);                l[i]--;                dc.pb(l[i]), dc.pb(r[i]);            }        }        sort(all(dc)), uni(dc);        fill_n(fen, n + 1, 0);        FOR(i, 0, q) {            if (op[i] == 1) {                int ll = lower_bound(all(dc), l[i]) - dc.begin();                int rr = lower_bound(all(dc), r[i]) - dc.begin();                upd(x[i], val[ll] ^ val[rr]);                upd(y[i] + 1, val[ll] ^ val[rr]);            }            else {                if (query(x[i]) == query(y[i])) {                    cout << "YES\n";                }                else {                    cout << "NO\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 << "ms\n";    return 0;}`

### Second solution

`t = int(raw_input())maxn = 101010xx = {}def getVal(v):    if v in xx:        return xx[v]    import random    val = random.randint(0, (1L << 100) - 1)    xx[v] = val    return valdef update(f, p, v):    while p < len(f):        f[p] ^= v        p += p & -pdef query(f, p):    r = 0    while p > 0:        r ^= f[p]        p -= p & -p    return rfor ____ in xrange(t):    n,q = map(int, raw_input().split())    f = [0 for __ in xrange(maxn)]    for ___ in xrange(q):        s = map(int, raw_input().split())        if s[0] == 1:            v = getVal(s[3]-1)^getVal(s[4])            update(f, s[2]+2, v)            update(f, s[1]+1, v)        else:            print "YES" * (query(f, s[1]+1) == query(f, s[2]+1)) or "NO"`