# HackerEarth Boring Not boring problem solution

In this HackerEarth Boring Not boring problem solution You are given two integers n(1 <= n <= 10^5) and m(1 <= m <= 10^5). Initially, you have an array a of n zeros, and m applied queries on it. Query is given as l r x, where you apply ai = ai xor x (xor denotes the operation bitwise XOR) for every i between l and r. But this problem seems boring, so Almas modify some queries interval. By modifying, he applied that operation on a subset of positions on a segment(maybe an empty subset). Now he asks you how many different arrays he could have after applying queries. Since the answer can be large, output by modulo 1000000007 (10^9 + 7).

## HackerEarth Boring Not boring problem solution.

`# include <bits/stdc++.h># include <ext/pb_ds/assoc_container.hpp># include <ext/pb_ds/tree_policy.hpp>using namespace __gnu_pbds;using namespace std; template<typename T> using ordered_set = tree <T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;#define _USE_MATH_DEFINES_#define ll long long#define ld long double#define Accepted 0#define pb push_back#define mp make_pair#define sz(x) (int)(x.size())#define every(x) x.begin(),x.end()#define F first#define S second#define lb lower_bound#define ub upper_bound#define For(i,x,y)  for (ll i = x; i <= y; i ++) #define FOr(i,x,y)  for (ll i = x; i >= y; i --)#define SpeedForce ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0)void setIn(string s) { freopen(s.c_str(),"r",stdin); }void setOut(string s) { freopen(s.c_str(),"w",stdout); }void setIO(string s = "") {    if (sz(s)) { setIn(s+".in"), setOut(s+".out"); } // for USACO}const double eps = 0.000001;const ld pi = acos(-1);const int maxn = 1e7 + 9;const int mod = 1e9 + 7;const ll MOD = 1e18 + 9;const ll INF = 1e18 + 123;const int inf = 2e9 + 11;const int mxn = 1e6 + 9; const int N = 1e5+5;                                         const int M = 22;const int pri = 997;const int Magic = 2101;const int dx[] = {-1, 0, 1, 0};const int dy[] = {0, -1, 0, 1};mt19937 gen(chrono::steady_clock::now().time_since_epoch().count()); int rnd (int l, int r) {    return uniform_int_distribution<int> (l, r)(gen);}int n, m;vector < int > g[N*4];struct GG {    //gauss    vector < int > basis;    void add(int x) {        for(auto &it:basis) if((x^it)<x) x^=it;        for(auto &it:basis) if((it^x)<it) it^=x;        if(x) basis.pb(x);    }        int get() {        int ans = 0;        for (auto x:basis) ans ^= x;        return ans;    }        void clear() {        basis.clear();    }};void add(int l, int r, int x, int v=1, int tl = 1, int tr =n) {    if(tl>r || l > tr) return;    if(tl >= l && tr <= r) {        g[v].pb(x);        return;    }        int tm = (tl+tr)>>1;    add(l, r, x, v<<1, tl, tm);    add(l, r, x, v<<1|1, tm+1, tr);}int pw[33];int calc(GG c, int v, int l, int r) {    for (auto x : g[v])        c.add(x);        if(l == r) {        return pw[c.basis.size()];    }        int md = (l+r)>>1;    return calc(c, v<<1, l, md)         * (ll)calc(c, v<<1|1, md+1, r) % mod;}void solve() {    cin >> n >> m;        for (int i = 1, l, r, x; i <= m; ++i) {        cin >> l >> r >> x;        add(l, r, x);    }        GG c;    cout << calc(c, 1, 1, n) << '\n';}int main () {    SpeedForce;    pw[0] = 1;        for (int i = 1; i < 33; ++i) {        pw[i] = (pw[i-1] * 2) % mod;    }        int T = 1;    //cin >> T;    while(T--) solve();        return Accepted;}`

### Second solution

`#include <bits/stdc++.h>using namespace std;typedef long long ll;const int MAX_N = 1e5 + 14, LG = 30, MOD = 1000000007;struct Gauss {    int a[LG];    Gauss() {        memset(a, 0, sizeof a);    }    void add(int x) {        for (int i = LG - 1; i >= 0; i--)            if (x >> i & 1)                if (a[i])                    x ^= a[i];                else {                    a[i] = x;                    break;                }    }    void operator+=(vector<int> &o) {        for (int i : o)            if (i)                add(i);    }    int cnt() {        return 1 << LG - count(a, a + LG, 0);    }};vector<int> node[1 << 18];int n;void add(int s, int e, int x, int l = 0, int r = n, int id = 1) {    if (s <= l && r <= e) {        node[id].push_back(x);        return;    }    if (e <= l || r <= s)        return;    int mid = (l + r) / 2;    add(s, e, x, l, mid, id * 2);    add(s, e, x, mid, r, id * 2 + 1);}int get_ans(int l = 0, int r = n, int id = 1, Gauss carry = Gauss()) {    carry += node[id];    if (l + 1 == r)        return carry.cnt();    int mid = (l + r) / 2;    return (ll) get_ans(l, mid, id * 2, carry) * get_ans(mid, r, id * 2 + 1, carry) % MOD;}int main() {    ios::sync_with_stdio(0), cin.tie(0);    int m;    cin >> n >> m;    for (int i = 0; i < m; ++i) {        int l, r, x;        cin >> l >> r >> x;        add(l - 1, r, x);    }    cout << get_ans() << '\n';}`