HackerEarth Teachers and students problem solution

In this HackerEarth Teachers and students problem solution There are N students in a class and each student has a number denoted by an array A. The teacher will select three students and give a task to each of them. The first student has to find a number of subsets of students having cumulative xor less than K. The second student has to find a number of subsets of students having cumulative xor equal to K. The third student has to find a number of subsets of students having cumulative xor greater than K. Let the three numbers returned by the selected students are cnt1, cnt2, and cnt3. You need to find the value of :

(cnt1 + cnt2) X 2 + (cnt2 + cnt3) X 2 + (cnt3 + cnt1) X 2 - (cnt1 X 2 + cnt2 X 2 + cnt3 X 2)

HackerEarth Teachers and students problem solution.

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
#define ff first
#define ss second
#define pb push_back
const ll M=1e9+7;
const ll inf=2e18;
string toString(ll x) {  stringstream ss; ss<<x; string s=ss.str(); return s; }
ll toInteger(string s){ stringstream ss(s); ll x=0; ss>>x; return x;}

ll power(ll x,ll n)
{
    if(n==0)
        return 1;
    else if(n%2 == 0)
        return power((x*x)%M,n/2);
    else
        return (x*power((x*x)%M,(n-1)/2))%M;
}

int main()
{
    ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
    ll t;
    cin>>t;
    while(t--)
    {
        ll n,k;
        cin>>n>>k;
        for(ll i=0;i<n;i++)
            cin>>k;
        ll val=power(2LL,n)%M;
        cout<<(val*val)%M<<"\n";
    }
    return 0;
}

Second solution

#include<bits/stdc++.h>
using namespace std;
/*#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
/*template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
*/typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<ll,ll> pl;
typedef pair<int,int> pii;

#define LOCAL 0
#define dbg(x) cout << #x << " is " << x << "\n"
#define gll(x) scanf("%d",&x)
#define gll2(x,y) scanf("%d%d",&x,&y)
#define gll3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define gllarr(arr,n) f(i,n) gll(arr[i]);
#define sz(x) ((int)x.size())
#define s(x) sort(x.begin(),x.end())
#define all(v) v.begin(),v.end()
#define rs(v) { s(v) ; r(v) ; }
#define r(v) {reverse(all(v));}
#define pb push_back
#define f(i,n) for(int i=0;i<n;i++)
#define fr(i,n) for(int i=n-1;i>=0;i--)
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define repr(i,a,b) for(int i=a;i>=b;i--)

const ll mod = (ll)1e9 + 7;
const ll inf = (ll)1e16;
const ld eps = 1e-12;
const ll N = (int)1e5 + 5;
const ll LOGN = 19;
const ld PI = 3.14159265358979323846;
inline ll mul(ll a, ll b, ll m = mod) { return (ll)(a * b) % m;}
inline ll add(ll a, ll b, ll m = mod) { a += b; if(a >= m) a -= m; if(a < 0) a += m; return a;}
inline ll power(ll a, ll b, ll m = mod) { if(b == 0) return 1; if(b == 1) return (a % m); ll x = power(a, b / 2, m); x = mul(x, x, m); if(b % 2) x = mul(x, a, m); return x;}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    if (LOCAL) {
        freopen("C:\\Users\\Dishant\\Desktop\\Collection-DEV c++\\input.txt", "r", stdin);
        freopen("C:\\Users\\Dishant\\Desktop\\Collection-DEV c++\\output.txt", "w", stdout);
    }
    int t;
    assert(cin>>t);
    assert(t >= 1 && t <= 20);
    while(t--){
        ll n, k;
        assert(cin>>n);
        assert(cin>>k);
        assert(n >= 1 && n <= (int)1e5);
        assert(k >= 1 && k <= (int)1e5);
        ll a[n];
        f(i, n) {
            assert(cin>>a[i]);
            assert(a[i] >= 1 && a[i] <= (int)1e5);
        }
        ll ans = power(2ll, n);
        ans = mul(ans, ans);
        cout<<ans<<"\n";
    }
    return 0;
}