# HackerEarth Special numbers problem solution

In this HackerEarth Special numbers problem solution, Alice is traveling to a special town with her friends. All the residents of this special town communicate through a special language. This special language consists of 4 special alphabets a, b, c and d. This language contains special words. These words are composed of these 4 special alphabets and are always even in length. The words are always palindromic in nature. The words are also assigned special numbers. A special number is the lexicographical rank of special word with respect to other special words.

You are given a special number. Your task is to determine the special word corresponding to that special number.

## HackerEarth Special numbers problem 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;    cin>>t;    while(t--){        int k;        cin>>k;        assert(k <= (int)1e9);        int len = 1;        ll curr = 0, next = 4;        while(curr + next < k) {            curr += next;            next *= 4ll;            len++;        }        k -= curr;        k--;        string temp;        int tmp = len;        while(tmp--) {            int to = k % 4;            k /= 4;            temp.pb(to + 'a');        }        string put = temp;        r(temp);        cout<<temp<<put<<endl;    }    return 0;}`

### Second solution

`#include<bits/stdc++.h>using namespace std;#define   pb              push_back#define   REP(i,n)        for(i=1;i<=n;i++)#define   FOR(i,a,b)      for(i=a;i<=b;i++)#define   all(v)          v.begin(),v.end()#define   F               first#define   S               second#define   vl              vector<LL>#define   itr             ::iterator it#define   lb              lower_bound#define   ub              upper_bound#define   LL              long long#define   ULL             unsigned long long#define   ret             return LL t,n,i,j,ans,k;LL a[100],b[45] ; int main(){    ios_base::sync_with_stdio(false);    cin.tie(NULL);    cin>>t ;     a[1] = 4 ;    LL p = 16 ;     for(i=2;i<=30;i++) a[i] = p + a[i-1],p = 4 * p ;    while(t--)    {  cin>>k ;        string ans ;        REP(i,30)        {  if(a[i]<k) continue ;          LL start,buck ;          buck = (a[i] - a[i-1])/4 ;                      if(a[i-1] + buck>=k) start = a[i-1] ,ans.pb('a') ;           else if(a[i-1] + 2 * buck>=k) start = a[i-1] + buck ,ans.pb('b') ;           else if(a[i-1] + 3 * buck>=k) start = a[i-1] + 2 * buck ,ans.pb('c') ;          else start = a[i-1] + 3 * buck ,ans.pb('d') ;                       while(ans.length()<i)          {  buck /= 4 ;                           if(start + buck>=k) ans.pb('a') ;              else if(start + 2 * buck>=k) start = start + buck,ans.pb('b') ;              else if(start + 3 * buck>=k) start = start + 2 * buck,ans.pb('c') ;             else start = start + 3 * buck,ans.pb('d') ;                             }           break ;        }        cout<<ans ;       reverse(all(ans)) ;        cout<<ans<<endl ;      }}`