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 ;  
    }
}