1328B_K-th_Beautiful_String.cpp

// Vidur Goel

//Codeforcees Handle: Vidurcodviz

#include<iostream>
#include<string>
#include<cmath>
#include<climits>
#include<algorithm>
#include<cstddef>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<stack>
#include<chrono>
#include<random>
#include<cassert>
#include<array>
#include<bitset>
#include<complex>
#include<cstring>
#include<functional>
#include<iomanip>
#include<map>
#include<numeric>
#include<queue>
#include<set>
#include<utility>
#include<string_view>
#include<deque>
#include<iterator>
#include<sstream>

using namespace std;
using namespace chrono;

void solve_array();
void solve_single();
void solve_mul();

typedef long long int ll;
typedef unsigned long long int ull;
typedef long double lld;
typedef vector<ll> vl;
typedef pair<ll,ll> pll;
typedef vector<pll> vpll;
typedef vector<vl> vvl;

#define make_it_fast() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define rept(i, a, n) for (ll i = (a); i < (n); i++)
#define all(x) (x).begin(), (x).end()
#define sor(x) sort(all(x))
#define lb lower_bound
#define ub upper_bound
#define pb push_back
#define ppb pop_back
#define mp make_pair
#define MOD 1000000007
#define MOD1 998244353
#define PI 3.141592653589793238462
#define mset multiset<ll> // it contains multiple instances of the same value in ascending order
#define rep(i,a,b) for(ll i=a;i<b;i++)
#define repd(i,a,b) for(ll i=b-1;i>=a;i--)
#define nn endl
#define setbits(n) __builtin_popcount(n)

ll seiv[1000001]={0};

string yup="YES";
string nope="NO";

ll lmin(vl arr){return *min_element(arr.begin(),arr.end());}
ll lmax(vl arr){return *max_element(arr.begin(),arr.end());}

ll fibonacci(ll n){ll a=0;ll b=1;ll c;if(n==0 || n==1){return n;}for(ll i=2;i<n+1;i++){c=a+b;a=b;b=c;}return c;}

ll sum(vl a){return accumulate(a.begin(),a.end(),0);}
void rev(vl &arr,ll n){rep(i,0,n){cin>>arr[i];}return;}
void prv(vl arr){rep(i,0,arr.size()){cout<<arr[i]<<" ";}cout<<nn;return;}

bool prime(ll n){rep(i,2,ceil(sqrt(n))){if(n%i==0){return false;}}return true;}
void seive(){seiv[0]=0;seiv[1]=1;for(ll i=2;i*i<1000001;i++){if(seiv[i]==0){seiv[i]=i;for(ll j=i*i;j<1000001;j=j+i){if(seiv[j]==0){seiv[j]=i;}}}}}

ll gcd(ll a,ll b){if(b==0){return a;}if(a>=b){return gcd(b,a%b);}else{return gcd(b,a);}}
ll expo(ll num,ll coef){ll res=1;while(coef!=0){if(coef%2==0){coef=coef/2;num=num*num;}else{coef=coef-1;res=res*num;}}return res;}
ll expo_mod(ll a,ll n,ll m){ll res=1;while(n!=0){if(n%2==0){a=a*a;a=a%m;n=n/2;}else{res=res*a;res=res%m;n--;}}return res;}

ll add_mod(ll a, ll b, ll m) {a = a % m; b = b % m; return (((a + b) % m) + m) % m;}
ll mul_mod(ll a, ll b, ll m) {a = a % m; b = b % m; return (((a * b) % m) + m) % m;}
ll sub_mod(ll a, ll b, ll m) {a = a % m; b = b % m; return (((a - b) % m) + m) % m;}

/*
    sqrt() in built function to give the square root in float/double
    cbrt() in built function to give the cube root in float/double
    abs() is used for the absolute value of a number
    swap() function in c++ used to swap value of two elements of the same data type.
    toupper() This function is used for converting a lowercase character to uppercase.
    tolower() This function is used for converting an uppercase character to lowercase.
    ceil() and floor() function
    sort(vect.begin(),vect.end(), greater<int>());
    reverse(vect.begin(), vect.end());
    count(first_iterator, last_iterator,x) – To count the occurrences of x in vector.
    find(first_iterator, last_iterator, x) – Returns an iterator to the first occurrence of x in vector and points to last address of vector ((name_of_vector).end()) if element is not present in vector
    
    maximium value long long can take 9, 223, 372, 036, 854, 775, 807
    2^63-1
    i.e, length of 19 only
    maximium value long long can take 18, 446, 744, 073, 709, 551, 615
    2^64-1
    i.e, length of 20 only

    __builtin_popcount(n) - we use this function to count the number of 1's (set bits) in the number in binary form
    __builtin_parity(n) - this is boolean function which return true if number of 1's in binary form of n are odd else returns false;
    __builtin_clz(n) - eg: Binary form of 16 is 00000000 00000000 00000000 00010000 therefore will return the number of the leading zeroes in n here answer will be 27
    __builtin_ctz(n) - eg: Binary form of 20 is 00000000 00000000 00000000 00010100 therefore will return the number of the trailing zeroes in n here answer will be 2
*/

bool mycompare(pll p1 ,pll p2){
    if(p1.first<p2.first){
        return true;
    }
    else if(p1.first==p2.first){
        return p1.second<p2.second;
    }
    else{
        return false;
    }
}

void solve_mul(){
    ll test;
    cin>>test;
    rep(i,0,test){
        solve_single();
    }
}

void solve_single(){
    ull n,k;
    cin>>n>>k;
    ull numa=n-2;
    ull numb=2;
    ull t=1;
    while(k>1){
        if(numb!=0){
            ll sum=0;
            if(numa==0){
                if(numb==1){
                    cout<<"b"<<nn;
                    return;
                }
                else{
                    cout<<"bb"<<nn;
                    return;
                }
            }
            else{
                if(numb==1){
                    sum+=n-t;
                }
                else{
                    sum+=((n-t)*(n-t-1))/2;
                }
            }
            if(sum<=k-1){
                cout<<"b";
                numb-=1;
                k-=sum;
                t++;
            }
            else{
                cout<<"a";
                numa-=1;
                k-=0;
                t++;
            }
        }
        else{
            while(t<=n){
                cout<<"a";
                t++;
            }
            k=0;
            cout<<nn;
            return;
        }
    }
    rep(i,0,numa){
        cout<<"a";
    }
    rep(i,0,numb){
        cout<<"b";
    }
    cout<<nn;
    return;
}

void solve_array(){
    ll n;
    cin>>n;
    vl arr(n,0);
    rev(arr,n);
}

int main(){
    make_it_fast();
    //seive();
    solve_mul();
    //solve_array();
    //solve_single();
    return 0;
}