# HackerEarth A digit in a sequence problem solution

In this HackerEarth A digit in a sequence problem solution Bob is writing a program that solves the following problem:

You are given the numbers a,b to display the infinite sequence of numbers a,a+b,a+2*b,...,a+N*b,a..(N+1)*b,... Bob has made a mistake, he namely forgot to draw a space for the division between the numbers, resulting in a long line of numbers. In order not to correct the mistake, Bob decided to find out how to find the kth digit in the formed line (numbering of digits begins with one).

Bob could not find a good solution, so he asks you for help.

## HackerEarth A digit in a sequence problem solution.

`#include<bits/stdc++.h>#define int long long using namespace std;int len;int l,r;int c[100],LL[100];int pos,st[100],val,cnt;int result; int lenNumber(long long x){     int len = 0;      while(x>0){          x/=10;          ++len;      }      return len; } void solve(){     int a, b, k, n;     cin >> a >> b >> k;     n = 1e9;    int x = lenNumber(a);    int y = lenNumber(a + b * n);    int L = 0;    int R = n;     for(int i=x;i<=y;++i){       int l = L;       int r = R;        while(l + 1 < r){            int m = (l + r) >> 1;            if(lenNumber(a + m * b)>i) r=m; else l=m;        }        for(int m=min(r+10,R);m>=max(L,r-10);--m)            if(lenNumber(a + m * b)==i)        {            c[i] = (m-L+1);            LL[i] = L;            L=m+1;            break;        }        if(L>R)break;     }  int ans = 0;    for(int i=x;i<=y;++i)     {         ans+=c[i] * i;         if(ans>=k)         {             ans-=c[i] * i;             k-=ans;              pos = k / i;              if(k % i == 0)              {                val = a + b * (LL[i] + pos - 1);                result = val % 10;              } else              {                val = a + b * (LL[i] + pos);                k%=i;                cnt = 0;                while(val>0)                {                    st[++cnt] = val % 10;                    val/=10;                }                reverse(st + 1,st + cnt + 1);                result = st[k];              }              cout<< result << '\n';              return;         }     } }   main()  {    ios_base::sync_with_stdio(false);    cin.tie(NULL);    cout.tie(NULL);        int t;        cin >> t;        while(t-->0){            solve();        }    return 0;  }`

### Second solution

`#include<bits/stdc++.h>#define int long longusing namespace std;int a, b, k;int n;int x, y;int h;int L, R;int m;int len;int i;int l, r, c[100], LL[100];int ans, pos, ANSWER;int st[100], val, kol;int f(int x){    int len = 0;    while(x>0)    {        x/=10;        ++len;    }    return len;}void solve(){    memset(c, 0, sizeof c);    cin>>a>>b>>k;    n = 1e9;    x=y=0;    h = a;    while(h>0)    {        ++x;        h/=10;    }    h = a + b * n;    while(h>0)    {        ++y;        h/=10;    }    L = 0;    R = n;    for(i=x;i<=y;++i)    {        l = L;        r = R;        while(l+1<r)        {            m = (l+r)>>1;            if(f(a + m * b)>i) r=m; else l=m;        }        for(m=min(r+10,R);m>=max(L,r-10);--m)            if(f(a + m * b)==i)            {                c[i] = (m-L+1);                LL[i] = L;                L=m+1;                break;            }        if(L>R)break;    }    ans = 0;    for(i=x;i<=y;++i)    {        ans+=c[i] * i;        if(ans>=k)        {            ans-=c[i] * i;            k-=ans;            pos = k / i;            if(k % i == 0)            {                val = a + b * (LL[i] + pos - 1);                ANSWER=val % 10;            } else            {                val = a + b * (LL[i] + pos);                k%=i;                kol = 0;                while(val>0)                {                    st[++kol] = val % 10;                    val/=10;                }                reverse(st+1,st+kol+1);                ANSWER=st[k];            }            cout<<ANSWER<<'\n';            return;        }    }    cout<<-1<<'\n';}main(){    ios_base::sync_with_stdio(false);    cin.tie(NULL);    cout.tie(NULL);    int t;    cin >> t;    while(t-->0){        solve();    }    return 0;}`