HackerEarth Base numbers problem solution

In this HackerEarth Base numbers problem solution you are given three integers a, b, and n. Your task is to determine the number of positive integers x related to [1,10^9] such that ax^x is an n-digit number of base b.

HackerEarth Base numbers problem solution.

#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <queue>
#include <map>
#include <string.h>
#include <math.h>
#include <stdio.h>
#include <deque>
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pii pair<int,int>
#define qi ios::sync_with_stdio(0)

bool debug=true;
template<typename T1,typename T2>ostream& operator<<(ostream& os,pair<T1,T2> ptt){
    os<<ptt.first<<","<<ptt.second;
    return os;
}
template<typename T>ostream& operator<<(ostream& os,vector<T> vt){
    os<<"{";
    for(int i=0;i<vt.size();i++){
        os<<vt[i]<<" ";
    }
    os<<"}";
    return os;
}

ll a,n,b;
inline double log(ll a,ll b){
    return log10(a)/log10(b);
}

inline bool ok(ll x,ll n){
    long double v=floor(x*log(x,b)+log(a,b))+1;
    if(v>=n){
        return true;
    }else{
        return false;
    }
}

void solve(){
    ll l=0,r=1e9;
    while(l<=r){
        ll m=(l+r)/2;
        if(ok(m,n)){
            r=m-1;
        }else{
            l=m+1;
        }
    }
    ll l2=0,r2=1e9;
    while(l2<=r2){
        ll m=(l2+r2)/2;
        if(ok(m,n+1)){
            r2=m-1;
        }else{
            l2=m+1;
        }
    }
    
    cout<<l2-l<<endl;
}

int main(int argc,char* argv[]){
    while(cin>>a>>n>>b){
        solve();
    }
    return 0;
}