 # JONBASE - Editorial

Contest

Setter: sayan_kashyapi

Tester: mishra_roshan , debrc

Editorialist: ritik0602

Easy

# PREREQUISITES:

modulo operator, log function

# PROBLEM:

Given a decimal number N and another number B, find the number of digits in it’s equivalent B base number where B ≥ 2.

# EXPLANATION

In order to convert a decimal number to a base B number, we keep on dividing that number with B and store the remainders of the diviions. The final ans is the reverse of the number obtained.
So the number of digits can be obtained by counting the number of divisions by B needed to reduce the number to zero.
Also, when N is negative, we can just get rid of the sign.

A quick Mathematical Formula:
The number of digits in base B representation of a decimal number N can be given as :
floor(log(N)/log(B)) + 1, N > 0
For N = 0, we can set the ans to zero as log(0) is not defined.

# TIME COMPLEXITY

Time complexity is O(logN) as we keep dividing the number N with B

# SOLUTIONS:

Setter's Solution
C++
``````//number of digits of an b base number of a decimal number
#include<bits/stdc++.h>
using namespace std;
int main()
{
// #ifndef ONLINE_JUDGE
// 	freopen("input.txt", "r", stdin);
// 	freopen("op.out", "w", stdout);
// #endif
// long long int t;
long long int t;
// srand(time(NULL));
cin >> t;
// t = 1;
while (t--)
{
long long int n, b, ans;
cin >> n >> b;
// srand(time(0));
// b = (rand() % 19) + 2;
// n = (rand() % 100000) + 1;
// cout << n << " " << b << endl;
ans = n ? log(abs(n)) / log(b) : 0;
cout << ++ans << endl;
}
return 0;
}
``````
Tester's Solution
Java
``````import java.util.*;

// Compiler version JDK 11.0.2

class Dcoder
{
public static void main(String args[])
{
Scanner  sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0){

long n=sc.nextLong();
long  b=sc.nextLong();

long  ans=(long)(n!=0?((Math.log(Math.abs(n)))/(Math.log(b))):0);
System.out.println(ans+1);       }
}
}

``````
Python
``````for _ in range(inp()):
n,b=minp()
n=abs(n)
if n==0:
print(1)
continue
count=0
while(n>0):
count+=1
n=n//b
print(count)

``````

Feel free to Share your approach.
Suggestions are welcomed as always had been.