Find the total count of numbers up to N digits in a given base B

Given two integers N and B, the task is to find the count of natural numbers of Base B up to N digits.

Examples:

Input: N = 2, B = 10
Output: 99
Explanation:
1, 2, 3, 4, 5, 6, 7, 8, 9 are 1 digit Natural numbers of Base 10.
10, 11, 12………99 are 2 digit Natural numbers of Base 10
So, total = 9 + 90 = 99

Input: N = 2, B = 16
Output: 255
Explanation:
There are a total of 240 two digit hexadecimal numbers and 15 one digit hexadecimal numbers.
Therefore, 240 + 15 = 255.

Approach: On observing carefully the count of numbers with N digits in base B is a geometric progression formed with the first term being (B – 1) and a common ratio of B.



Therefore,

Nth term = Number of natutal numbers of N digits in Base B = (B – 1) * BN – 1

Finally, count of all natural numbers in Base B up to N digits can be found out by iterating a loop from 1 to N and calculating the sum of ith term using the above formula.

Below is the implementation of the above approach:

C++

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// C++ implementation to find the count
// of natural numbers upto N digits
  
#include <bits/stdc++.h>
  
using namespace std;
  
// Function to return the count of
// natural numbers upto N digits
int count(int N, int B)
{
    int sum = 0;
  
    // Loop to iterate from 1 to N
    // and calculating number of
    // natural numbers for every 'i'th digit.
    for (int i = 1; i <= N; i++) {
        sum += (B - 1) * pow(B, i - 1);
    }
    return sum;
}
  
// Driver Code
int main()
{
    int N = 2, B = 10;
    cout << count(N, B);
  
    return 0;
}

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Java














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// Java implementation to find the count 


// of natural numbers upto N digits 


  


class GFG{ 


  


// Function to return the count of 


// natural numbers upto N digits 


static int count(int N, int B) 


{ 


    int sum = 0; 


  


    // Loop to iterate from 1 to N 


    // and calculating number of 


    // natural numbers for every 'i'th digit. 


    for (int i = 1; i <= N; i++){ 


        sum += (B - 1) * Math.pow(B, i - 1); 


    } 


    return sum; 


} 


  


// Driver Code 


public static void main(String[] args) 


{ 


    int N = 2, B = 10; 


    System.out.print(count(N, B)); 


} 


} 


  


// This code is contributed by gauravrajput1 



















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Python3














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# Python3 implementation to find the count 


# of natural numbers up to N digits 


  


from math import pow


  


# Function to return the count of 


# natural numbers upto N digits 


def count(N, B): 


    sum = 0


  


    # Loop to iterate from 1 to N 


    # and calculating number of 


    # natural numbers for every 'i'th digit. 


    for i in range(1, N+1): 


        sum += (B - 1) * pow(B, i - 1) 


    return sum


  


# Driver Code 


if __name__ == '__main__': 


    N = 2


    B = 10


    print(int(count(N, B))) 


  


# This code is contributed by Bhupendra_Singh 



















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C#














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// C# implementation to find the count 


// of natural numbers upto N digits 


using System; 


using System.Collections.Generic; 


class GFG{ 


  


// Function to return the count of 


// natural numbers upto N digits 


static int count(int N, int B) 


{ 


    int sum = 0; 


  


    // Loop to iterate from 1 to N 


    // and calculating number of 


    // natural numbers for every  


    // 'i'th digit. 


    for(int i = 1; i <= N; i++) 


    { 


       sum += (int)((B - 1) * Math.Pow(B, i - 1)); 


    } 


    return sum; 


} 


  


// Driver Code 


public static void Main(String[] args) 


{ 


    int N = 2, B = 10; 


      


    Console.Write(count(N, B)); 


} 


} 


  


// This code is contributed by amal kumar choubey 



















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Output:


99





Time Complexity: O(N)


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