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 = 99Input: 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 natural 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++ 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;
} |
// 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 |
# 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 |
// 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 |
<script> // Javascript implementation to find the count // of natural numbers upto N digits // Function to return the count of // natural numbers upto N digits function count(N, B)
{ var sum = 0;
// Loop to iterate from 1 to N and
// calculating number of natural
// numbers for every 'i'th digit.
for ( var i = 1; i <= N; i++)
{
sum += (B - 1) * Math.pow(B, i - 1);
}
return sum;
} // Driver code var N = 2, B = 10;
document.write(count(N, B)); // This code is contributed by Ankita saini </script> |
99
Time Complexity: O(N)
Auxiliary Space: O(1)