All possible N digits and base B, but excluding numbers with leading zeros

Here we will see a problem where we have N and base B. Our task is to count the number of N digits in base B without leading 0's. So if N is 2 and B is 2, then there will be four numbers 00, 01, 10, and 11. So only two of the numbers are valid for this part. They are 10, 11, without leading 0's.

If the base is B, then there are 0 to B-1 different numbers. So B^N different N-digit numbers (including leading 0s) can be generated. If we ignore the first number 0, then there are B^(N-1) numbers. So the total number of N digits without leading 0 is B^N - B^(N-1)

Algorithm

countNDigitNum(N, B)

Begin
   total := B<sup>N</sup>
   with_zero := B<sup>N-1</sup>
   return BN &ndash; B<sup>N-1</sup>
End

Example## The Chinese translation of # is:

Example

#include <iostream>
#include <cmath>
using namespace std;
int countNDigitNum(int N, int B) {
   int total = pow(B, N);
   int with_zero = pow(B, N - 1);
   return total - with_zero;
}
int main() {
   int N = 5;
   int B = 8;
   cout << "Number of values: " << countNDigitNum(N, B);
}

Output

Number of values: 28672

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