Question

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2 Decimal to N-Base Conversion
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Problem statement
Instr
N-base notation is a system for writing numbers which uses only n
different symbols. These symbols are the first n symbols from the
given notation list (including the symbol for 0). Decimal to n-base
notations are {0:0, 1:1, 2:2, 3:3, 4:4, 5:5, 6:6, 7:7, 8:8, 9:9, 10:A,
11:B and so on up to 35:Z).
Implement the following function:
char* DectoNBase(int n, int num);
Now
2.
GLAR
The function accepts positive integers 'n' and 'num'. Implement the
function to calculate n-base equivalent of 'num' and return the same
as a string
Steps:
1. Divide the decimal number num by n. Treat the division as
an integer division,
2. Write down the remainder (in n-base notation).
DEBIT
ಸಿಪಿ
695
G
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POLISI
DIPE
3. Divide the quotient again by n. Treat the division as an
Integer division
2
3
4
5
6
GEN
4. Repeat step 2 and 3 until quotient is 0.
5. The n-base value is the sequence of the remainders from the
last to first
Assumption: 1< n == 36
Example:
Input:
12
13
13
num: 718
Output:
ABA​

Answer 1

Answer:

I cannot understand

write properly

Answer 2

Answer:

#include <iostream>

#include <math.h>

using namespace std;

// function to print number of

// digits

void findNumberOfDigits(long n, int base)

{

// Calculating log using base

// changing property and then

// taking it floor and then

// adding 1.

int dig = (int)(floor( log(n) /

log(base)) + 1);

// printing output

cout << "The Number of digits of "

<< "Number " << n << " in base "

<< base << " is " << dig;

}

// Driver method

int main()

{

// taking inputs

long n = 1446;

int base = 7;

// calling the method

findNumberOfDigits(n, base);

return 0;

}