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bagirrra123 [75]
3 years ago
10

Binary is a base-2 number system instead of the decimal (base-10) system we are familiar with. Write a recursive function PrintI

nBinary(int num) that prints the binary representation for a given integer. For example, calling PrintInBinary(5) would print 101. Your function may assume the integer parameter is non-negative. The recursive insight for this problem is to realize you can identify the least significant binary digit by using the modulus operator with value 2. For example, given the integer 35, mod by 2 tells you that the last binary digit must be 1 (i.e. this number is odd), and division by 2 gives you the remaining portion of the integer (17). What 's the right way to handle the remaining portion
Computers and Technology
1 answer:
Paladinen [302]3 years ago
7 0

Answer:

In C++:

int PrintInBinary(int num){

if (num == 0)  

 return 0;  

else

 return (num % 2 + 10 * PrintInBinary(num / 2));

}

Explanation:

This defines the PrintInBinary function

int PrintInBinary(int num){

This returns 0 is num is 0 or num has been reduced to 0

<em> if (num == 0)  </em>

<em>  return 0;  </em>

If otherwise, see below for further explanation

<em> else </em>

<em>  return (num % 2 + 10 * PrintInBinary(num / 2)); </em>

}

----------------------------------------------------------------------------------------

num % 2 + 10 * PrintInBinary(num / 2)

The above can be split into:

num % 2 and + 10 * PrintInBinary(num / 2)

Assume num is 35.

num % 2 = 1

10 * PrintInBinary(num / 2) => 10 * PrintInBinary(17)

17 will be passed to the function (recursively).

This process will continue until num is 0

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koban [17]

Answer:

<em>The program written in Python is as follows:</em>

def solution(N):

     concat = ""

     for i in range(1,N+1):

           if not(i%2 == 0 or i%3 ==0 or i%5 == 0):

                 print(str(i))

           else:

                 if i%2 == 0:

                       concat= concat+"Codility"

                 if i%3 == 0:

                       concat= concat+"Testers"

                 if i%5 == 0:

                       concat= concat+"Coders"

                 print(concat)

                 concat = ""

N = int(input("Enter a positive integer: "))

solution(N)

Explanation:

This line declares the function

def solution(N):

This line initializes a variable named concat to an empty string

     concat = ""

This line iterates from 1 to the input integer

     for i in range(1,N+1):

<em>This line checks if the current number of iteration is divisible by 2,3 or 5, if no, the number is printed</em>

           if not(i%2 == 0 or i%3 ==0 or i%5 == 0):

                 print(str(i))

<em>If otherwise</em>

           else:

<em>This lines checks if current number is divisible by 2; if yes the string "Codility" is concatenated to string concat</em>

                 if i%2 == 0:

                       concat= concat+"Codility"

<em>This lines checks if current number is divisible by 3; if yes the string "Testers" is concatenated to string concat</em>

<em>                 </em> if i%3 == 0:

                       concat= concat+"Testers"

<em>This lines checks if current number is divisible by 2; if yes the string "Coders" is concatenated to string concat</em>

                 if i%5 == 0:

                       concat= concat+"Coders"

<em>The concatenated string is printed using this line</em>

                 print(concat)

This variable concat is intialized back to an empty string

                 concat = ""

The main method starts here

N = int(input("Enter a positive integer: "))

This line calls the defined function solution

solution(N)

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