#include #include // This library is for math operation int main() { int n; //number of cases int x; //number in base 10 scanf("%i", ??); // Loop for the number of cases while (??){ scanf("%i", &??); int size = log2 (??) + 1; //size of the array. log2: logarithm base 2 int bin[??]; // this array will contain the digits of the binary number // Convert from base 10 to base 2 int i = ??; while (??) { bin[i] = x%2; // reminder x /= 2; // update x ??; } // Print binary number i = ??; while (??) { printf("%i", bin[i]); ?? ; } printf("\n"); ??; } return 0; }
#include #include // This library is for math operation int main() { int n; //number of cases int x; //number in base 10 scanf("%i", ??); // Loop for the number of cases while (??){ scanf("%i", &??); int size = log2 (??) + 1; //size of the array. log2: logarithm base 2 int bin[??]; // this array will contain the digits of the binary number // Convert from base 10 to base 2 int i = ??; while (??) { bin[i] = x%2; // reminder x /= 2; // update x ??; } // Print binary number i = ??; while (??) { printf("%i", bin[i]); ?? ; } printf("\n"); ??; } return 0; }
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
Related questions
Question
Fill in the code
![#include <stdio.h>
#include <math.h> // This library is for math operation
int main(){
int n; //number of cases
int x; //number in base 10
scanf("%i", ??);
// Loop for the number of cases
while (??){
scanf("%i", &??);
int size = log2(??) + 1; //size of the array. log2: logarithm base 2
int bin[??]; // this array will contain the digits of the binary number
// Convert from base 10 to base 2
int i = ??;
while (??){
bin[i]
x /= 2; // update x
= x%2; // reminder
??;
}
// Print binary number
i = ??;
while (??){
printf("%i", bin[i]);
??;
}
printf("\n");
??;
return 0;
}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb4fbdee7-4dd6-4b70-b473-f3d2f2c3fe84%2F02e85b27-dc23-498c-bca7-4976842734c4%2Ff057kkt_processed.png&w=3840&q=75)
Transcribed Image Text:#include <stdio.h>
#include <math.h> // This library is for math operation
int main(){
int n; //number of cases
int x; //number in base 10
scanf("%i", ??);
// Loop for the number of cases
while (??){
scanf("%i", &??);
int size = log2(??) + 1; //size of the array. log2: logarithm base 2
int bin[??]; // this array will contain the digits of the binary number
// Convert from base 10 to base 2
int i = ??;
while (??){
bin[i]
x /= 2; // update x
= x%2; // reminder
??;
}
// Print binary number
i = ??;
while (??){
printf("%i", bin[i]);
??;
}
printf("\n");
??;
return 0;
}
![Task 1: Defining an array
This task consists in converting a number from base 10 to base 2. The user will input the n
integer numbers. The program must return the n binary numbers.
Step 1: Divide the given number 13 repeatedly by 2 until
you get '0'as the quotient
13 + 2 = 6 (Remainder 1)
6 + 2 = 3 (Remainder 0)
3 + 2 = 1 (Remainder 1)
1 + 2 = 0 (Remainder 1)
Step 2: Write the remainders in the reverse 1 10 1
order
:. 1310
= 1101,
(Decimal)
(Binary)
10
arr I4] =
1
1.2° t0-2*t 1.2+ O•2"= 10
Hint:
You must store each digit of the binary number in an array and then print them. The length of
the array must be equal to the number of digits of the binary number.
DO NOT USE ANY BUILT-IN FUNCTION](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb4fbdee7-4dd6-4b70-b473-f3d2f2c3fe84%2F02e85b27-dc23-498c-bca7-4976842734c4%2Fpoppms_processed.png&w=3840&q=75)
Transcribed Image Text:Task 1: Defining an array
This task consists in converting a number from base 10 to base 2. The user will input the n
integer numbers. The program must return the n binary numbers.
Step 1: Divide the given number 13 repeatedly by 2 until
you get '0'as the quotient
13 + 2 = 6 (Remainder 1)
6 + 2 = 3 (Remainder 0)
3 + 2 = 1 (Remainder 1)
1 + 2 = 0 (Remainder 1)
Step 2: Write the remainders in the reverse 1 10 1
order
:. 1310
= 1101,
(Decimal)
(Binary)
10
arr I4] =
1
1.2° t0-2*t 1.2+ O•2"= 10
Hint:
You must store each digit of the binary number in an array and then print them. The length of
the array must be equal to the number of digits of the binary number.
DO NOT USE ANY BUILT-IN FUNCTION
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