1. FizzBuzz 2.0 by CodeChum Admin Remember the game of FizzBuzz from the last time? Well, I thought of some changes in the game, and also with the help of loops as well. Firstly, you'll be asking for a random integer and then loop from 1 until that integer. Then, Implement these conditions in the game: • print "Fizz" if the number is divisible by • print "Buzz" if the number is divisible by • print "Fizzbuzz" if the number is divisible by both 3 and 5 • print the number itself if none of the above conditions are met
1. FizzBuzz 2.0 by CodeChum Admin Remember the game of FizzBuzz from the last time? Well, I thought of some changes in the game, and also with the help of loops as well. Firstly, you'll be asking for a random integer and then loop from 1 until that integer. Then, Implement these conditions in the game: • print "Fizz" if the number is divisible by • print "Buzz" if the number is divisible by • print "Fizzbuzz" if the number is divisible by both 3 and 5 • print the number itself if none of the above conditions are met
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Program C
![main.c
Test Cases
Executions
1. FizzBuzz 2.0
Fincludecstdie.h
nainO
int n:
by CodeChum Admin
main.c:20:5: error: 'else' without a previous 1f
20 |
else if("nS--e");
Remember the game of FizzBuzz from the last
time? Well, I thought of some changes in the
game, and also with the help of loops as well.
Firstly, you'll be asking for a random integer
and then loop from 1 until that integer. Then,
printf("Enter n: );
scanf("Xd,&n");
int ni - 0:
Expected Output
implement these conditions in the game:
while (nenn1):{
• print "Fizz" if the number is divisible by
Enter n: 15
if (A33-0 nXS-0);
{
11
12
• print "Buzz" if the number is divisible by
13
printf("fizzbuzz");
Fizz
14
• print "FizzBuzz" if the number is
15
divisible by both 3 and 5
16
else if("n3-0");
Buzz
• print the number itself if none of the
17
Fizz
printf("fizz");
above conditions are met
18
19
20
Fizz
21
else if("AS-0");
Input
Buzz
22
11
1. An integer
23
printf(buzz");
24
13
25
Output
26
14
27
FizzBuzz
The first line will contain a message prompt to
28
input the integer.
The succeeding lines will contain either a string
or an integer.
O Test Case 2
Enter n: 15
1
2
O Test Case 3
Fiz
4
Buzz
Fiz
Buzz](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff4a50dfd-6dc6-45b7-897d-682dcecf7909%2Fcbd5a41e-b447-4ac5-8975-4540a89433f8%2Fknnps8a_processed.png&w=3840&q=75)
Transcribed Image Text:main.c
Test Cases
Executions
1. FizzBuzz 2.0
Fincludecstdie.h
nainO
int n:
by CodeChum Admin
main.c:20:5: error: 'else' without a previous 1f
20 |
else if("nS--e");
Remember the game of FizzBuzz from the last
time? Well, I thought of some changes in the
game, and also with the help of loops as well.
Firstly, you'll be asking for a random integer
and then loop from 1 until that integer. Then,
printf("Enter n: );
scanf("Xd,&n");
int ni - 0:
Expected Output
implement these conditions in the game:
while (nenn1):{
• print "Fizz" if the number is divisible by
Enter n: 15
if (A33-0 nXS-0);
{
11
12
• print "Buzz" if the number is divisible by
13
printf("fizzbuzz");
Fizz
14
• print "FizzBuzz" if the number is
15
divisible by both 3 and 5
16
else if("n3-0");
Buzz
• print the number itself if none of the
17
Fizz
printf("fizz");
above conditions are met
18
19
20
Fizz
21
else if("AS-0");
Input
Buzz
22
11
1. An integer
23
printf(buzz");
24
13
25
Output
26
14
27
FizzBuzz
The first line will contain a message prompt to
28
input the integer.
The succeeding lines will contain either a string
or an integer.
O Test Case 2
Enter n: 15
1
2
O Test Case 3
Fiz
4
Buzz
Fiz
Buzz
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