#include <stdio.h> int main(int argc, char *argv[]) { int i; for(i = 1; i <= 100; i++) { if(i % (3*5) == 0) printf("FizzBuzz\n"); else if(i % 5 == 0) printf("Buzz\n"); else if(i % 3 == 0) printf("Fizz\n"); /* default */ else printf("%d\n", i); } return 0; }
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Don't waste time interviewing developers who aren't suitable for the job. Find out quickly how well someone can code. With Interview Zen, you set your own programming challenges and watch how candidates solve them online.
Capture the candidate's thought process and problem solving speed. See how they structure and revise their code over time. Evaluate their programming ability before you invest time in a face-to-face interview.
How It Works
Create Create the questions you'd like candidates to answer. Be as creative and job-specific as you like.
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Review Collaboratively review the recorded interviews in your own time. Save time and money identifying the best developers.
Demo
Write a program that prints the numbers from 1 to 100. But for multiples of three print "Fizz" instead of the number and for the multiples of five print "Buzz". For numbers which are multiples of both three and five print "FizzBuzz". (Thanks to Jeff Atwood)