In the source code all commands writing to stdout (puts and printf) are commented out because they are slow.
for (unsigned int i = 1; i <= N; i++) {
bool div_three = i % 3 == 0;
bool div_five = i % 5 == 0;
if (div_three && div_five) {
puts("FizzBuzz");
}
else if (div_three) {
puts("Fizz");
}
else if (div_five) {
puts("Buzz");
}
else {
printf("%i\n", i);
}
}This is probably the most intuitive and straight-forward way to implement FizzBuzz. But it has an issue: The modulo operator is expensive.
So how can we do better? Let's take a look at the second algorithm.
char *strings[3] = {"Fizz", "Buzz", "FizzBuzz"};
int num_to_string[15] = {-1, -1, 0, -1, 1, 0, -1, -1, 0, 1, -1, 0, -1, -1, 2};
for (unsigned int i = 0; i <= N; i+=15) {
for (unsigned int j = 1; j <= 15 && i+j <= N; j++) {
if (num_to_string[j-1] == -1) {
printf("%i\n", i + j);
}
else {
puts(strings[num_to_string[j-1]]);
}
}
}Since 15 is the smallest number which is divisible by 3 and 5, every multiple of 15 is also divisible by both 3 and 5. Hence the same pattern will occur every 15 iterations. This means we can save some computation time by computing this pattern once and storing it in memory.
For N = 1000000000 these are the results of the benchmark (without printing) running it on an Intel(R) Core(TM) i5-3320M:
TIME IN fizzbuzz1: 10.57s
TIME IN fizzbuzz2: 4.09s