C programming. The FizzBuzz program [closed]

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Closed 2 years ago.

I had a quiz and I wrote this code:

Print Fizz if it is divisible by 3 and it prints Buzz if it is divisible by 5. It prints FizzBuss if it is divisible by both. Otherwise, it will print the numbers between 1 and 100.

But after I arrived home, I wondered if could have writen it with less code. However, I could not come out with a shorter code. Can I do it with a shorter code? Thanks.

This is what I wrote and I think it works well. But can I have done it with less code.

#include <stdio.h>

int main(void)
{
    int i;
    for(i=1; i<=100; i++)
    {
        if(((i%3)||(i%5))== 0)
            printf("number= %d FizzBuzz\n", i);
        else if((i%3)==0)
            printf("number= %d Fizz\n", i);
        else if((i%5)==0)
            printf("number= %d Buzz\n", i);
        else
            printf("number= %d\n",i);

    }

    return 0;
}

回答1:

You could also do:

#include <stdio.h>

int main(void)
{
    int i;
    for(i=1; i<=100; ++i)
    {
        if (i % 3 == 0)
            printf("Fizz");
        if (i % 5 == 0)
            printf("Buzz");
        if ((i % 3 != 0) && (i % 5 != 0))
            printf("number=%d", i);
        printf("\n");
    }

    return 0;
}

A few lines shorter, and a lot easier to read.

回答2:

I'm not sure when you'd start calling it unreadable, but there's this.

#include <stdio.h>

int main(void)
{
   int i = 1;
   for (; i<=100; ++i) {
      printf("number= %d %s%s\n", i, i%3?"":"Fizz", i%5?"":"Buzz");
   }
   return 0;
}

回答3:

If a number is divisible by both 3 and 5, then it's divisible by 15, so:

for each number 1 to 100:
    if number % 15 == 0:
        print number, "fizzbuzz"
    else if number % 5 == 0:
        print number, "buzz"
    else if number % 3 == 0:
        print number, "fizz"
    else:
        print number

Other than that, you probably won't get it much shorter, at least in a conventional language like C (and I'm assuming you don't want the normal code-golf style modifications that make your code unreadable).

You could also get the whole thing into two lines if you packed the entire main function onto a single large line, but I would hope you wouldn't be after that sort of trickery either.

You can possibly get it faster (though you should check all performance claims for yourself) with something like:

static const char *xyzzy[] = {
    "",     "",     "fizz", "",     "buzz",
    "fizz", "",     "",     "fizz", "buzz",
    "",     "fizz", "",     "buzz", "fizzbuzz",
    // Duplicate those last three lines to have seven copies (7x15=105).
};
for (int i = 1; i <= 100; i++)
    printf ("%d %s\n", i, xyzzy[i-1]);

As an aside, that array of char pointers is likely to be less space-expensive than you think, thanks to constant amalgamation - in other words, it will be likely that there will only be one of each C string.

As I say, whether it's faster should be tested. In addition, your original specs only called for the shortest code so it may be irrelevant.

回答4:

I would say that modulo is expensive while comparisons are cheap so only perform the modulo once. That would yield something like this.

int i;
for( i = 0; i!=100; ++i ) {
    bool bModThree = !(i % 3);
    bool bModFive = !(i % 5);

    if( bModThree || bModFive ) {
        if( bModThree ) {
            printf( "Fizz" );
        }
        if( bModFive ) {
            printf( "Buzz" );
        }
    } else {
        printf( "%d", i );
    }

    printf( "\n" );
}

回答5:

i would write something like that

    main(){
  if (i % 3 == 0){
  cout<<"Fizz";
  }
  if (i % 5 == 0){
  cout<<"Buzz";
  }
  // So if both are true, it will print “FizzBuzz” and augment the two strings
    }

回答6:

This one avoids some code repetition but requires a temporary variable char t

void FizzBuzz( ) {
    char t = 0;
    for (unsigned char i = 1; i <= 100; ++i, t = 2) {
        (i % 3) ? --t : printf("Fizz");
        (i % 5) ? --t : printf("Buzz");
        if (!t) printf("%d", i);
        printf("\n");   
    }
}

回答7:

#include <stdio.h>

char const * template[] = {
  "%i",
  "Buzz",
  "Fizz",
  "FizzBuzz"
};
const int __donotuseme3[] = { 2, 0, 0 };
const int __donotuseme5[] = { 1, 0, 0, 0, 0 };
#define TEMPLATE(x) (template[__donotuseme3[(x) % 3] | __donotuseme5[(x) % 5]])

int
main(void) {
  int i;
  for (i = 1; i <= 100; i++) {
    printf(TEMPLATE(i), i);
    putchar('\n');
  }
  return 0;
}

回答8:

void main()
{
    int i = 0;
    char h[4];

    while (++i <= 100)
    {
        sprintf(h, "%d", i);
        printf("%s%s%s\n", i%3 ? "" : "fizz", i%5 ? "" : "buzz", (i%3 && i%5) ? h: "");
    }
}

回答9:

You can do it using a String:

String s="";
if(num%3==0)
   s+="fizz";
if(num%5==0)
   s+="buzz";
if(s.length()==0)
   s+=num+"";

回答10:

I'd go with a helper function :-)

#include <stdio.h>

int fbindex(int n) {
    int i = 0;
    if (n % 3 == 0) i += 1;
    if (n % 5 == 0) i += 2;
    return i;
}

int main(void) {                                 
    const char *fb[] = {"%d\n", "Fizz\n", "Buzz\n", "FizzBuzz\n"};
    for (int i = 1; i <= 100; i++) printf(fb[fbindex(i)], i);                
}

回答11:

Obfuscated form of Mr Lister's answer

main(int i){while(i++<100){printf("number= %d %s%s",i,i%3?"":"Fizz",i%5?"":"Buzz");}}