Fizz Buzz.. Crackle Pop! (Generalized Fizz Buzz)

Question

We've all heard of the ol' Fizz Buzz problem, but what happens when you try to apply it with more factors? Fizz Buzz Crackle Pop!

The Challenge

Write a full program that takes an integer input n, then n tuples containing an integer and a string, then another integer (>1) k, as such:

 n int1 str1 int2 str2 (...) intn strn k

You can take this line from either command line or STDIN.

Then, for all integers 1 to k, if it is divisible by any of int1, int2...intn, output all corresponding strs in the input order, followed by a newline. If it isn't, then just output the integer, followed by a newline.

For example, with input

3 2 Fizz 3 Buzz 5 Crackle 10

we get

1
Fizz
Buzz
Fizz
Crackle
FizzBuzz
7
Fizz
Buzz
FizzCrackle

But with input (notice the order change)

3 3 Buzz 2 Fizz 5 Crackle 10

we get

1
Fizz
Buzz
Fizz
Crackle
BuzzFizz
7
Fizz
Buzz
FizzCrackle

Optional trailing newline is acceptable.

Shortest code in bytes wins.

Edits:

Obviously I've missed out a lot, sorry.

• Inputs from console and STDIN, anything else gets +5 bytes (:c) anywhere
• Full programs please.
• Assume non-empty strings for strs
• No guarantees for uniqueness of ints

Sample C++ program (limited at 20 because I'm lazy):

#include <iostream>
#include <string>
using namespace std;

int main() {
  string names[20];
  int mods[20], n, max;
  cin >> max >> n;
  for (int i=0; i<n; i++) {
    cin >> mods[i] >> names[i];
  }
  for (int i=1; i<=max; i++) {
    bool found = false;

    for (int j=0; j<n; j++) {
      if (i % mods[j] == 0) {
        found = true;
        cout << names[j];
      }
    }
    if (!found)
     cout << i;
    cout << endl;
  }

  return 0;
}

Answer 1

JavaScript (ES6), 90 bytes

Generates a leading newline.

f=(a,i=a.pop())=>i?f(a,i-1)+`
`+(a.map((_,j)=>++j>a[0]|i%a[j*2-1]?'':a[j*2]).join``||i):''

Test

f=(a,i=a.pop())=>i?f(a,i-1)+`
`+(a.map((_,j)=>++j>a[0]|i%a[j*2-1]?'':a[j*2]).join``||i):''

console.log(f([3, 2, 'Fizz', 3, 'Buzz', 5, 'Crackle', 10]))
console.log(f([3, 3, 'Buzz', 2, 'Fizz', 5, 'Crackle', 10]))

Answer 2

05AB1E, 28 bytes

#¦¤U¨2ôøVXLvY`sysÖÏJDg1‹iy},

Try it online!

Or with a different input format:

05AB1E, 16 bytes

Lv²y³ÖÏJDg1‹iy},

Try it online!

Answer 3

Python 2, 98 bytes

lambda b:[''.join(''if x%b[1+i*2]else b[2+i*2]for i in range(b[0]))or`x`for x in range(1,b[-1]+1)]

Try it online!

Answer 4

C++, 194 bytes

#include <iostream>
#define p std::cout<<
int main(int c,char**a){c=2*atoi(a[1])+2;int x,f,i,n=atoi(a[c]);for(x=1;x<=n;x++){f=0;for(i=2;i<c;i+=2)if(x%atoi(a[i])<1)f=1,p a[i+1];if(!f)p x;p'\n';}}

Ungolfed:

#include <iostream>

int main(int c, char **a) {
    c = 2 * atoi(a[1]) + 2;
    int x, f, i, n = atoi(a[c]);
    for (x = 1; x <= n; x++) {
        f = 0;
        for (i = 2; i < c; i += 2)
            if (x % atoi(a[i]) < 1) f = 1, std::cout << a[i+1];
        if (!f) std::cout << x;
        std::cout << '\n';
    }
}

Answer 5

PHP, 99 bytes

Based on primo´s FizzBuzz answer: õ is chr(245), a bit inverted newline.

for(;$i++<($a=$argv)[$z=$argc-1];){for($k=$s="";$z>$k+=2;)$s.=[$a[$k+1]][$i%$a[$k]];echo$s?:$i,~õ;}

ignores the first argument; run with -nr.

Answer 6

JavaScript (ES6), 105 97 bytes

g=(m,k,i=1)=>i<-~k?([...m.keys()].filter(j=>i%j<1).map(j=>m.get(j)).join``||i)+"\n"+g(m,k,i+1):""

Takes in a map of pairs m (integer, string) and an integer k. Comes with a trailing newline.

Here's a non-recursive version (105 bytes), but doesn't yield a trailing newline.

m=>k=>[...Array(k).keys()].map(x=>[...m.keys()].filter(j=>-~x%j<1).map(j=>m.get(j)).join``||x+1).join`\n`

Try it online!

Answer 7

Java, 331 bytes

Because Java.

import java.util.*;class A{A(int c,String x){i=c;v=x;}int i;String v;void x(String[]x){ArrayList<A>l=new ArrayList();int n=0;for(;++n<x.length-1;)l.add(new A(Integer.valueOf(x[n++]),x[n]));n=Integer.valueOf(x[n]);for(int i=1;i++<n;){String o="";boolean y=1>0;for(A a:l)if(i%a.i==0){y=1<0;o+=a.v;}if(y)o+=i;System.out.println(o);}}}

This is the full class required for this. However, in order to run it, you must call the method x on an existing instance of A. For the sake of testing, I have provided a command-line runnable class below, which is partially ungolfed.

import java.util.*;
class A{
A(int c,String x){i=c;v=x;}
int i;
String v;
void x(String[]x){
ArrayList<A>l=new ArrayList();
int n=0;
for(;++n<x.length-1;)
l.add(new A(Integer.valueOf(x[n++]),x[n]));
n=Integer.valueOf(x[n]);
for(int i=1;i++<n;){
String o="";
boolean y=1>0;
for(A a:l)if(i%a.i==0){y=1<0;o+=a.v;}
if(y)o+=i;
System.out.println(o);
}
}
public static void main(String[] args) {
new A(0,"").x(args);
}
}

Answer 8

stacked, 85 bytes

args rev...2*nsgroup rev 2 chunk@s~>{!s[1#]map s[0#n|]map keep''join:[n]\¬if out}map

Try it online! Alternatively, 86 bytes:

args behead...@k sgroup 2 chunk@s k~>{!s[1#]map s[0#n|]map keep''join:[n]\¬if out}map

Answer 9

dc, 121 bytes

?dstsw?[rdlt:Y:Rlt1-dst0<q]dsqx?sb1sm[ln;Y;RnlP1+sP]ss[lmn]sF[1sn0sP[lnd;Ylmr%0=s1+dsnlw!<g]dsgxlP0=Flm1+dsm10Plb!<i]dsix

Takes input on 3 separate lines, the first line containing the integer n, the second housing the int str tuples with the strings enclosed in square brackets ([]), and the third line consists of the integer k. For example, 3 2 Fizz 3 Buzz 5 Crackle 10 could be input as:

3
3 [Buzz] 2 [Fizz] 5 [Crackle]
10

Try it online!

Or taking input in a different order:

dc, 118 bytes

?dstsw[dlt:Y:Rlt1-dst0<q]dsqxsb1sm[ln;Y;RnlP1+sP]ss[lmn]sF[1sn0sP[lnd;Ylmr%0=s1+dsnlw!<g]dsgxlP0=Flm1+dsm10Plb!<i]dsix

This takes input in a different order, but on a single line in the format

k [str1] int1 [str2] int2 (...) [strn] intn n

For example, 3 2 Fizz 3 Buzz 5 Crackle 10 would be input as:

10 [Buzz] 3 [Fizz] 2 [Crackle] 5 3

Try it online!