6 ms·

Don't know it well enough to code it from scratch at work. Have an idea. Will post something by tonight. Just finished a Data Structures class and still wrappi

by mipnix 16y ago

Don't know it well enough to code it from scratch at work. Have an idea. Will post something by tonight. Just finished a Data Structures class and still wrapping my head around Trees and hashing. Thank you, kindly.

mipnix 16y ago
Ok. Had to get this out quick... Back to work. Don't know if it will compile <code> #include<iostream> using namespace std; int main() { for(int i = 1; i <= 100; ++i) { if(i%3==0 &&i%5==0) cout <<"Fizzbuzz "; else if(i % 3==0) cout <<"FIzz "; else if(i%5==0) cout<<"buzz "; else cout << i << " "; } return 0; } </code>
- RiderOfGiraffes 16y ago
  #include<iostream> using namespace std; int main() { for(int i = 1; i <= 100; ++i) { if(i%3==0 && i%5==0) cout <<"Fizzbuzz "; else if(i%3==0) cout <<"FIzz "; else if(i%5==0) cout<<"buzz "; else cout << i << " "; } return 0; } Cool. Assuming you knocked that out as quickly as you suggest it shows that you've got good basic skills. Design and control of larger tasks/projects is untested, but it looks like you're well ahead of the basics. Now here's a question: Do you find anything "unsatisfactory" about that code?
  - mipnix 16y ago
    Besides not knowing how to post it properly...? The looping sequence, I imagine, doesn't need the if else queries. Probably simpler way to code it. I am definitely untested with larger projects.
    - DEADBEEF 16y ago
      Something like... (Pseudocode) for( i= 1,100 ) { if( i%3==0 ) { print( 'Fizz' ) } if( i%5==0 ) { print( 'Buzz' ) } print( '\n' ) } There's no need to treat FizzBuzz as a special case, as if the number is divisible by both it will have already met the conditions of the previous two statements. I wonder if there's some magic you could sprinkle in to the iterator so it only iterates through numbers which are divisible by 3/5, ignoring the rest? It probably wouldn't speed the operation up much (if at all) in this case, in a more complex real world scenario though it's best to look at every angle.
      - RiderOfGiraffes 16y ago
        You haven't printed any of the other numbers. Read the spec more carefully. You still need to print 1, 2, 4, etc.

by mipnix 16y ago

mipnix 16y ago

Ok. Had to get this out quick... Back to work. Don't know if it will compile <code> #include<iostream> using namespace std; int main() { for(int i = 1; i <= 100; ++i) { if(i%3==0 &&i%5==0) cout <<"Fizzbuzz "; else if(i % 3==0) cout <<"FIzz "; else if(i%5==0) cout<<"buzz "; else cout << i << " "; } return 0; } </code>

RiderOfGiraffes 16y ago
#include<iostream> using namespace std; int main() { for(int i = 1; i <= 100; ++i) { if(i%3==0 && i%5==0) cout <<"Fizzbuzz "; else if(i%3==0) cout <<"FIzz "; else if(i%5==0) cout<<"buzz "; else cout << i << " "; } return 0; } Cool. Assuming you knocked that out as quickly as you suggest it shows that you've got good basic skills. Design and control of larger tasks/projects is untested, but it looks like you're well ahead of the basics. Now here's a question: Do you find anything "unsatisfactory" about that code?
- mipnix 16y ago
  Besides not knowing how to post it properly...? The looping sequence, I imagine, doesn't need the if else queries. Probably simpler way to code it. I am definitely untested with larger projects.
  - DEADBEEF 16y ago
    Something like... (Pseudocode) for( i= 1,100 ) { if( i%3==0 ) { print( 'Fizz' ) } if( i%5==0 ) { print( 'Buzz' ) } print( '\n' ) } There's no need to treat FizzBuzz as a special case, as if the number is divisible by both it will have already met the conditions of the previous two statements. I wonder if there's some magic you could sprinkle in to the iterator so it only iterates through numbers which are divisible by 3/5, ignoring the rest? It probably wouldn't speed the operation up much (if at all) in this case, in a more complex real world scenario though it's best to look at every angle.
    - RiderOfGiraffes 16y ago
      You haven't printed any of the other numbers. Read the spec more carefully. You still need to print 1, 2, 4, etc.