₦airaland Forum

Beaf:
It is really sad and upsetting that anyone would come up with such a cheap challenge.
Please dudes, either make sure to come up with brain crackers, or just ask for coding help.

Another surface-thinker. Don't be ignorant. Read the thread. How many people actually got what I wanted?

Mobinga:
Another surface-thinker, How many people actually got what I wanted?

Nailed!

//recursive solution
public int multiples7(int i)
{
if(i%7==0 && i<350)
{
System.out.print(i + ' ');
i*=7;
multiples7(i);
}
}

okechukwu.diei:
//recursive solution
public int multiples7(int i)
{
if(i%7==0 && i<350)
{
System.out.print(i + ' ');
i*=7;
multiples7(i);
}
}

There are a few errors I made. The method should return 'void' and not 'int' and the second boolean expression in the if statement should b 'i<=350'.

Good, but doesn't it check if every positive integer below 350 has a modulus 7 => 0?

Mobinga:
Nice. I think this is faster though tedious

System.out.println("0/n7/n14, " 

Since it doesn't loop at all. It just prints the string.


We should be doing code-benchmarks more often

Ladies and gentlemen i believe we have ourselves a winner

Anyway for now i think i'll just stand by the side, watch the winning code and rewrite it in Python

Mobinga:
Good, but doesn't it check if every positive integer below 350 has a modulus 7 => 0?

@Mobinga it doesn't. The % operator is used to evaluate the integer remainder of a modulus division. This trick can be used to find multiples of 7 cos a multiple of 7 will evaluate to 0 if it's modulo 7. For instance 14 modulo 7 is 0 whereas 15 modulo 7 is 1. Thus 15 is not a multiple of 7 cos its modulo doesn't evaluate to 0. The second part of the conditional if statement i <= 350 ensures that the statement within the if block will be executed in 50 recursive calls. Now carefully examine my code and you'll discover that if it's implemented you'll get an infinite loop. This is because in the call to the recursive method within the if block i forgot to increment the argument i by 1. Hence, the same argument is always passed to the recursive method and i is ALWAYS less than 350. The simplest way to prevent this ugly situation is presented bellow:
public void multiple7(int i)
{
if (i%7==0 && i<=350)
{
System.out.format("%d3", i);
/* notice i used System.out.format() & not println(). This is Java's implementation of the printf() function in C programming language. */
multiple7(i+1);
/* or pass argument i as ++i using pre-order increment operator */
}
} nted you'll get an infinite loop. This is because in the call to the recursive method within the if block i forgot to increment the argument i by 1. Hence, the same argument is always passed to the recursive method and i is ALWAYS less than 350. The simplest way to prevent this ugly situation is presented bellow:
public void multiple7(int i)
{
if (i%7==0 && i<=350)
{
System.out.format("%d3", i);
/* notice i used System.out.format() & not println(). This is Java's implementation of the printf() function in C programming language. */
multiple7(i+1);
/* or pass argument i as ++i using pre-order increment operator */
}
}

@mobinga dnt forget to add the line of code i*=7; BEFORE calling the recursive method multiple7(i+1). Also, in the statement System.out.format(), modify the format string "%d3" to "%d3 %n \r\n". If u know C\C++ then I don't need to explain the format string.