C Programmers, please help with small program

[Aznguy1872]

This is my second time asking for help again and I don't really like to but sometimes I just get stuck. If you guys could give me advice that would be great. So the program criteria is:

Write a program that prints all the perfect #'s less than 10,000.

A perfect # is one where the sum of its divisors is equal to the #, ex. 6=1 +2+3

this is what I got so far. I am just confused but i think what I got so far could be right.

int main (void) {
int i, j,SUM;
for (i=1;i<10000, i++)
for (j=1;j<i, j++)
{
}
if(i % j) == 0 {
SUM+= j
}

 

[EagleKeeper]

You need to come up with a thought process for solving such problems.

Without it; programming will never be your forte.

 

[Aznguy1872]

yeah.. I am an industrial engineer. Programming is not too important in this major. I've never had C programming expereince before this course.. SO kind of rough on me...

 

[EagleKeeper]

It is the logic of tackling a problem that you need to get a handle on; not the programming of the language at this point.

 

[JetBlack69]

6
28
496
8128


That should be your output.

 

[Aznguy1872]

ah hrm, u are right eagle keeper, tackling the problem. Thats where I usually struggle, is I don't know how to approach it... hrm 6,28,496, 8128. Shouldn't 1 be one also?

 

[JetBlack69]

Not according to mathworld

http://mathworld.wolfram.com/PerfectNumber.html

 

[Aznguy1872]

oh wait nm 1 shouldn't be one... oopz

 

[Aznguy1872]

i was just thinking about it, haha. and realize it, thanks

 

[Aznguy1872]

with my code, I think i need 2 if statements right? But at least am I going in the right direction?

 

[JetBlack69]

Oh, and your code is very close. You are on the right track.

EDIT: In fact, I copy and pasted your code, made some modifications that you are thinking of to print out the numbers in my original post.

 

[minofifa]

how did you figure that out JetBlack?

I was thinking about this, and i would write two functions to call:
- one would determine the prime factors of "i"
- the other would add the prime factors determined above

Then if tthat sum equals i, print it out. Finally, i would call those two above functions in a for loop that increments "i" from 0 to 10000.

My way seem a bit inefficient but i think it would work.

 

[Aznguy1872]

You guys on here are too smart for me...

 

[Aznguy1872]

So this is the code I have now.. I am at work and not at a compiler so I can't check what its printing....

int main (void) {
int i, j,SUM=0;
for (i=1;i<10000, i++)
for (j=1;j<i, j++)
{
}
if(i % j) == 0 {
SUM+= j
}
printf(" %d", SUM);
return 0;
}

 

[JetBlack69]

Originally posted by: minofifa
how did you figure that out JetBlack?

I was thinking about this, and i would write two functions to call:
- one would determine the prime factors of "i"
- the other would add the prime factors determined above

Then if tthat sum equals i, print it out. Finally, i would call those two above functions in a for loop that increments "i" from 0 to 10000.

My way seem a bit inefficient but i think it would work.

That's pretty much what Azneguy1872's code does. His "%" function determines if the numbers are factors and then it adds it to SUM.

 

[JetBlack69]

Originally posted by: Aznguy1872
So this is the code I have now.. I am at work and not at a compiler so I can't check what its printing....

int main (void) {
int i, j,SUM=0;
for (i=1;i<10000, i++)
for (j=1;j<i, j++)
{
}
if(i % j) == 0 {
SUM+= j
}
printf(" %d", SUM);
return 0;
}

That's good, but do you always want to increment SUM?

EDIT: Also, do you always want to print SUM?

 

[Aznguy1872]

oh... but doesn't incrementing that mean that if the modulus of i and j = 0, it will add that # to SUM?

 

[Aznguy1872]

maybe im interpreting that incorrectly. Dang, i wish I was at a compiler

 

[Aznguy1872]

oh its in the loop.... is that the problem?

EDIT: Thanks for advice btw

 

[JetBlack69]

Originally posted by: Aznguy1872
oh... but doesn't incrementing that mean that if the modulus of i and j = 0, it will add that # to SUM?

Yes, but once j == i, then you increment i and start the process over again, do you want to still add to SUM then? SUM will still contain the results from the previous iteration.

 

[JetBlack69]

Also, your brackets are wrong so nothing is getting done in the second for loop, but since you are not at a compiler, it's hard to debug.

 

[Aznguy1872]

yeah, i justed used a SSH so was able to get to a compiler, yea i got some errors.

 

[beggerking]

NM

 

[Aznguy1872]

How do my brackets look now? I am still getting some crazy errors though.

int main (void) {
int i, j,SUM=0;
for (i=1;i<10000, i++) {
for (j=1;j<i, j++) {

}
if(i % j) == 0 {
SUM+= j
}
}
printf(" %d", SUM);
return 0;
}

 

[JetBlack69]

Originally posted by: beggerking
NM

1+ 2 + 4 + 7 + 14