Can't think of how to continue this prog....

[Unregistered]

I can't figure out how to continue the prog....it is supposed to find all perfect numbers from 1 to 100 and i'm supposed to make it "efficient". Here is what i have so far...not much...but i'm stuck now...

#include <stdio.h>

main()
{
int findPerfect(int num);
void displayPerfect(int num);
printf("This program displays a list of all perfect numbers from 1 to 100\n\n");
displayPerfect();
}

void displayPerfect(bool findPerfect())
{
int counter;
for(counter=1;counter<=100;counter++);
{
if(findPerfect(counter))
{
return (printf("%d\n",counter));
}
}
}

bool findPerfect(int num)
{

}

[golfinguy4]

Think about what a perfect # is (any number squared). Then, display the squares of all number from 1 to 10 because 10^2=100.

[zahari]

Listen man I think you don't know what a perfect is!!!
6 is the first != 1 because 2*3=6 %)).
....
But in long type they are a few (three or four)!!!

So this is a geeky way to find them but if you are obstinate
you'll reach success. (I've just written the code in some seconds.)

#include <stdio.h>

long find_next_perfect(long k) {
long i,sum=1,numb=k+1;
int ok=0;

while (!ok) {
for (i=2;i<=numb/2;i++) sum*=i;
ok=sum==numb;
sum=1; numb++;
}
return --numb;
}

int main() {
long p=0;

while (1) { /* forever uuuh yea */
p=find_next_perfect(p);
printf("%ld\n",p);
}
return 0;
}

[pinko_liberal]

Originally posted by golfinguy4
Think about what a perfect # is (any number squared). Then, display the squares of all number from 1 to 10 because 10^2=100.

A perfect number is one which is equal to the sum of its divisors less than it
e.g, 6= 1 + 2 + 3
28 = 1 + 2 + 4 + 7 + 14

[Prelude]

>it is supposed to find all perfect numbers from 1 to 100 and i'm supposed to make it "efficient".
Well, since you asked for something efficient and we know that there are only two perfect numbers from 1 to 100, we can either hard code it or use a lookup table:

#include <stdio.h>

int main ( void )
{
  printf ( "%d %d\n", 6, 28 );
  return 0;
}

#include <stdio.h>

static int perfect[] = { 6, 28, -1 };

int main ( void )
{
  int i;
  for ( i = 0; perfect[i] != -1; i++ )
    printf ( "%d ", perfect[i] );
  printf ( "\n" );
  return 0;
}

-Prelude

[quzah]

I love that solution. I really do.

Quzah.