Solving Perfect Number Program in C

[Dili]

I need to write a program to check whether a number is perfect
eg: 6=1+2+3 , 28=1+2+4+7+14
i wrote one, but it doesn't work. also I've seen programs written with different preprocessors, we are required to write it on stdio.h and math.h


#include<stdio.h>
#include<math.h>

int main()
{
int a,b,c,d,i;
scanf("%d",&a);
{
for(i=1;i<a;i++)
b=a%i;
if(b==0)
{
c=0;
c=c+i;
}
if(c==a)
printf("perfect\n");
}
return 1;
}


 

[arunbg]

Few mistakes in variable initialization and placing of parentheses. Here is the modified correct version. Make sure you understand the logic.

#include<stdio.h>
#include<math.h>

int main()
{
int a,b,c,d,i;
scanf("%d",&a);c=0;
for(i=1;i<a;i++)
{
b=a%i;
if(b==0)
{
c=c+i;
} }
if(c==a)
printf("perfect\n");
return 1;
}

 

[CRGreathouse]

Heh. Here's a version you can't turn in, but which is far faster; it uses the Euclid identity of perfect numbers.

#include<stdio.h>
#include<math.h>

int is_prime(n) {
  int i,root;
  if (!(n&1))
    return n == 2;
  if (n%3 == 0)
    return n == 3;
  if (n < 25)
    return n > 1;
  root = (int)sqrt(n);
  for (i = 5; i <= root; i += 6)
    if (n%i == 0 || n%(i+2) == 0)
      return 0;
  return 1;
}

int main() {
  int even_part, odd_part;
  scanf("%d",&odd_part);
  even_part = 1;
  while (!(odd_part&3)) {
    odd_part = odd_part >> 2;
    even_part = even_part << 2;
  }
  if (!(odd_part&1)) {
    odd_part = odd_part >> 1;
    even_part = even_part << 1;
  }
  if (even_part << 1 == odd_part + 1 && is_prime(odd_part))
    printf("perfect\n");
  return 0;
}

Even faster and even cheatier, for 32-bit machines:

int main() {
  int n;
  scanf("%d",&n);
  if (n == 6 || n == 28 || n == 496 || n == 8128 || n == 33550336)
    printf("perfect\n");
  return 0;
}

 

[Dili]

Thank you very much arunbg. I am not good at C programming.
I do now understand the placing of brackets and c=0.
If no bother can you explain what is the difference between return 1 and return 0
Ive used them but with no real understanding
Thanks again

 

[arunbg]

int main() {
int n;
scanf("%d",&n);
if (n == 6 || n == 28 || n == 496 || n == 8128 || n == 33550336 || n == 8589869056)
printf("perfect\n");
return 0;
}

Lol...nice one.

 

[CRGreathouse]

If no bother can you explain what is the difference between return 1 and return 0
Ive used them but with no real understanding

Don't worry about it. These are error codes that the program returns, but they won't display or do anything unless they're checked by an outside program. 0 is supposed to mean "terminated normally" and anything other than 0 is supposed to mean "terminated with some kind of error".