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Author Topic: Problem with multiple functions  (Read 2649 times)

Bill

  • Guest
Problem with multiple functions
« on: October 23, 2004, 01:43:24 PM »
Hi i wrote this piece of code. It's ment to take in a user input and didive it by 2 to get all it's factors. Then it is ment to divide the factors by 1-100 to check it is a prime number then display the prime factors. Can someone please help me.
Code: [Select]

#include <stdio.h>

int collection(int a);
int factor (int b);

int main()
{
int a;
printf("Please enter a number: ");
scanf("%d",&a);
collection (a);
return 0;
}

int collection(int a)
{
int b,i;
for(i=1; i<=100; i++)
{
b=a/i;
factor (b);
//printf("  %lf \n",b);
if(a%i==0)
{
// printf("Display the factors %d %t %t %d\n",b,i);

}
}
return 0;
}

int factor(int b)
{
float k,c;
for(k=1; k<=100; k++)
{
c=b/k;
printf(" %d \n",c);
}
return 0;
}

wbhart

  • Guest
Primes
« Reply #1 on: May 01, 2005, 01:28:26 AM »
Your friend Nazmul must be in the same computing class.

Let's look at what your program actually does.

It gets an integer "a" from the user.

For each integer i from 1 to 100, it divides a by i and rounds down to the nearest integer, and calls the result b.

Then, each time, it calls the routine, factor(b). This divides b by each number from 1 to 100 and displays ..... something.

Is that what you wanted to do?

Where is your routine for checking whether a number is prime? How does your program actually tell if a number is a factor or not? Or does it just blindly divide through regardless of whether it is dividing by a factor or not.

When writing a program like this, you need to think about what steps your program is going to need to undertake, before writing the code.

E.g:

for each factor of the number a, if the factor is prime, display the factor and divide the number by that factor and go again, until we have only 1 left.

Hope that helps.

Bill.