Answer / karna

//all devisors of a number
#include<stdio.h>
#include<conio.h>
void main()
{
int n,i;
printf("Enter the number");
scanf("%d",&n);
for(i=1;i<=n;i++)
{
if((n%i)==0)
{
printf("%d\n",i);
}
else;
}
getch();
}

Answer / vclingisetty@gmail.com

#include<stdio.h>
#include<conio.h>
void main()
{
int n,i;
clrscr();
printf("enter a number\n");
scanf("%d",&n);
for(i=1;i<=n;i++)
{
if(n%i==0)
printf("%d",i);
}
getch();
}

Answer / veluri.haritha

BETTER CODE:


void main()
{
int i,n;
clrscr();
printf("Enter the number:");
scanf("%d",&n);
printf("\n\n The divisor's of %d are:",n);
for(i=1;i<=n/2;i++)
{
if(n%i==0)
printf(" %d ",i);
}
printf(" %d ",n);
getch();
}

This is better logic than the above program's.
For any number all factors will exit up to n/2 after n/2
only n is the factor.So, it is enough to execute the loop
for n/2 cycles . By this speed of execution increases.

EVEN BETTER LOGIC IS AVAILABLE 'U' TRY THIS...........
BY
V.HARITHA, B.TECH 1st YEAR..
student of "MADHAVI INFOTECH SOFTWARE TRAINING"
ANANTHAPUR,A.P,INDIA.

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