The program is to find all the prime numbers falls inside the user defined range. A prime number is a one, whose divisors are 1 and the number itself.
Logic: This is advanced version of the previous program. Here, user need to enter two numbers as the lower and upper limits for the iteration loop to find the prime number in between. The outer for loop traces the iteration till the limit, wherein each of iteration inner for loop checks the present number is prime or not, with the prime number’s logic. If it is, it prints out the present number.
In this program, both the lower limit and the upper limit are variables, and so is flexible. These three programs show, how we can upgrade the logic to make the code flexible one.
The similar growth can be seen in the case of finding the perfect numbers.
Program to find the prime numbers between a given range
#include<stdio.h>
void main()
{
int i, prime, lim_up, lim_low, n;
clrscr();
printf(“\n\n\t ENTER THE LOWER LIMIT…: “);
scanf(“%d”, &lim_low);
printf(“\n\n\t ENTER THE UPPER LIMIT…: “);
scanf(“%d”, &lim_up);
printf(“\n\n\t PRIME NUMBERS ARE…: “);
for(n=lim_low+1; n<lim_up; n++)
{
prime = 1;
for(i=2; i<n; i++)
if(n%i == 0)
{
prime = 0;
break;
}
if(prime)
printf(“\n\n\t\t\t%d”, n);
}
getch();
}
Download exe and source code here.
thank you
Thanks for your logic dude which helped me a lot in finding my little
mistake in my program.
thanx a lot.. pls help me wid a prog to convert a num to its binary form..
thanks
nice program dude……..
nice programme by revathi!! its best one
#include
void main()
{
int i,n,a,b,j=0;
printf(“enter range in which you want to find out the prime numbers seperated by space: “);
scanf(“%d%d”,&a,&b);
if(a<0||b<0)
{
printf("number is negative");
goto there;
}
printf("prime number in entered range are=\n");
for(n=a;n<=b;n++)
{
for(i=2;i<=(n/2);i++)
{
if (n%i==0)
j++;
}
if(j==0)
printf("%d ",n);
j=0;
}
there:
}
good
//sudheer.k mulapadava
class A
{
public static void main(String[] args)
{
int count=0;int sum=0;
for (int i=2;i<100 ;i++ )
{
for (int j=2;j<i ;j++ )
{
if(i%j==0)
count++;
}
if(count==0)
System.out.println("Prime numbers are :"+i);
else
{
sum=sum+i;
}
count=0;
}
System.out.println(" Sum of non Prime numbers are :"+sum);
}
}