Find the sum of all the prime numbers

A friend of mine who is learning to program in C, was having diffculty in writing a program to find the sum of all the prime numbers between a specified limit (in her case 3-60). I helped her out with the program, and I thought I'll post it here which might help others.

Code:

 /*
 ** Program to find the sum of all numbers between 3 and 60
 ** @author : Pradeep
 ** @date : 11/29/2006
 */
 
 #include <stdio.h>
 
 void main(void)
 {
     unsigned int i,j,s=0,is_prime;
 
     for(i=3;i<=60;i++)
     {
         is_prime = 1; // Assuming that current value of i is prime
         
         // Checking for prime
         for(j=2;j<=(i/2);j++)
         {
             if(i%j==0) // Not prime, set is_prime to 0 and break
             {
                 is_prime = 0;
                 break;
             }
         }
 
         if(is_prime) // If the number is prime, sum it up
         {
             s += i;
             // optionally you can print the prime numbers too
             printf("%d ",i);
         }
     }
 
     printf("\n\nThe sum of the prime numbers = %d",s);
 }

 

well this code is good but it ll take more time(for computational work)
please check this code the 2nd for loop

Code:

 /* ** Program to find the sum of all numbers between 3 and 60 ** @author : Pradeep ** @date : 11/29/2006 */  
#include <stdio.h>  
#include<math.h> 
void main(void)
  { 
    printf("\n please enter the upper limit");
   int n;
    scanf("%d",&n);\\now n can be given by user
    unsigned int i,j,s=0,is_prime;     
    for(i=3;i<=n;i++) 
    {   
       is_prime = 1; // Assuming that current value of i is prime           
       // Checking for prime    
       for(j=2;j<=sqrt(i);j++)  
       {  
           if(i%j==0) // Not prime, set is_prime to 0 and break   
          {
            is_prime = 0;           
            break;       
           }  
       }     
     if(is_prime) // If the number is prime, sum it up     
      {  s += i;             // optionally you can print the prime numbers too 
            printf("%d ",i);  
       }  
   }   
   printf("\n\nThe sum of the prime numbers = %d",s);
 } 

//see though your program will work proper but if we have to sum up prime nos. between 3 to n
//where n verey big(though within the limits of int )
// then the checkin condition from to n/2 will take lot of time
// suppose ur n==100 then it will check from n=2 to n=50
//so the no. of checking loops required for 100 will be 48
//where as if we go from n=2 to n= 10 we only require 8 loops so it will be much faster
// now think for the case of n=1000 wouldnt it will take lot of time

 

I understood your point, but I see you have used sqrt() ! How does that help?? And we need to check whether it got a factor execpt for itself and 1 or not.

 

yes i got you!
the thing is we dont need to go more than the sqrt of the no. which to be checked as prime if it is not divisibe by any no. upto sqrt(n) than there wont be any factor above that you can check it for any no.

 

Its not using square root anywhere, and neither does it require too. Could you please tell me why does it need to use square root??

 

good point shabbir

 

For example, sqrt(60) = 7.46, that means we will loop 7 times, but 60 is divisible by 30 also.
The program works fine with sqrt, but what is the logic?

 

Interesting, so we will find the factor of n within any of the number in sqrt(n) if not then its a prime number.

 

let me tell you the logic now
it is very simple
look if k is any no.
then k can be represented as k=i*j
where i and are k's factor
now i varies from 1to k
and j varies from k to 1
but for prime no. we check for i=2 to k/2
acordingly j will varies from (k/2) to 2
now if 2 is a factor of k
then automatically k/2 is also a factor and so on
that is when i increases, j decreases
so we continue like this then there is condition that both i and j becomes equal
so i*i=k
so i=sqrt(k)
but if k is not a square of any no. then i increases upto a integer just less than sqrt iof k
thats all what i did
now try this problem
for a given no. print all its prime factors that is
if input is 12 out put should be 2*2*3
using the hint that the lowest factor of any no. other than one is always prime
try it

 

thank u sir
but ma name is niraj

 

why do we need a square root

 

Excellent explaination. Now any one can understand either he is new or experience

 

for(j=2;j<=sqrt(i);j++) :cryin:

is similar to

for(j=2;j*j <= i;j++) :p

so, no sqrt in here since it is really slow. At least if was some years ago.

 

for getting the prime number try

http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes

No devisions in here