Plz help me to write this program..

X-Man · #1 07-May-2004, 06:59

Hi All,

Who can i write a program that tests a number if its "prime" or not

regards

machinated · #2 07-May-2004, 07:25

You must start on your own and come up with something and then ask specific questions when you are stuck. can't write your code for you

Hint: prime numbers are not naturally divisible by anything except themselves and 1.
Hint: all the prime numbers are odd except 2.
Examples: 2(only even prime number), 3, 5, 7, (not 9), 11 ....

X-Man · #3 07-May-2004, 07:47

Hi

Thanks machinated,

Thats My program

________________________

CPP / C++ / C Code:

#include<stdio.h>

int main()
{
int a;

printf ("Enter a number to test it: ");
sancf ("d%",a);

if (a%2 !! a%3 !! a%5 == 0)  printf ("The number is not prime");
else printf ("The number is prime");


}

_________________

What do you say?

Regards

dsmith · #4 07-May-2004, 08:00

Hi X-man, welcome to GIDForums.

First of all let me direct you to this thread. Therein is a lengthy discussion of prime numbers with various solutions and code snippets.

As for your code, I think that you are on the right track, but it is not complete. 49 is at least one number I can think of off the top of my head that would be shown as prime by your program. I think using what you have and some of the algorithms shown in that other thread, you should be able to solve this thing quite easily.

Good Luck,
d

aaroncohn · #5 07-May-2004, 09:38

Quote:

Originally Posted by X-Man

CPP / C++ / C Code:

if (a%2 !! a%3 !! a%5 == 0)  printf ("The number is not prime");

"!!" is not an operator... go to this webpage for a list of valid C++ operators.

tay · #6 07-May-2004, 11:28

i help u code 1 example of prime number
actually there is many way to write
but is almost similar
i not sure my code is short enough and efficiency
you try to understand it
you can reference but try not copy, cause programming is not just have a solution
they is many solution, is either efficiency or not.

CPP / C++ / C Code:

int main()
{
int a;
int is_prime=0;  // this is counter to check whether is any num and mod to 0
cin>>a;  // receive a digit from user
cout<<endl; // next line

for (int i=2; i<a; i++)  // check any num can be divided 
{                            // i must be start at 2 to (a-1)  e.g a=9, so i=2;i<9;i++
if(a%i==0)  //check any number can be mod to 0
is_prime++;  
}
if(is_prime==0)  // mean no any num can mod to 0
cout<<a <<" It is prime number "<<endl;

else // mean is_prime is >0
cout<<"Not prime number"<<endl;

X-Man · #7 09-May-2004, 09:25

Hi all,

Thanks for your replay

I wrote the program befor i see your replys

Here is my "program"

CPP / C++ / C Code:

#include<stdio.h> 

int main() 
{ 


int a,b,num; 

b = 0 ; 

printf ("Enter a number: "); 
scanf ("%d",&num); 


for ( a = 2 ; a < 1000 ; a ++) 
{ 

if (a == num ) continue; 

if ( num%a == 0 ) b = 1; 



} 

if (b==1) printf ("Its not prime"); 
if (b==0) printf ("Its prime"); 

}

Max Payne · #8 10-May-2004, 01:37

I don't see why there is a need for a 'for' loop of 1000 when you have the number to validate. Well, you may not looking forward for effeciancy, but it is a good practice!

WaltP · #9 10-May-2004, 02:05

Quote:

Originally Posted by X-Man

Hi all,

Thanks for your replay

I wrote the program befor i see your replys

Here is my "program"

CPP / C++ / C Code:

#include<stdio.h> 

int main() 
{ 
    int a,b,num; 
    b = 0 ; 

    printf ("Enter a number: "); 
    scanf ("%d",&num); 
    for ( a = 2 ; a < 1000 ; a ++)
    { 
        if (a == num ) continue; 
        if ( num%a == 0 ) b = 1; 
    } 

    if (b==1) printf ("Its not prime"); 
    if (b==0) printf ("Its prime"); 
}

Fisrt off, watch your formatting. This program should be formatted like above.

The for loop should definitely be changed. Following are progressively better loops:
1) Don't loop to 1000. The number is your absolute max:
for ( a = 2 ; a <= num ; a ++)

2) Thinking about the loop, you also obviously only have to go thru half the numbers:
for ( a = 2 ; a < num/2 ; a ++)

3) Not so obviously, the real limit is the square root of the number:
for ( a = 2 ; a < sqrt(num) ; a ++)
For this you need to add the math header file.

4) Check for 2 first, the the loop can start at 3 and skip all even numbers:

CPP / C++ / C Code:

if (num == 2)
{
    prime = 1;
}
else
{
    for ( a = 3 ; a < sqrt(num) ; a+=2)
    {
    }
}

Finally, inside the loop all you need is:

CPP / C++ / C Code:

for ( a = 3 ; a < sqrt(num) ; a+=2)
{ 
    if ( num%a == 0 )
    {
        b = 1; 
        break;  // no need to continue thru the loop so leave
    }
}

So the entire program can become:

CPP / C++ / C Code:

#include<stdio.h> 

int main() 
{ 
    int a,b,num; 
    b = 0 ; 

    printf ("Enter a number: "); 
    scanf ("%d",&num); 
    if (num == 2)
    {
        prime = 1;
    }
    else
    {
        for ( a = 3 ; a < sqrt(num) ; a+=2)
        {
            if ( num%a == 0 )
            {
                b = 1; 
                break;  // no need to continue thru the loop so leave
            }
        }
    }    for ( a = 2 ; a < 1000 ; a ++)

    if (b==1) printf ("Its not prime"); 
    if (b==0) printf ("Its prime"); 

    return 0;    // don't forget the return, main() is an int function
}

WaltP · #10 10-May-2004, 10:07

Oops, the line prime = 1; should be b = 1; in the final code above.

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Plz help me to write this program..

Thanks All