Program to calculate prime and perfect numbers

[Alenik1989]

i am a begginer in c++ programming and im stuck at using loops so plz help me!!!

i need to write a program that analyze one or more # (between 1 and 1000) entered by a user. Then determin if it is a prime number or perfect number.
And i have to use the loops at a bigginer level. so plz help me to get through this with a simple program!!!!!!!!!!!!TANX

[Ganon11]

We'd be glad to help you through it - so what do you have so far?

[Alenik1989]

ok i have the code for finding the devisor of any number!!!!!i just cant find how to write a code that show the number entered by user is a prime number?and also how to get the sum of factors for perfect number??????TANX FOR YOU HELPS
so far i have this basic code~~~~dont laugh!!!

#include<iostream>
include<cmath>
using namespace std;
int main()
{
int num,root,dev;
const int SENTINEL=-1;                                //constant and integers
cout<<"ENTER #<enter -1 to stop>:";
 
    cin>>num;                                   //# entered by user
    while(num!=SENTINEL)                 
{
    root=pow(num,0.5);                       //second root of that #
    cout<<"devisors are: 1"<<" "<<num<<" ";
    for (root;1<root;root--)
{
        if(num%root==0)                                  //the loop and conditions
{
    dev=num/root;
    cout<<root<<" "<<dev<<" ";
 
}
       else
    continue;                
}
    cout<<endl;
    cin>>num;
}
    return 0;
}

[ajrattan05]

Heres a hint you should try using the mod operator"%"..
It will give you single digits for perfect number evaluation.
And prime is repeated use of operator "mod" until remainder is not equal to zero

[Alenik1989]

Tanx but i still need to fin a code for adding the devisors (except the # itself).
and a code for determining a entered number as a prime or perfect or just a number!!!!

[Banfa]

For determining perfect numbers in the users input the most efficient method would be to use a look-up table. This is beacuse there are only 5 perfect numbers with-in range for a 32-bit int and only 8 in range of a 64 bit int.

Perfect numbers are very rare and you can easily look up a list of the lower ones on the internet.

[sicarie]

Tanx but i still need to fin a code for adding the devisors (except the # itself).
and a code for determining a entered number as a prime or perfect or just a number!!!!

Why do you want to add the divisors? Is that part of the assignment?

Also, given the site's rules on homework, what have you tried on this?

[Banfa]

Why do you want to add the divisors? Is that part of the assignment?

Part of the definition of a perfect number which is for a number n

if 2n = SUM(All divisors of n) then n is perfect.

if 2n < SUM(All divisors of n) then n is deficient.

if 2n > SUM(All divisors of n) then n is abundent.

[sicarie]

Part of the definition of a perfect number which is for a number n

if 2n = SUM(All divisors of n) then n is perfect.

if 2n < SUM(All divisors of n) then n is deficient.

if 2n > SUM(All divisors of n) then n is abundent.

Ah. Okay. (Sorry, next time I'll research the familiar-sounding terms ;). Well that shouldn't be too difficult, could use an array or list to hold any of the divisors...

[Alenik1989]

Part of the definition of a perfect number which is for a number n

if 2n = SUM(All divisors of n) then n is perfect.

if 2n < SUM(All divisors of n) then n is deficient.

if 2n > SUM(All divisors of n) then n is abundent.

Tanx Banfa that helped alot. the only thing i need is a code for determining prime numbers in the users inputs.

[Banfa]

Part of the definition of a perfect number which is for a number n

if 2n = SUM(All divisors of n) then n is perfect.

if 2n < SUM(All divisors of n) then n is deficient.

if 2n > SUM(All divisors of n) then n is abundent.

Eeeek I have detected an error the correct forumlas are

if 2n = SUM(All divisors of n) then n is perfect.

if 2n > SUM(All divisors of n) then n is deficient.

if 2n < SUM(All divisors of n) then n is abundent.

[Alenik1989]

ok i have the code for finding the devisor of any number!!!!!i just cant find how to write a code that show the number entered by user is a prime number?and also how to get the sum of factors for perfect number??????TANX FOR YOU HELPS
so far i have this basic code~~~~dont laugh!!!

#include<iostream>
include<cmath>
using namespace std;
int main()
{
int num,root,dev;
const int SENTINEL=-1;                                //constant and integers
cout<<"ENTER #<enter -1 to stop>:";
 
    cin>>num;                                   //# entered by user
    while(num!=SENTINEL)                 
{
    root=pow(num,0.5);                       //second root of that #
    cout<<"devisors are: 1"<<" "<<num<<" ";
    for (root;1<root;root--)
{
        if(num%root==0)                                  //the loop and conditions
{
    dev=num/root;
    cout<<root<<" "<<dev<<" ";
 
}
       else
    continue;                
}
    cout<<endl;
    cin>>num;
}
    return 0;
}

I wrote a code to determine a entered number as a prime number, ijust wanna know is it a good code or not? and also how can i relate this code with the code above with a loop to make a perfect code????
here is my code for primes!!!!!>>>>

#include <iostream> 
using namespace std;
int main ()
{
    int num, range=1000;
    int n;
   const int SENTINEL=-1;
    cout<<"enter numbers between 1 and 1000 <enter -1 to exit> :";
    cin>>num;                      //entered number by user
    if(0<num && num<=range)               
    {                                                  //LOOps and conditions
    while(num!=SENTINEL)
    {
         for ( n=2; n < num; n++ )
 
            if (num%n==0 )
                break;                             //formula to determine a prime number
            if ( n==num )        
                cout << num<<" is prime"<< endl;
 
    cin>>num;
    }
    }
    else
    cout<<"invalid Number, number should be between 1 and 1000";    
    return 0;
}

[Laharl]

It's not the most optimized code ever written, but does the job. For a quick optimization, think of what property all primes but 2 have and consider only numbers with that property.

[Alenik1989]

can anybody give me a loop to connect the two code above?
i mean, user can input as many numbers as he/she desires and then get that these numbers are prime or perfect. and when they input a negative # or over 1000 it give error but not a input failure?
TANX

[sicarie]

can anybody give me a loop to connect the two code above?
i mean, user can input as many numbers as he/she desires and then get that these numbers are prime or perfect. and when they input a negative # or over 1000 it give error but not a input failure?
TANX

Once again, given the site's policy on homework, what did you get when you tried?

[Alenik1989]

#include <iostream> 
using namespace std;
int main ()
{
    int num, range=1000;
    int n;
   const int SENTINEL=-1;
    cout<<"enter numbers between 1 and 1000 <enter -1 to exit> :";
    cin>>num;                      //entered number by user
    if(0<num && num<=range)               
    {                                                  //LOOps and conditions
    while(num!=SENTINEL)
    {
         for ( n=2; n < num; n++ )
 
            if (num%n==0 )
                break;                             //formula to determine a prime number
            if ( n==num )        
                cout << num<<" is prime"<< endl;
 
    cin>>num;
    }
    }
    else
    cout<<"invalid Number, number should be between 1 and 1000";    
    return 0;
}

[/quote]

ok, considering above code!!!!!when i tried it, it determines ane number between 1 and 1000 as prime or not, correctly. but when a negative or a # greater than 1000 entered it gives an input failure and exites the program.
now i just want somebody to help me to find a loop that when an invalid # has been entered it gives an error but then loop to the first line (whithout input failure). Gives the user another chance to enter the number!!!!!

TANX

[Ganon11]

You will have to do exactly that: create a loop that continually asks the user for input until it is valid input.

[Alenik1989]

for this code i added { while((num<0)||(1001<num)) } inorder to avoid input failure when user enteres a #r<0 or #>1001. BUt when i ran the program it just worked for the 1st #. for example if a user enters 1002 or negative # it worked great, but when the user enters 1002 2 1002 the program output was like

{{{{{{{ENTER #<enter -1 to stop>:1002 2 1002
ENTER #<enter -1 to stop>:devisors are: 1 2
devisors are: 1 1002 6 167 3 334 2 501
}}}}}}}}}}}}(it avoid the first 1002 but executed the 2nd one)
how can i correct it?????

#include<iostream>
include<cmath>
using namespace std;
int main()
{
int num,root,dev;
const int SENTINEL=-1;                                //constant and integers
cout<<"ENTER #<enter -1 to stop>:";
 
 
while((num<0)||(1001<num))
{
    cin>>num;                                   //# entered by user
    while(num!=SENTINEL)    
}             
{
    root=pow(num,0.5);                       //second root of that #
    cout<<"devisors are: 1"<<" "<<num<<" ";
    for (root;1<root;root--)
{
        if(num%root==0)                                  //the loop and conditions
{
    dev=num/root;
    cout<<root<<" "<<dev<<" ";
 
}
       else
    continue;                
}
    cout<<endl;
    cin>>num;
}
    return 0;
}

[Alenik1989]

You will have to do exactly that: create a loop that continually asks the user for input until it is valid input.

but how? i tried that on above post, but it did'nt work !!!!!Any idea Why????

[Alenik1989]

tanx y all. I got it
i just added
while(num)
{
------
------
if(num==-100)
break;
else
{
----------------
----------
}

return 0;