Question

How can you write code to find a prime factor. I can write code to find factors of a number,but how do you determine if it is a prime factor? Number is 600 billion plus, so need to do it with code.language c++

Answer 1

int main()
{
int n,num,i;
printf("enter the num");
scanf("%d",&num);
n=num;
printf("\n");
for(i=2;i<=n/2;i++)
{
if(n%i==0)
{
prime(i);
}
}
}

void prime(int x)
{
int i,f=0;
for(i=2;i<=x/2;i++)
{
if(x%i==0)
{
f=1;
break;
}
}
if(f==0)
printf(" %d",x);

}

Answer 2

//Prime Number
#include<iostream>
#include<conio.h>
using namespace std;

int main()
{
int n,i,k;

k=0;

cout<<" Please Enter n: ";
cin>>n;

for(i=2;i<n;i++)
{
if(n%i==0)
k=1;

}

if(k==0)
cout<<"prime";
else
cout<<"not prime" ;

getch();
return 0;

}

Answer 3

I give up!

Answer 4

is mine correct?

Answer 5

I did that once, using C++, doesn't matter, what matter is the logic:

This is what I basically did:

//To determine if the number entered is prime or not.
bool Primos::primo(unsigned num)
{
//
long j = static_cast<long>(sqrt(static_cast<float> (num)) );

if(num < 0)
return false;

if(num == 1)
return false;

if(num == 2)
return true;

if(num % 2 == 0)
return false;

for(int i = 3; i <= j; i += 2)
{
if(!(num % i))
return false;
}

return true;

}

and another function to display in screen if what prime or not:

// Prints the actual prime number and those that are not too to the screen and to an archive of output.
unsigned imprime(ofstream& salida)
{
Primos unPrimo;

unsigned numero = entraNumero();

if (numero > 0)
{

if ( unPrimo.primo(numero) == true )
{
cout << "\n\n\t El " << numero << " es un n'umero primo."
<< "\n\n\n\t" ;

salida << "\n\n\t El " << numero << " es un n'umero primo."
<< "\n\n\n\t" ;

system("pause");
}

else
{
cout << "\n\n\t El " << numero << " no es un n'umero primo."
<< "\n\n\n\t" ;

salida << "\n\n\t El " << numero << " no es un n'umero primo."
<< "\n\n\n\t" ;

system("pause");
}

}

return numero;
}

Of course I am omitting other functions and code, since I did this using C++, in a OO way, with class header file, definition file and implementation , etc etc...

Those are the main things anyways..

Answer 6

First, you find the primes in the range of a the number and store them in a vector or something. Second you iterate through them and see which divide into the number, resulting in another member of the vector.

Answer 7

yeah, that what I just didm only that U not necesarily need to stored them on a vector, lost, array, I stored them on an output text file as they were show up on the screen

Answer 8

Usually two methods to find the prime numbers:
1st: Traditional method,
This function returns true if n is prime, or false otherwise.
#include <math.h>
bool isPrime (int n)
{
if (n<=1) return false;
if (n==2) return true;
if (n%2==0) return false;
int m=sqrt(1.0*n);

for (int i=3; i<=m; i+=2)
if (n%i==0)
return false;

return true;
}

2nd, Sieve method,
This function will return an array, which represent ith number if prime number if it's true, otherwise it's false. This function mainly returns all the prime numbers from 1 to n
vector<bool> sieve(int n)
{
vector<bool> prime(n+1, true);
prime[0]=false;
prime[1]=false;
int m=sqrt(1.0*n);

for (int i=2; i<=m; i++)
if (prime[i])
for (int k=i*i; k<=n; k+=i)
prime[k]=false;

return prime;
}

Answer 9

#include<iostream.h>
int fact(int x);
void main()
{
char y;
do
{
int x;
cout<<"plz,enter no."<<endl;
cin>>x;
cout<<fact(x)<<endl;
cout<<"do u want to try again(y/n)?";
a: cin>>y;
if (y!='y' && y!='n')
{
cout<<"plz enter y/n";
goto a;
}
}
while(y=='y');
}
int fact(int x)
{
int i,f=1;
for(i=1;i<=x;i++)
f*=i;
return f;
}


OR use this

#include <iostream>
using namespace std;


int main()

{

int nLoopCount = 0; //declare variables
int nFirst ;


cout << "Enter Number to be factored:\n"; //get info to and from the user
cin >> nLoopCount; //if the user enters a huge number this will keep
//their machine from going into a long loop
if(nLoopCount > 36000){
nLoopCount = 0;
cout << "Number to be factored must be less than 36001:\n";
}

for (int j=1; j <= nLoopCount; j++){ //loops through int and stops a point set by nLoopCount
for (int i=1; i <=nLoopCount; i++){ //loops through the other number in the factors
nFirst = j*i; //multiplies the 2 numbers together nLoopCount times



if(nFirst <= nLoopCount && nFirst > nLoopCount-1){ //sort out the numbers we want


cout << j <<" * "<< i <<" = "<< nFirst << " sum of factors = "<< j+i <<
" difference of factors = " << j-i << " , " << i-j << "\n"; //display result

}
}
}
return 0;
}