How to factorize a number? - c

Question

I've been asked to factorize a number and show it in a specific way .
e.g: 100 = 2^2*5^2
This is the C++ code I've used so far with no dice , unfortunately:
#include <stdio.h>
#include <math.h>
//IsPrime indicates whether a given number is or is not prime.
bool IsPrime(long long n)
{
int j = 3;
if (n == 2)
{
return true;
}
else if (n % 2 == 0)
{
return false;
}
else
{
for (j = 3; j <= sqrt(n); j += 2)
{
if (n%j == 0)
{
return false;
}
}
}
return true;
}
int main(void)
{
long long n_orig,n, i=3 , primecount=0;
scanf("%lld", &n_orig);
n = n_orig;
if (n == 1)
{
printf("1");
return 0;
}
if (IsPrime(n))
{
printf("%lld", n);
return 0;
}
if (n % 2 == 0)
{
while (n >= 2 && n % 2 == 0)
{
primecount++;
n = n / 2;
}
if (primecount == 1)
{
printf("2*");
}
else
{
printf("2^%lld*", primecount);
}
}
primecount = 0;
n = n_orig;
while (i <= n/2)
{
if (IsPrime(i))
{
while (n >= i && n % i == 0)
{
primecount++;
n = n / i;
}
}
n = n_orig;
if (primecount == 0)
{
i++;
continue;
}
if (primecount == 1)
{
printf("%lld*", i);
}
else
{
printf("%lld^%lld*", i, primecount);
}
primecount = 0;
i+=2;
}
printf("\b");
return 0;
}
Using this code I was able to generate a few test cases, though when I uploaded my answer to the website where the codes are presumably evaluated , out of 7 test cases (which I cannot know what they exactly are) , I pass 3 , fail 3 and exceed time limit (the one that hasn't even been declared in the question) in one case. I'd really appreciate some help , and please be noob-friendly!
Also , I don't really wanna know if my answer could be improved in some way , my top priority right now is understanding why MY own code doesn't work as intended.
P.S : Usage of iostream and arrays is not allowed.
Thanks in advance.

Answer 1

Try this:
#include <stdio.h>
#include <math.h>
unsigned long long PrintMultiplicity(unsigned long long n,unsigned long long factor)
{
unsigned int count = 0;
while (n%factor == 0)
{
count++;
n /= factor;
}
if (count > 0)
{
printf("%llu^%u",factor,count);
if (n > 1)
printf("*");
}
return n;
}
void PrintFactorization(unsigned long long n)
{
unsigned long long factor;
unsigned int add;
printf("%llu = ",n);
n = PrintMultiplicity(n,2);
n = PrintMultiplicity(n,3);
// Check only factors that are adjacent to multiples of 6
for (factor = 5, add = 2; factor <= sqrt(n); factor += add, add = 6-add)
n = PrintMultiplicity(n,factor);
if (n > 1)
printf("%llu^1",n);
printf("\n");
}
int main()
{
unsigned long long n;
scanf("%llu",&n);
PrintFactorization(n);
return 0;
}

Answer 2

You need to perform some fine optimisations. Do not invoke isPrime() method for each value, instead consider a different approach so that irrelevant values can be ignored altogether at the very beginning.
Get the list of relevant primes numbers that comes under n, using Sieve of Eratosthenes concepts.
Start from the lowest prime value from the list, divide n to get intermediate values as
n / lowest_prime_that_perfectly_divide_n.
Continue doing this by checking with next higher prime value till n becomes 1. This way you would have count of each dividing factors.

Answer 3

You do not need prime tests, and lists of primes or prime wheels are only needed for acceleration. A simple program listing all prime factors is
#include <stdio.h>
#include <math.h>
int main(void)
{
long long n_orig,n,k;
scanf("%lld", &n_orig);
n = n_orig;
k=2;
while(k*k<=n) {
while(0==n%k) {
n = n/k;
printf("%lld ",k);
}
k++;
}
if(n>1) printf("%lld ",n);
printf("\n");
return 0;
}
This does not generate the required output format, but that can easily added to it.