Finding the largest prime multiple of a compound number in c

Question

I accept a composite number as input. I want to print all of its factors, as well as the largest simple coefficient of this number. I wrote the following code. It works fine up to number 51. But if you enter any number greater than 51, the wrong output is displayed. How can I fix my code?

#include<stdio.h>
void main()
{
 int i, j, b=2, c;
 printf("\nEnter a composite number: ");
 scanf("%d", &c);
 printf("Factors: ");

 for(i=1; i<=c/2; i++)
 {
  if(c%i==0)
  {
   printf("%d ", i);
   for(j=1; j<=i; j++)
   {
    if(i%j > 0)
    {
     b = i;
    }
    if(b%3==0)
     b = 3;
    else if(b%2==0)
     b = 2;
    else if(b%5==0)
     b = 5;
   }
  }
 }
 printf("%d\nLargest prime factor: %d\n", c, b);
}

Answer 1

This is a little spoiler, so if you want to solve it yourself, do not read it yet :). I will try to give hints in order of sequence so that you can read each hint in order, and if you need more tips, proceed to the next hint, etc.

# 1: n, n/divisor n. , 100/2 = 50 0, 2 100. , 50 100.

# 2 Hint # 1, , = 2 * <= n . , 100, 10 (10 * 10 - <= 100), , №1, . :

100 / 2 = 50, so 2 and 50 are factors
100 / 5 = 20, so 5 and 20 are factors
100 / 10 = 10, so 10 is a factor

# 3 n, n, , n/. sieve , .

# 4 C:

bool factors[100000];

void getprimefactors(int n) {
  // 0 and 1 are not prime
  if (n == 0 || n == 1) return;

  // find smallest number >= 2 that is a divisor of n (it will be a prime number)
  int divisor = 0;
  for(int i = 2; i*i <= n; ++i) {
    if (n % i == 0) {
      divisor = i;
      break;
    }
  }
  if (divisor == 0) {
    // we didn't find a divisor, so n is prime
    factors[n] = true;
    return;
  }

  // we found a divisor
  factors[divisor] = true;
  getprimefactors(n / divisor);
}

int main() {
  memset(factors,false,sizeof factors);
  int f = 1234;
  getprimefactors(f);
  int largest;
  printf("prime factors for %d:\n",f);
  for(int i = 2; i <= f/2; ++i) {
    if (factors[i]) {
      printf("%d\n",i);
      largest = i;
    }
  }
  printf("largest prime factor is %d\n",largest);
  return 0;
}

:

---------- Capture Output ----------
> "c:\windows\system32\cmd.exe" /c c:\temp\temp.exe
prime factors for 1234:
2
617
largest prime factor is 617
> Terminated with exit code 0.

Answer 2

, , , , .

, , . , ?

Answer 3

, , 2,3 5. , , 2, 3 5. 7, 11, 13, 17, 19 , , , .

Answer 4

, , (, , 100 000). :

#include <cmath>

void addfactor(int n) {
    printf ("%d\n", n);
}

int main()
{
    int d;
    int s;
    int c = 1234567;
    while (!(c&1)) {
        addfactor(2);
        c >>= 1;
    }
    while (c%3 == 0) {
        addfactor(3);
        c /= 3;
    }
    s = (int)sqrt(c + 0.5);
    for (d = 5; d <= s;) {
        while (c % d == 0) {
            addfactor(d);
            c /= d;
            s = (int)sqrt(c + 0.5);
        }
        d += 2;
        while (c % d == 0) {
            addfactor(d);
            c /= d;
            s = (int)sqrt(c + 0.5);
        }
        d += 4;
    }
    if (c > 1)
        addfactor(c);
    return 0;
}

addfactor - , . .

This is significantly faster than other code snippets posted here. For random input, such as 10597959011, my code will take about 2000 bits of operations plus another 1000 to re-compose divisors, and others billions of operations. This is the difference between “instant” and minute in this case.

Answer 5

Simplification of dcp answer (iterative way):

#include <stdio.h>
void factorize_and_print(unsigned long number) {
  unsigned long factor;
  for(factor = 2; number > 1; factor++) {
    while(number % factor == 0) {
      number = number / factor;
      printf("%lu\n",factor);
    }
  }
}

/* example main */
int main(int argc,char** argv) {
  if(argc >= 2) {
    long number = atol(argv[1]);
    factorize_and_print(number);
  } else {
    printf("Usage: %s <number>%<number> is unsigned long", argv[0]);
  }
}

Note. There is a parsing error for a number that is not getting the correct number in argv.