# Prime factors code in C and Python

Today someone spoke about prime factors and I thought it was something interesting to code. I needed to practice my C so I’ve coded it in C. Here it goes:

```#include <stdio.h> #define VMAX 25   int main() { /*Prime numbers < 100 */ int primes[VMAX] = {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97}; int factors[VMAX] = {}; /*Prime factors*/ int repeat[VMAX] = {}; /*Store times each factor repeats*/ int a, i, n=0;   printf("Type a number to find it's prime factors: \n"); scanf("%d", &a);   /*Start to divide the number by prime numbers while a is not 1*/ for (i=0; i<VMAX; i++){ while ((a != 1) && (a % primes[i] == 0)){   a = a / primes[i]; factors[i] = primes[i]; /*The prime factor is the one being used*/ n++; /*Count how many times each prime is used*/ } if ((a % primes[i] == 0) ) n = 0; repeat[i] = n; n = 0; /*Reinitialize counter*/   } printf("Prime factors:\n"); for (i=0; i<VMAX; i++){   /*Print all factors in the array != 0*/ if (factors[i] !=0){ /*Since x ^ 0 = 1, we have to print x ^ 1 instead*/ if (repeat[i] == 0) printf("%d ^ 1\n", factors[i]); else printf("%d ^ %d\n", factors[i], repeat[i]); } }   return 0;   }```

I really like python but unfortunately I haven’t had the time to code anything with python. Now it’s learning C , then C++ and so on. But this time I wanted to code this in python also. I’m a bit out of practice and my brain was thinking in C. But here it goes.

```#! /usr/bin/env/python VMAX = 25   #Prime numbers < 100 primes = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97] factors = *VMAX #Prime factors repeat = *VMAX #Store times each factor repeats i = 0 n = 0 k = 0   a = raw_input("Type a number to find it's prime factors:")   #Start to divide the number by prime numbers while a is not 1 for i in range(VMAX):   while ((a != 1) & (int (a) % primes[i] == 0)):   a = int (a) / primes[i]   print primes[i] factors[i] = primes[i] # The prime factor is the one being used   n = n + 1 #*Count how many times each prime is used print n if (int (a) % primes[i] == 0): n = 0 #Reinitialize counter repeat[k] = n n = 0 k = k + 1   print "Prime factors:"   #Print all factors in the array != 0 for i in range(VMAX):   if (factors[i] !=0): #Since x ^ 0 = 1, we have to print x ^ 1 instead if (repeat[i] == 0): print factors[i] else: print "%s ^ %s " % (factors[i], repeat[i])```

1. wayne - 20 de Fevereiro de 2010 @ 18:09

VMAX = 25

#Prime numbers < 100
primes = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97]
factors = *VMAX #Prime factors
repeat = *VMAX #Store times each factor repeats
i = 0
n = 0
k = 0
spacer = 0
a = raw_input("Type a number to find it's prime factors:")

#Start to divide the number by prime numbers while a is not 1
for i in range(VMAX):

while ((a != 1) & (int (a) % primes[i] == 0)):

a = int (a) / primes[i]

factors[i] = primes[i] # The prime factor is the one being used

n = n + 1 #*Count how many times each prime is used

# formatting
if(spacer == 0):
print
spacer = spacer + 1
else:
if(n == 1):
print
print
print primes[i],
print n,
print " ",

if (int (a) % primes[i] == 0):
n = 0 #Reinitialize counter
repeat[k] = n

n = 0
k = k + 1
print
print
print "Prime factors:"

#Print all factors in the array != 0
for i in range(VMAX):

if (factors[i] !=0):
#Since x ^ 0 = 1, we have to print x ^ 1 instead
if (repeat[i] == 0):
print factors[i],
else: print "%s ^ %s " % (factors[i], repeat[i]),

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