Solves problem 34. Average running time, could be optimized with

memoization

Still have to perform some good work on prime numbers! 

Signed-off-by: Julien Lengrand-Lambert <julien@lengrand.fr>

README.markdown

@@ -41,6 +41,7 @@ Should be used in order to help future reuse of code :)
 27 - Find a quadratic formula that produces the maximum number of primes for consecutive values of n. - too long <br />
 29 - How many distinct terms are in the sequence generated by ab for 2  a  100 and 2  b  100? - < 1 sec <br />
 30 - Find the sum of all the numbers that can be written as the sum of fifth powers of their digits. - < 3 sec <br />
+34 - Find the sum of all numbers which are equal to the sum of the factorial of their digits. - 30 sec  <br />
 36 - Find the sum of all numbers less than one million, which are palindromic in base 10 and base 2. - 0.933 <br />
 42 - Using words.txt, a 16K text file containing nearly two-thousand common English words, how many are triangle words? - < 1 sec <br />
 45 - Find the next triangle number that is also pentagonal and hexagonal. - < 1 sec <br /> 
@@ -52,7 +53,6 @@ Should be used in order to help future reuse of code :)
 
 26 - Find the value of d < 1000 for which 1/d contains the longest recurring cycle. <br />
 28 - What is the sum of both diagonals in a 1001 by 1001 spiral? <br />
-34 - Find the sum of all numbers which are equal to the sum of the factorial of their digits.  <br />
 35 - How many circular primes are there below one million? <br />
 
 **WARNING : Spoil inside for those who want to solve problems by themselves :)**

e_34.py

@@ -11,14 +11,55 @@ Find the sum of all numbers which are equal to the sum of the factorial of their
 
 Note: as 1! = 1 and 2! = 2 are not sums they are not included.
 ''' 
-def sum_power():
+def find_limit():
     """
-    Finds
+    Returns the max number of digits we have to loop through to find the solution
     """
+    nb_digits = 2
+    max_num = fact(9) * nb_digits 
     
+    while len(str(max_num)) >= nb_digits:
+        nb_digits += 1
         
-    return 1
+    return nb_digits - 1
+
+def fact(value):
+    """
+    Returns value! 
+    """
+    if value == 0:
+        return 1
+    return value * fact(value - 1)      
+
+def digsum(num):
+    """
+    Returns the sum of factorial of digits of num
+    ex : digsum(123, 4) = fact(1) + fact(2) + fact(3)
+    """
+    val = 0
+    for el in str(num):
+        val += fact(int(el))
+    
+    return val
+
+def sum_fact():
+    """
+    Finds the sum of all the numbers that can be written as the sum of the factorial of their digits.
+    """
+    max_dig = find_limit()
+    max_val = fact(9) * max_dig
+    print "Max dig is : %d" %(max_dig)
+    sumf = 0    
+    
+    cpt = 3
+    while cpt <= max_val:    
+        if cpt ==  digsum(cpt):
+            sumf += cpt
+        
+        cpt +=1
+
+    return sumf
 
 if __name__ == '__main__':
     # The major problem in there is to find the upper limit.
-    print "Answer : %d " % (sum_power())
+    print "Answer : %d " % (sum_fact())