Solves problem 30. Good running time

Still have to perform some good work on prime numbers! 

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

README.markdown

@@ -40,6 +40,7 @@ Should be used in order to help future reuse of code :)
 25 - What is the first term in the Fibonacci sequence to contain 1000 digits? - 0.741 <br />
 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 />
 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 /> 
@@ -50,7 +51,8 @@ Should be used in order to help future reuse of code :)
 **In progress: **
 
 26 - Find the value of d < 1000 for which 1/d contains the longest recurring cycle. <br />
-30 - Find the sum of all the numbers that can be written as the sum of fifth powers of their digits. <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_30.py

@@ -16,14 +16,47 @@ The sum of these numbers is 1634 + 8208 + 9474 = 19316.
 
 Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.
 '''
-
+def find_limit(power):
-
-def largest():
     """
-    Returns
+    Returns the max number of digits we have to loop through to find the solution, for a given power
     """
+    nb_digits = 2
+    max_num = 9**power * nb_digits 
+    
+    while len(str(max_num)) >= nb_digits:
+        nb_digits += 1
+        
+    return nb_digits - 1
+    
+def digsum(num, power):
+    """
+    Returns the sum of power of digits of num
+    ex : digsum(123, 4) = 1**4 + 2**4 + 3**4
+    """
+    val = 0
+    for el in str(num):
+        val += int(el) ** power
+    
+    return val
+    
+def sum_power(power):
+    """
+    Finds the sum of all the numbers that can be written as the sum of power powers of their digits.
+    """
+    max_dig = find_limit(power)
+    max_val = 9**power * max_dig
+    print "Max dig is : %d" %(max_dig)
+    sump = 0
+    
+    cpt = 2
+    while cpt <= max_val:        
+        if cpt ==  digsum(cpt, power):
+            sump += cpt
+        
+        cpt +=1
 
-    return 1
+    return sump
 
 if __name__ == '__main__':
-    print "Answer : %d " % (largest())
+    # The major problem in there is to find the upper limit.  
+    print "Answer : %d " % (sum_power(5))

e_34.py

@@ -0,0 +1,24 @@
+#!/usr/bin/env python 
+'''
+Created on 8 feb. 2012
+
+@author: Julien Lengrand-Lambert
+
+DESCRIPTION: Solves problem 34 of Project Euler
+145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.
+
+Find the sum of all numbers which are equal to the sum of the factorial of their digits.
+
+Note: as 1! = 1 and 2! = 2 are not sums they are not included.
+''' 
+def sum_power():
+    """
+    Finds
+    """
+
+
+    return 1
+
+if __name__ == '__main__':
+    # The major problem in there is to find the upper limit.  
+    print "Answer : %d " % (sum_power())