From afcf2ab0c578d8fa2a03ed4d1686b90cb5011d85 Mon Sep 17 00:00:00 2001
From: Julien Lengrand-Lambert <julien@lengrand.fr>
Date: Fri, 13 Jan 2012 15:26:24 +0100
Subject: [PATCH] (Finally) solves problem 10. The code is slow !

Prepares problem 11
---
 e_10.py      | 17 +++++++++++++++--
 e_10_nope.py | 37 +++++++++++++++++++++++++++++++++++++
 e_11.py      | 20 ++++++++++++++++++++
 3 files changed, 72 insertions(+), 2 deletions(-)
 create mode 100755 e_10_nope.py
 create mode 100755 e_11.py

diff --git a/e_10.py b/e_10.py
index 133fa51..2ef902b 100755
--- a/e_10.py
+++ b/e_10.py
@@ -13,8 +13,21 @@ def sum_primes(value):
     """
     Returns the sum of prime number below value
     """
+    prime_list = [2] 
+    curr_val = 3
+    ptr = False
+    while( curr_val < value):
+        for primes in prime_list:
+            if (curr_val % primes == 0):
+                ptr = True # multiple of at least one value
+        if not ptr:
+            prime_list.append(curr_val)
+            print "%d%% / %d" % (((prime_list[-1] / float(value)) * 100), prime_list[-1])
+
+        curr_val +=1
+        ptr = False
     
-    return 1
+    return sum(prime_list)
 
 if __name__ == '__main__':
-    print "Answer : %d" % (sum_primes())
+    print "Answer : %d" % (sum_primes(2000000))
diff --git a/e_10_nope.py b/e_10_nope.py
new file mode 100755
index 0000000..1b06c74
--- /dev/null
+++ b/e_10_nope.py
@@ -0,0 +1,37 @@
+#!/usr/bin/env python 
+"""
+ ##---
+ # jlengrand
+ #Created on : Fri Jan 13 11:42:09 CET 2012
+ #
+ # DESCRIPTION : Solves problem 10 of Project Euler
+ The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
+ Find the sum of all the primes below two million.
+ ##---
+"""
+def sum_primes(value):
+    """
+    Returns the sum of prime number below value
+    """
+    prime_list = [2] 
+    curr_val = 3
+    ptr = False
+    while( curr_val <= value):
+        while(not ptr):
+            for primes in prime_list:
+                if primes < pow(curr_val, 0.5):
+                    if (curr_val % primes == 0):
+                        ptr = True # multiple of at least one value
+            if not ptr:
+                prime_list.append(curr_val)
+                print "%d%% / %d" % (((prime_list[-1] / float(value)) * 100), prime_list[-1])
+                ptr = True
+                       
+        curr_val +=1
+        ptr = False
+
+    #print prime_list
+    return sum(prime_list)
+
+if __name__ == '__main__':
+    print "Answer : %d" % (sum_primes(20000))
diff --git a/e_11.py b/e_11.py
new file mode 100755
index 0000000..cd166df
--- /dev/null
+++ b/e_11.py
@@ -0,0 +1,20 @@
+#!/usr/bin/env python 
+"""
+ ##---
+ # jlengrand
+ #Created on : Fri Jan 13 15:24:59 CET 2012
+ #
+ # DESCRIPTION : Solves problem 11 of Project Euler
+ 2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
+ What is the sum of the digits of the number 2^1000?
+ ##---
+"""
+def sum_power_2(value):
+    """
+    Returns the sum of the digits of 2^value
+    """
+    
+    return 1
+
+if __name__ == '__main__':
+    print "Answer : %d" % (sum_power_2(1000))