From aa3240015b7a973c7dd0704676322f95ecd009e4 Mon Sep 17 00:00:00 2001
From: Correl Roush <correl@gmail.com>
Date: Wed, 6 Apr 2011 23:29:56 -0400
Subject: [PATCH] Completed exercise 031

---
 e031.py | 22 ++++++++++++++++------
 1 file changed, 16 insertions(+), 6 deletions(-)

diff --git a/e031.py b/e031.py
index 6b1f162..cf58638 100644
--- a/e031.py
+++ b/e031.py
@@ -10,11 +10,21 @@ It is possible to make £2 in the following way:
 How many different ways can £2 be made using any number of coins?
 """
 
+def combinations(amount, denominations):
+    total = 0
+    if not denominations:
+        return total
+    denominations = sorted(denominations, reverse=True)
+    for i in range(len(denominations)):
+        d = denominations[i]
+        n = 1
+        while d * n <= amount:
+            if d * n == amount:
+                total += 1
+            total += combinations(amount - (d * n), denominations[i+1:])
+            n += 1
+    return total
+
 if __name__ == '__main__':
-    coins = [1, 2, 5, 10, 20, 50, 100, 200]
     coins = [200, 100, 50, 20, 10, 5, 2, 1]
-    combinations = 0
-    for start in range(len(coins)):
-        current = start
-        total = 0
-        break
+    print combinations(200, coins)