#!/usr/bin/env python from common import memo #denominations of currency values = [200, 100, 50, 20, 10, 5, 2, 1] #recursive function to find permutations of the currency passed in def permutations(denomination, maxNum = 200): count = 0 for v in values: #if the number is greater than the value or max number to go to. if v > denomination or v > maxNum: continue #if they are equal then add 1 otherwise continue down permutations if v == denomination: count += 1 else: count += permutations(denomination - v, v) return count #attach memory function to recursive function permutations = memo(permutations) print permutations(200)
#!/usr/bin/python #get the memo function from common from common import memo def fib(n): if n == 2 or n == 1: return 1 else: return fib(n-1) + fib(n-2) #attach memory function to fibinoci function fib = memo(fib) #how many digits to look for numDigits = 1000 #largest number given in problem count = 12 #loop while the number of digits is less than 1000 while(len(str(fib(count))) < 1000): count += 1 #print result print count
def sumFact(num):
global facts
total = 0
for i in str(num):
total += facts[int(i)]
return total
#value of factorials
facts = [1]
for i in range(1,10):
facts.append(facts[i-1] * i)
#holds the chain
nums = {}
#attach a memory function
sumFact = memo(sumFact)
#count of the number of numbers that have chain = 60
count = 0
#for all numbers less than 1,000,000
for i in range(0, 1000000):
#add the number to the chain
nums[i] = 0
j = sumFact(i)
#while the chain number isn't in the chain already
while j not in nums:
nums[j] = 0
j = sumFact(j)
#if its length is 60
if len(nums) == 60:
count += 1
