Euler Project question 14 in python way

标签：io   os   ar   for   sp   on   art   ef   bs   

# This Python file uses the following encoding: utf-8
# The following iterative sequence is defined for the set of positive integers:

# n → n/2 (n is even)
# n → 3n + 1 (n is odd)

# Using the rule above and starting with 13, we generate the following sequence:

# 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
# It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.

# Which starting number, under one million, produces the longest chain?

# NOTE: Once the chain starts the terms are allowed to go above one million.
import time
start = time.time()
def chain_len(n):
    len = 1
    while n != 1:
        if n % 2 == 0:
            n /= 2
            len += 1
        else:
            n = 3*n + 1
            len += 1
        return len
max_len = 1
max_len_num = 1
for i in xrange(1, 1000000):
    if chain_len(i) > max_len:
        max_len = chain_len(i)
        max_len_num = i
print "%s produces the longest chain and cost %s seconds" % (max_len_num, time.time() - start)

Euler Project question 14 in python way

标签：io   os   ar   for   sp   on   art   ef   bs   

原文地址：http://www.cnblogs.com/cyberpaz/p/4036496.html