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Project Euler: Problem 14: Longest Collatz sequence
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.

Answer: 837799（共有525个步骤）
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import time
startTime = time.clock()

def f(x):
    c = 0  #计算考拉兹序列的个数
    while x != 1:
        if x%2 == 0:  #若为偶数
            x = x//2
            c += 1
        else:  #若为奇数
            x = x*3+1
            c += 1
    if x == 1:
        c += 1  #数字1也得算上
        return c

chainItemCount = 0
startingNumber = 0
for i in range(1, 1000000):
    t = f(i)
    if chainItemCount < t:
        chainItemCount = t
        startingNumber = i

print(‘The number %s produces the longest chain with %s items‘ % (startingNumber, chainItemCount))

print(‘Time used: %.2d‘ % (time.clock()-startTime))