Infinite Iterators:

Iterator	Arguments	Results	Example
	start, [step]	start, start+step, start+2*step, ...	count(10) --> 10 11 12 13 14 ...
	p	p0, p1, ... plast, p0, p1, ...	cycle('ABCD') --> A B C D A B C D ...
	elem [,n]	elem, elem, elem, ... endlessly or up to n times	repeat(10, 3) --> 10 10 10

Iterators terminating on the shortest input sequence:

Iterator	Arguments	Results	Example
	p, q, ...	p0, p1, ... plast, q0, q1, ...	chain('ABC', 'DEF') --> A B C D E F
	data, selectors	(d[0] if s[0]), (d[1] if s[1]), ...	compress('ABCDEF', [1,0,1,0,1,1]) --> A C E F
	pred, seq	seq[n], seq[n+1], starting when pred fails	dropwhile(lambda x: x<5, [1,4,6,4,1]) --> 6 4 1
	iterable[, keyfunc]	sub-iterators grouped by value of keyfunc(v)
	pred, seq	elements of seq where pred(elem) is true	ifilter(lambda x: x%2, range(10)) --> 1 3 5 7 9
	pred, seq	elements of seq where pred(elem) is false	ifilterfalse(lambda x: x%2, range(10)) --> 0 2 4 6 8
	seq, [start,] stop [, step]	elements from seq[start:stop:step]	islice('ABCDEFG', 2, None) --> C D E F G
	func, p, q, ...	func(p0, q0), func(p1, q1), ...	imap(pow, (2,3,10), (5,2,3)) --> 32 9 1000
	func, seq	func(*seq[0]), func(*seq[1]), ...	starmap(pow, [(2,5), (3,2), (10,3)]) --> 32 9 1000
	it, n	it1, it2, ... itn splits one iterator into n
	pred, seq	seq[0], seq[1], until pred fails	takewhile(lambda x: x<5, [1,4,6,4,1]) --> 1 4
	p, q, ...	(p[0], q[0]), (p[1], q[1]), ...	izip('ABCD', 'xy') --> Ax By
	p, q, ...	(p[0], q[0]), (p[1], q[1]), ...	izip_longest('ABCD', 'xy', fillvalue='-') --> Ax By C- D-

Combinatoric generators:

Iterator	Arguments	Results
	p, q, ... [repeat=1]	cartesian product, equivalent to a nested for-loop
	p[, r]	r-length tuples, all possible orderings, no repeated elements
	p, r	r-length tuples, in sorted order, no repeated elements
	p, r	r-length tuples, in sorted order, with repeated elements
product('ABCD', repeat=2)		AA AB AC AD BA BB BC BD CA CB CC CD DA DB DC DD
permutations('ABCD', 2)		AB AC AD BA BC BD CA CB CD DA DB DC
combinations('ABCD', 2)		AB AC AD BC BD CD
combinations_with_replacement('ABCD', 2)		AA AB AC AD BB BC BD CC CD DD

python源码：https://docs.python.org/2/library/itertools.html?module-itertools

应用举例：

解题思路：利用permutations函数可以直接得到序列列表，再用enumerate确定第1000000个是谁

for i, v in enumerate(permutations(range(10)), 1):    if i == 1000000:        print v        break #(2, 7, 8, 3, 9, 1, 5, 4, 6, 0)