探索高效算法：寻找所有和为N的组合方案

6  

Much faster of other proposed solutions:

其他提出的解决方案要快得多:

def iter_fun(sum, deepness, myString, Total):
    if deepness == 0:
        if sum == Total:
            print myString
    else:    
        for i in xrange(min(10, Total - sum + 1)):
            iter_fun(sum + i,deepness - 1,myString + str(i),Total) 

def fixed_sum_digits(digits, Tot):
    iter_fun(0,digits,"",Tot) 


fixed_sum_digits(6,20)

Still some room for speeder code but then the code would be boring to be read!

仍然有一些空间用于调速器代码但是代码将无聊被阅读!

2  

Using itertools and permutations:

使用itertools和permutations:

>>> from itertools import product
>>> l = []
>>> for digits in product('0123456789', repeat=6):
...     if sum(map(int, digits)) == 20:
...             l.append(digits)
...
>>> len(l)
35127
>>> l[1234]
('0', '1', '9', '0', '5', '5')

Seems to be a bit faster that eumiro's:

似乎比eumiro快一点:

>>> stm = """l = []
... for digits in product('0123456789', repeat=6):
...     if sum(map(int, digits)) == 20:
...             l.append(digits)
... """
>>> timeit.timeit(stm, setup="from itertools import product", number=3)
10.368315935134888
>>> timeit.timeit("['{0:06}'.format(i) for i in xrange(1000000) if sum(map(int,str(i))) == 20]", number=3)
14.926225900650024

-1  

You can use numpy.

你可以使用numpy。

import numpy
a=[1,2,1,6,7,3]
print(numpy.cumsum(a)[-1])