EularProject 43: 带条件约束的排列组合挑选问题

Sub-string divisibility
Problem 43
The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9 in some order, but it also has a rather interesting sub-string divisibility property.

Let d1 be the 1st digit, d2 be the 2nd digit, and so on. In this way, we note the following:

d2d3d4=406 is divisible by 2
d3d4d5=063 is divisible by 3
d4d5d6=635 is divisible by 5
d5d6d7=357 is divisible by 7
d6d7d8=572 is divisible by 11
d7d8d9=728 is divisible by 13
d8d9d10=289 is divisible by 17
Find the sum of all 0 to 9 pandigital numbers with this property.

Answer:
16695334890
python code:

from functools import reduce

def helpfunc1(x,y):
    return 10*x+y

def helpfunc2(lst):
    return reduce(helpfunc1,lst)

def Descriminent(k,value):
    if k==3:
        if value%17==0:
            return True
        else:
            return False
    if k==4:
        if value%13==0:
            return True
        else:
            return False
    if k==5:
        if value%11==0:
            return True
        else:
            return False
    if k==6:
        if value%7==0:
            return True
        else:
            return False
    if k==7:
        if value%5==0:
            return True
        else:
            return False
    if k==8:
        if value%3==0 and (value//10)%2==0:
            return True
        else:
            return False
    return True

def func(result,charlist):
    total=0
    length=len(result)
    if length>2 and length<9:
        if Descriminent(length,helpfunc2(result[0:3]))==False:
            return 0
    if length==10:
        return helpfunc2(result)
    if len(charlist)>0:
        for i in range(0,len(charlist)):
            resultNext=result.copy()
            charlistNext=charlist.copy()
            resultNext.insert(0,charlistNext[i])
            del charlistNext[i]
            total+=func(resultNext,charlistNext)
    return total

charlist=[0,1,2,3,4,5,6,7,8,9]
result=[]
print(func(result,charlist))

解题思路：使用递归
提高效率点：从后往前递归
计算时间<1s

由于可以整除2的数比整除17的多的太多了，其它的一样道理。因此相比于从后往前。从前往后递归的话无用功将成阶乘阶数添加。

EularProject论坛上的一个非递归版本号

digits = ['1','2','3','4','5','6','7','8','9','0']
divisors = [13, 11, 7, 5, 3, 2, 1]
res = []
res1 = []

j = 1
while j*17 < 1000:
    N = j*17
    Nstr = str(N)
    if len(set(Nstr)) == len(Nstr):
        if N < 100: res.append('0' + str(N))
        else: res.append(str(N))
    j += 1

for div in divisors:
    for Nstr in res:
        for d in digits:
            if d not in Nstr:
                Newstr = d + Nstr[:2]
                if int(Newstr)%div == 0:
                    res1.append(d + Nstr)
    res = res1
    res1 = []

tot = 0
for Nstr in res:
    print(Nstr)
    tot += int(Nstr)
print(tot)