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)