Re: how many nested for can we utilize?

[Patrol Sun]

Of course We needn't 100 levels,but I use the exec function can concise the
code. See the attachment.

2008/8/17 Fredrik Lundh <[EMAIL PROTECTED]>

> Nick Dumas wrote:
>
> A good quote I read (I can't remember who it was from, though) is "If
>> you need more than three levels of indentation, then something is
>> seriously wrong with your code." Possibly Guido himself? Anyway. If
>> you've got 100 levels of for, you're probably making things way harder
>> than they need to be.
>>
>
> assuming 100 levels of for and 2 items in each sequence, you'll end up with
> 1267650600228229401496703205376 iterations through the inner loop.  assuming
> each iteration takes a picosecond, it'll take approx 40 billion years to run
> the program.
>
> </F>
>
>
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>

#=========================================================================
#
# Python Source File -- Created with SAPIEN Technologies PrimalScript 4.1
#
# NAME: Problem 52
#
# AUTHOR: patrol , ustc
# DATE  : 2008-8-7
#
# COMMENT: 
# 
# URL:http://projecteuler.net/index.php?section=problems&id=52
# Result:142857
#=========================================================================
import sys
import string
def isPro52Num(n):
        s=[]
        for i in range(1,7):
                s.append(str(i*n))
        for i in s[0]:
                ll = len(string.join(string.split(s[0],i),''))
                for j in range(1,6):
                        if ll!=len(string.join(string.split(s[j],i),'')):
                                return False
        return True
def f(n):
        index=[]
        num=[]
        digitsList=[0]+range(2,10)
        for i in range(n):
                index.append(1)
                num.append(1)
        num.append(1)
        stringCode='for index[0] in digitsList:\n' 
        for i in range(1,n):
                stringCode += '\t'*(i+0)+'num[%d] = num[%d]*10 + 
index[%d]\n'%(i-1,i-2,i-1) \
                +'\t'*(i+0)+'for index[%d] in digitsList:\n'%i
        stringCode +='\t'*n+'num[%d] = num[%d]*10 + index[%d]\n'%(n-1,n-2,n-1) \
        +'\t'*n+'if isPro52Num(num[%d]):\n'%(n-1) \
        +'\t'*(n+1)+'print num[%d]\n'%(n-1) \
        +'\t'*(n+1)+'sys.exit()\n'
        exec(stringCode)

length = 1
while True:
        f(length)
        length += 1


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