python连续数字_大量N个连续数字的Python和

I need to get the largest sum of n consecutive digits in a range of a large number.

For example, the range could be 5^150000, within this range I want to find out the largest sum of 50,000 consecutive digits.

The approach I have taken of using two loops seems to never terminate. I will appreciate any input.

The code:

count = 0

tempsum = 0

summ = 0 # variables init

oldlist = ''

newlist = ''

num = str(3**2209) # for example

for digit in num: # go over all the digits in the number

while count < 12 and len(num) >= 12 : # check in 12-digits blocks while possible

oldlist += num[count] # update old list with new digit

tempsum += int(num[count]) # add current digit to sum

count += 1

if tempsum > summ: # check if new sum is larger than old one

summ = tempsum # update sum

newlist = oldlist # update the sequence list

oldlist = ''

count = 0

tempsum = 0

num = num[1:] # slice the 'first' digit of the number

print(newlist, summ) # print sequence and desired sum

解决方案

You don't need two loops.

First, let's put all the digits in a list:

>>> a = list(map(int, str(5**150000)))

Then calculate the sum of the first 50000 digits:

>>> maximum = current = sum(a[:50000])

>>> current

225318

Now, let's loop through the list, removing the lowest digit from the sum and adding the next one 50000 digits ahead during each iteration:

>>> for i in range(0, len(a)-50000):

... current = current - a[i] + a[i+50000]

Check if that new sum is larger than the previous one, and if so, make it the new "interim maximum":

... if current > maximum: maximum = current

...

Once the loop exits, maximum contains the maximum value:

>>> maximum

225621

Let's put it all into a function, so no copying mistakes occur:

def maxdigitsum(number, digits):

l = list(map(int, str(number)))

maximum = current = sum(l[:digits])

for i in range(0, len(l)-digits):

current = current - l[i] + l[i+digits]

if current > maximum: maximum = current

return maximum