Python总和N个连续数字的大数

Question

我需要获得n大数范围内的最大连续数字总和.

例如,范围可以是5^150000,在此范围内,我想找出50,000个连续数字的最大总和.

我使用两个循环的方法似乎永远不会终止.我会很感激任何意见.

代码:

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

Answer 1

你不需要两个循环.

首先,让我们将所有数字放在一个列表中:

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

然后计算前50000位的总和:

>>> maximum = current = sum(a[:50000])
>>> current
225318

现在,让我们循环遍历列表,从总和中删除最低位数,并在每次迭代期间添加下一个50000位数字:

>>> for i in range(0, len(a)-50000):
...     current = current - a[i] + a[i+50000]

检查新总和是否大于前一个,如果是,请将其设为新的"临时最大值":

...     if current > maximum: maximum = current
...

循环退出后,maximum包含最大值:

>>> maximum
225621

让我们将它全部放入一个函数中,这样就不会出现复制错误:

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