python - 如何在数字的恒定总和等于9 python的数字序列中找到特定数字？

Question

Enter the natural number N (1 <= N <= 1,000,000). Print to the screen the Nth number of the sequence: 9 18 27 36 45 54 63 72 81 90 108 117 (they all have their sum of digit equal to 9)

N = 9+(int(input())-1)*9
def sumnum(N):
    sum = 0
    while N > 0:
        d = N%10
        N = N//10
        sum += d
sumnum(N)
while sum != 9:
    N+=1
    sumnum(N)
print(N)

Here's my code and it got this error

TimeLimitError: Stopped on Line 4

Answer 1

Your sumnum function doesn't return anything, so it never changes sum (or any other value) outside of the function. Normally you'd get an error trying to compare an undefined variable, but sum is actually the name of a builtin function, so that's what you're comparing 9 to.

Here's a simple approach: iterate through all numbers in a while loop, and check the sum of each one's digits by converting it to a string and summing the int values of its characters. Each time you find one whose sum is 9, decrement N by one. When it hits zero, print the current number.

>>> N = 12
>>> i = 0
>>> while True:
...     i += 1
...     if sum(int(d) for d in str(i)) == 9:
...         N -= 1
...     if N == 0:
...         print(i)
...         break
...
117

Here's a more code-golf-y approach using a filtered generator with itertools.count() (less efficient for big numbers because it builds a list of the sequence up to N instead of just printing the last element):

>>> import itertools
>>> list(zip(range(N), (i for i in itertools.count() if sum(int(d) for d in str(i)) == 9)))[-1][-1]
117

Answer 2

This worked for me!

def getSum(n):
    sum = 0
    for digit in str(n):
        sum += int(digit)
    return sum

for i in range(0, 1000000):
    n = i
    sum = getSum(n)
    if sum == 9:
        print(f"{n} -- {sum}")

Answer 3

Performing arithmetic on strings can work but it's very inefficient. This may help (and performs very well). Generates a list of all numbers in the inclusive range 1 -> 1,000,000 whose digits sum to 9. Of course, the range can be limited to 900,000 because that's the highest number less than one million that has this particular property.

We can also take advantage of the fact that a number whose digits sum to 9 are also multiples of 9

import time

def sumdigits(n):
    r = 0
    while n > 0:
        r += n % 10
        n //= 10
    return r

lon = []
start = time.perf_counter()

for i in range(9, 900001, 9):
    if sumdigits(i) == 9:
        lon.append(i)
        
end = time.perf_counter()
print(len(lon))
print(f'Duration = {end-start:.4f}')

Output:

2002
Duration = 0.0531

Answer 4

Numbers whose digits add to 9 are multiples of 9 and vice versa. So the Nth number whose digits add to 9 is N*9. So all you need is this:

N = int(input("Enter N: "))
print (N*9)

You can add range checking for N being between 1 and 1,000,000 if you like.