[Baekjoon/Python] No.4673 Self Numbers

No.4673 Self Numbers

 

Problmes

In 1949 the Indian mathematician D.R. Kaprekar discovered a class of numbers called self-numbers. For any positive integer n, define d(n) to be n plus the sum of the digits of n. (The d stands for digitadition, a term coined by Kaprekar.) For example, d(75) = 75 + 7 + 5 = 87. Given any positive integer n as a starting point, you can construct the infinite increasing sequence of integers n, d(n), d(d(n)), d(d(d(n))), .... For example, if you start with 33, the next number is 33 + 3 + 3 = 39, the next is 39 + 3 + 9 = 51, the next is 51 + 5 + 1 = 57, and so you generate the sequence

33, 39, 51, 57, 69, 84, 96, 111, 114, 120, 123, 129, 141, ...

The number n is called a generator of d(n). In the sequence above, 33 is a generator of 39, 39 is a generator of 51, 51 is a generator of 57, and so on. Some numbers have more than one generator: for example, 101 has two generators, 91 and 100. A number with no generators is a self-number. There are thirteen self-numbers less than 100: 1, 3, 5, 7, 9, 20, 31, 42, 53, 64, 75, 86, and 97.

Write a program to output all positive self-numbers less than 10000 in increasing order, one per line.

Example

1

3

5

7

9

20

31

42

53

64

|

| <-- a lot more numbers

|

9903

9914

9925

9927

9938

9949

9960

9971

9982

9993

 

Solution

def self_num():
    arr=list()
    res=0
    for i in range(1, 10001):
        if i < 10:
            res = i+i
            arr.append(res)
        elif i < 100:
            res = i+(i//10)+(i%10)
            arr.append(res)
        elif i < 1000:
            res = i+(i//100)+((i%100)//10)+((i%100)%10)
            arr.append(res)
        elif i < 10000:
            res = i+(i//1000)+((i%1000)//100)+(((i%1000)%100)//10)+(((i%1000)%100)%10)
            if res <= 10000:
                arr.append(res)
    arr.sort()
    arr1 = [i for i in range(1,10001)]
    notSelf = [item for item in arr1 if item not in arr]
    for each in notSelf:
        print(each)
 
self_num()