PROGRAMMING:Narcissus number (20 points)
Narcissus number refers to an n-bit positive integer (n ≥ 3), the sum of the n-th power of the numbers in each bit of narcissus number is equal to itself.
For example: 153 = 1 & times; 1× 1+5× 5× 5+3× 3× 3。
This problem requires the preparation of procedures to calculate the number of all n-bit Narcissus.
###Input format:
The input gives a positive integer n (3 ≤ n ≤ 5) in one line
###Output format:
Output all n-bit Narcissus numbers in ascending order, each number occupies one line.
###Input example:
Here is a set of inputs. For example:
```in
three
```
###Output example:
The corresponding output is given here. For example:
```out
one hundred and fifty-three
three hundred and seventy
three hundred and seventy-one
four hundred and seven
```
answer:If there is no answer, please comment
```
t=int(input())
start=10**(t-1)
end=10**t-1
for i in range(start,end+1):
if i==sum([int(j)**len(str(i)) for j in str(i)]):
print(i)
```
For example: 153 = 1 & times; 1× 1+5× 5× 5+3× 3× 3。
This problem requires the preparation of procedures to calculate the number of all n-bit Narcissus.
###Input format:
The input gives a positive integer n (3 ≤ n ≤ 5) in one line
###Output format:
Output all n-bit Narcissus numbers in ascending order, each number occupies one line.
###Input example:
Here is a set of inputs. For example:
```in
three
```
###Output example:
The corresponding output is given here. For example:
```out
one hundred and fifty-three
three hundred and seventy
three hundred and seventy-one
four hundred and seven
```
answer:If there is no answer, please comment
```
t=int(input())
start=10**(t-1)
end=10**t-1
for i in range(start,end+1):
if i==sum([int(j)**len(str(i)) for j in str(i)]):
print(i)
```