Hướng dẫn how do you make python not divisible? - làm thế nào để bạn làm cho python không chia hết?

Xem thảo luận

Cải thiện bài viết

Lưu bài viết

Xem thảo luận

Cải thiện bài viết

Lưu bài viết

Đọc

Example:

Bàn luận: We check if the number is not divisible by 2 and 3 using the and clause, then outputs the number. 

Python3

Chúng ta có thể nhập một tập hợp số nguyên và kiểm tra số nguyên nào trong phạm vi này, bắt đầu với 1 không chia hết cho 2 hoặc 3, bằng cách kiểm tra phần còn lại của số nguyên với 2 và 3. & nbsp;

Phương pháp 1: Chúng tôi kiểm tra xem số không chia hết cho 2 và 3 bằng mệnh đề và điều khoản, sau đó xuất số. & NBSP;

>>> def isDivis(divisor,num): return not num%divisor >>> isDivis(2,12) True >>> isDivis(2,21) False >>> isDivis(3,21) True >>> isDivis(3,14) False 5>>> def isDivis(divisor,num): return not num%divisor >>> isDivis(2,12) True >>> isDivis(2,21) False >>> isDivis(3,21) True >>> isDivis(3,14) False 6>>> def isDivis(divisor,num): return not num%divisor >>> isDivis(2,12) True >>> isDivis(2,21) False >>> isDivis(3,21) True >>> isDivis(3,14) False 7>>> def isDivis(divisor,num): return not num%divisor >>> isDivis(2,12) True >>> isDivis(2,21) False >>> isDivis(3,21) True >>> isDivis(3,14) False 8

max_num ____10 >>> def isDivis(divisor,num): return not num%divisor >>> isDivis(2,12) True >>> isDivis(2,21) False >>> isDivis(3,21) True >>> isDivis(3,14) False 1

>>> def isDivis(divisor,num): return not num%divisor >>> isDivis(2,12) True >>> isDivis(2,21) False >>> isDivis(3,21) True >>> isDivis(3,14) False 2>>> def isDivis(divisor,num): return not num%divisor >>> isDivis(2,12) True >>> isDivis(2,21) False >>> isDivis(3,21) True >>> isDivis(3,14) False 0 >>> def isDivis(divisor,num): return not num%divisor >>> isDivis(2,12) True >>> isDivis(2,21) False >>> isDivis(3,21) True >>> isDivis(3,14) False 4

Input: 10 Output: Numbers not divisible by 2 and 3 1 5 79>>> def isDivis(divisor,num): return not num%divisor >>> isDivis(2,12) True >>> isDivis(2,21) False >>> isDivis(3,21) True >>> isDivis(3,14) False 5Numbers not divisible by 2 and 3 1 5 7 11 13 17 191

>>> def isDivis(divisor,num): return not num%divisor >>> isDivis(2,12) True >>> isDivis(2,21) False >>> isDivis(3,21) True >>> isDivis(3,14) False 9 num1=int(input("Enter your number:")) if(num1%3==0): print("{} is divisible by 3".format(num1)) else: print("{} is not divisible by 3".format(num1)) 0>>> def isDivis(divisor,num): return not num%divisor >>> isDivis(2,12) True >>> isDivis(2,21) False >>> isDivis(3,21) True >>> isDivis(3,14) False 0 num1=int(input("Enter your number:")) if(num1%3==0): print("{} is divisible by 3".format(num1)) else: print("{} is not divisible by 3".format(num1)) 2

Output:

Độ phức tạp về thời gian: O (1) O(1)

Không gian phụ trợ: O (1) O(1)

Làm thế nào để bạn không sử dụng chia sẻ trong Python?: We traverse the odd numbers starting with 1 since even numbers are divisible by 2. So, we increment the for a loop by 2, traversing only odd numbers and check, which one of them is not divisible by 3. This approach is better than the previous one since it only iterates through half the number of elements in the specified range.

Mã python sau đây minh họa điều này: & nbsp;

Python3

max_num ____10 Numbers not divisible by 2 or 3 : 1 5 7 11 13 17 19 23 25 29 31 35 370

>>> def isDivis(divisor,num): return not num%divisor >>> isDivis(2,12) True >>> isDivis(2,21) False >>> isDivis(3,21) True >>> isDivis(3,14) False 5>>> def isDivis(divisor,num): return not num%divisor >>> isDivis(2,12) True >>> isDivis(2,21) False >>> isDivis(3,21) True >>> isDivis(3,14) False 6Numbers not divisible by 2 or 3 : 1 5 7 11 13 17 19 23 25 29 31 35 373>>> def isDivis(divisor,num): return not num%divisor >>> isDivis(2,12) True >>> isDivis(2,21) False >>> isDivis(3,21) True >>> isDivis(3,14) False 8

Numbers not divisible by 2 or 3 : 1 5 7 11 13 17 19 23 25 29 31 35 375 Numbers not divisible by 2 or 3 : 1 5 7 11 13 17 19 23 25 29 31 35 376Numbers not divisible by 2 or 3 : 1 5 7 11 13 17 19 23 25 29 31 35 377 Numbers not divisible by 2 or 3 : 1 5 7 11 13 17 19 23 25 29 31 35 378>>> def isDivis(divisor,num): return not num%divisor >>> isDivis(2,12) True >>> isDivis(2,21) False >>> isDivis(3,21) True >>> isDivis(3,14) False 6>>> def isDivis(divisor,num): return not num%divisor >>> isDivis(2,12) True >>> isDivis(2,21) False >>> isDivis(3,21) True >>> isDivis(3,14) False 4(a % 2) != 01 num1=int(input("Enter your number:")) if(num1%3==0): print("{} is divisible by 3".format(num1)) else: print("{} is not divisible by 3".format(num1)) 7(a % 2) != 03

num1=int(input("Enter your number:")) if(num1%3==0): print("{} is divisible by 3".format(num1)) else: print("{} is not divisible by 3".format(num1)) 3 num1=int(input("Enter your number:")) if(num1%3==0): print("{} is divisible by 3".format(num1)) else: print("{} is not divisible by 3".format(num1)) 4 Numbers not divisible by 2 or 3 : 1 5 7 11 13 17 19 23 25 29 31 35 376 num1=int(input("Enter your number:")) if(num1%3==0): print("{} is divisible by 3".format(num1)) else: print("{} is not divisible by 3".format(num1)) 6 Input: 10 Output: Numbers not divisible by 2 and 3 1 5 74 num1=int(input("Enter your number:")) if(num1%3==0): print("{} is divisible by 3".format(num1)) else: print("{} is not divisible by 3".format(num1)) 8>>> def isDivis(divisor,num): return not num%divisor >>> isDivis(2,12) True >>> isDivis(2,21) False >>> isDivis(3,21) True >>> isDivis(3,14) False 0

Input: 10 Output: Numbers not divisible by 2 and 3 1 5 79>>> def isDivis(divisor,num): return not num%divisor >>> isDivis(2,12) True >>> isDivis(2,21) False >>> isDivis(3,21) True >>> isDivis(3,14) False 5 5

Output:

Độ phức tạp về thời gian: O (1) O(1)

Không gian phụ trợ: O (1) O(1)