Variable n contains an integer number. Which of the below Python statements correctly find the number of odd digits of n? (Example: 1234987 has 4 odd digits: 1, 3, 9 and 7, so program will find 4.) Lütfen birini seçin: a. cnt = 0 k = 1 while k <= n: if n % 2 == 1: cnt += 1 k += 1 print(n, "has", cnt, "odd digits") b. cnt = 0 k = n while k > 0: if k % 2 == 1: cnt += 1 k //= 10 print(n, "has", cnt, "odd digits") c. cnt = 0 k = n while k > 0: if n % 2 == 1: cnt += 1 k //= 10 print(n, "has", cnt, "odd digits") d. cnt = 0 k = 1 while k < n: if k % 2 == 1: cnt += 1 k += 1 print(n, "has", cnt, "odd digits") e. cnt = 0 k = n while k > 0: if k % 10 == 1: cnt += 1 k //= 10 print(n, "has", cnt, "odd digits")
Variable n contains an integer number. Which of the below Python statements correctly find the number of odd digits of n? (Example: 1234987 has 4 odd digits: 1, 3, 9 and 7, so program will find 4.) Lütfen birini seçin: a. cnt = 0 k = 1 while k <= n: if n % 2 == 1: cnt += 1 k += 1 print(n, "has", cnt, "odd digits") b. cnt = 0 k = n while k > 0: if k % 2 == 1: cnt += 1 k //= 10 print(n, "has", cnt, "odd digits") c. cnt = 0 k = n while k > 0: if n % 2 == 1: cnt += 1 k //= 10 print(n, "has", cnt, "odd digits") d. cnt = 0 k = 1 while k < n: if k % 2 == 1: cnt += 1 k += 1 print(n, "has", cnt, "odd digits") e. cnt = 0 k = n while k > 0: if k % 10 == 1: cnt += 1 k //= 10 print(n, "has", cnt, "odd digits")
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
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Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Question
Variable n contains an integer number.
Which of the below Python statements correctly find the number of odd digits of n?
(Example: 1234987 has 4 odd digits: 1, 3, 9 and 7, so
Lütfen birini seçin:
a. cnt = 0 k = 1 while k <= n: if n % 2 == 1: cnt += 1 k += 1 print(n, "has", cnt, "odd digits")
b. cnt = 0 k = n while k > 0: if k % 2 == 1: cnt += 1 k //= 10 print(n, "has", cnt, "odd digits")
c. cnt = 0 k = n while k > 0: if n % 2 == 1: cnt += 1 k //= 10 print(n, "has", cnt, "odd digits")
d. cnt = 0 k = 1 while k < n: if k % 2 == 1: cnt += 1 k += 1 print(n, "has", cnt, "odd digits")
e. cnt = 0 k = n while k > 0: if k % 10 == 1: cnt += 1 k //= 10 print(n, "has", cnt, "odd digits")
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