当前位置:网站首页>[weekly pit] information encryption + [answer] positive integer factorization prime factor
[weekly pit] information encryption + [answer] positive integer factorization prime factor
2022-07-06 20:19:00 【Crossin's programming classroom】
Zero basis python Introductory tutorial :python666.cn
Hello everyone , Welcome to Crossin Programming classroom of !
Today, I'll give you one of my favorite topics : Information encryption
Of course , This problem is the fur of information encryption .
Please listen to :
A yuan and Qiqi pass notes in class , But I don't want to be found by others . So they agreed on an encryption method :
Only for English or Pinyin in the information content ( That's the letter ) To encrypt , Other symbols remain intact
Each letter moves backward 5 position , Become ciphertext , such as a->f,k->p
If backward 5 More than z, Just fold back to the beginning of the letter , such as x->c
Please complete the encryption and decryption of this rule with a piece of code .
The reference sample :
【 encryption 】
Input :there is a vegetable in your teeth
Output :ymjwj nx f ajljyfgqj ns dtzw yjjym
【 Decrypt 】
Input :onz sn mzf izt
Output :jiu ni hua duo
Additional questions
Can you decrypt this text :
ild gdpsh sxktgvts xc p ldds, pcs x--
x iddz iwt dct athh igpktats qn,
pcs iwpi wph bpst paa iwt sxuutgtcrt.
Detailed answers and reference codes will be given in the next column , You can also refer to the code in the message of other students .
I look forward to your answers , I hope you can complete the whole series .
Simple code can be submitted directly in the message , Long code is recommended paste.ubuntu.com or
codeshare.io And other code sharing sites , Just copy the code and save it , You can get a share address , Very convenient .
For past questions, click the collection at the beginning of the article “ Every Monday ” Enter the view .
【 answer 】 Positive integer factorization prime factor
The original title is :【 Every Monday 】 Positive integer factorization prime factor
One idea of this question is :
from 2 Start trying to divide the original number N
If you can divide , The original number is updated to N//i, And the output i; If you can't divide it, keep trying , until N Divide by itself
If there is a prime factor before , The last remaining i Is the last prime factor ; without , It shows that the original number is a prime number
The reason why there is no need to judge the divisor i Is it a prime number , Because i It's from 2 Start trying one by one , If it can be divided by a smaller number , Then it has been dealt with before . So what can be calculated i It must be the prime factor .
In addition, here is a small optimization point : Every time you try, you don't have to start from 2 Start , You can start with the previous prime factor .
Reference code :
n = input(' Enter the positive integer to be decomposed :')
start = 2
is_prime = True
while n > 1: # Keep breaking down n, Until it breaks down into 1
for index in range(start, n + 1) :
if n % index == 0: # n Can be index to be divisible by
n //= index # n=n//index
if n == 1:
if is_prime:
print(f'{index} Prime number ')
else:
print(index)
else: # index It must be prime
print(f'{index} * ', end='')
is_prime = False
break
start = index
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