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CS507 Assignment No. 03  Solution and Discussion

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start=0 evensum=0 oddsum=0 counteven=0 countodd=0 print("Enter the upper Limit:") upperLimit=input() print print("Even numbers are:") for num in range(start, upperLimit+1): if num%2==0: evensum=evensum+num counteven=counteven+1 print(num) print print("Sum of Even numbers:") print(evensum) print print("Total number of Even numbers within the range are:") print(counteven) print print("Odd numbers are:") for num1 in range(start, upperLimit+1): if num1%2!=0: oddsum=oddsum+num1; countodd=countodd+1 print(num1) print print("Sum of Odd numbers:") print(oddsum) print print("Total number of Odd numbers within the range are:") print(countodd)
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@zareen said in FRL101 Assignment 1 Solution and Discussion:

Read the profile description of the searched freelancer carefully and note down the main points
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7+ Year Experience in designing and web app development. PHP, MySQL using as backend Working in different cat like (Taxi booking, food ordering, on Demand services, dating ,social finance, health and fitness apps. Approach work (understanding, developing, as per client, meeting deadline effectivily) Expert in (Performance, Security, Design) Review (4.9) Member since Mar,2016 Different packages offering in mobile apps
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Use the caesar cipher method to encrypt the word “ACTIVATION” and show the complete process step by step.

A B C D E F G H I K L M N O P Q R S T U V W X Y Z

Chose a number form 1 to 26 .the number will represent the key or the “shift” value that will be used to shift the standard Alphabet.

KEY= 4

A B C D E F G H I K L M N O P Q R S T U V W X Y Z

W X Y Z A B C D E F G H I K L M N O P Q R S T U V

Now bottom row of ALPHABE will now by known as Cipher ALPHABET.

KEY=4

Message =”ACTIVATION”

Plain : A B C D E F G H I K L M N O P Q R S T U V W X Y Z

Cipher: W X Y Z A B C D E F G H I K L M N O P Q R S T U V

Ciper text : WYPERWPEKI

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CS507 Assignment No. 03 Solution and Discussion

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