Extensible Hypertext Markup Language.
Answer:
The output of this question is 21. As show in the image
The explanation is given in below
Explanation:
Let first write the question
C=1
sum = 0
while(C<10):
C=C+3
sum=sum + C
print(sum)
Now Focus on
while(C<10):
C=C+3
sum=sum + C
The value of C is initially 1
C=1+3
Sum= 0+4
In second loop the value of C will become 4
c=4+3
sum=4+7
In third loop the value of C will be 7
c=7+3
sum=11+10
so the answer is 11+10=21
Answer:
#include <iostream>
using namespace std;
int main() {
int k;
double d;
string s;
cin >> k >> d >> s;
cout << s << " " << d << " " << k << "\n" << k << " " << d << " " << s; }
Explanation:
k is int type variable that stores integer values.
d is double type variable that stores real number.
s is string type variable that stores word.
cin statement is used to take input from user. cin takes an integer, a real number and a word from user. The user first enters an integer value, then a real number and then a small word as input.
cout statement is used to display the output on the screen. cout displays the value of k, d and s which entered by user.
First the values of k, d and s are displayed in reverse order. This means the word is displayed first, then the real number and then the integer separated again by EXACTLY one space from each other. " " used to represent a single space.
Then next line \n is used to produce a new line.
So in the next line values of k, d and s are displayed in original order (the integer , the real, and the word), separated again by EXACTLY one space from each other.
The program along with the output is attached.
Answer:
This is not true
Explanation:
The optimal Huffman code is used to encrypt and compress text files. It uses fixed-length code or variable-length code for encryption and compression of data.
The professor's character code is similar to Huffman's variable-length coding which uses variable length od binary digits to represent the word strings. The file size of the text file above is;
= 6 x 1 + 2 x 2 + 3 x 2 + 2 x 2 + 8 x 1 = 28 bits
This would be the same for both cases.
The encrypt would be the problem as the encoded and decoding of the characters B and E may cause an error.