Python

1) Assume there is a variable , h already associated with a positive integer value. Write the code necessary to count the number of perfect squares whose value is less than h , starting with 1 . (A perfect square is an integer like 9 , 16 , 25 , 36 that is equal to the square of another integer (in this case 3*3 , 4*4 , 5*5 , 6*6 respectively).) Assign the sum you compute to a variable q For example, if h is 19 , you would assign 4 to q because there are perfect squares (starting with 1 ) that are less than h are: 1 , 4 , 9 , 16 .



2) Assume there are two variables , k and m , each already associated with a positive integer value and further assume that k 's value is smaller than m's. Write the code necessary to compute the number of perfect squares between k and m . (A perfect square is an integer like 9 , 16 , 25 , 36 that is equal to the square of another integer (in this case 3*3 , 4*4 , 5*5 , 6*6 respectively).) Associate the number you compute with the variable q . For example, if k and m had the values 10 and 40 respectively, you would assign 3 to q because between 10 and 40 there are these perfect squares: 16 , 25, and 36 ,.



3) In the following sequence, each number (except the first two) is the sum of the previous two numbers: 0, 1, 1, 2, 3, 5, 8, 13, ... . This sequence is known as the Fibonacci sequence.

We speak of the i'th element of the sequence (starting at 0)-- thus the 0th element is 0, the 1st element is 1, the 2nd element is 1, the 3rd element is 2 and so on. Given the positive integer n , associate the nth value of the fibonacci sequence with the variable result . For example, if n is associated with the value 8 then result would be associated with 21.

h = 19  # h is already assigned a positive integer value 19

i = 1  # used to calculate perfect square

q = 0   # stores the sum of the perfect squares

while q < h:  

# this loop executes until value of number of perfect squares is less than h

   q = q + (i*i)  # to calculate the sum of perfect squares

   i = i + 1  # increments i by 1 at every iteration

print(q) # displays the sum of perfect squares

Output:

30

Explanation:

Here we see that the sum of the perfect square is printed in the output i.e. 30. If we want to assign value 4 to q then we should alter the code as  following:  

h = 19

i = 1

sum_of_ps = 0

q=0

while sum_of_ps < h:

   sum_of_ps = sum_of_ps + (i*i)

   q = q + 1

   i = i + 1  

print(q)  

If you want to take input from the user and also display the perfect squares in output you can use the following code:

h = int(input('Enter the value of h: ')) #takes value of h from user

i = 1

sum_of_ps = 0

q=0

while sum_of_ps < h:

   sum_of_ps = sum_of_ps + (i*i)

   print(i*i) #displays the perfect squares of h

   q = q + 1

   i = i + 1    

print(sum_of_ps)   #displays sum of perfect squares

print(q) #displays number of perfect squares whose value is less than h

Output:

1

4

9

16

30

4

# k and m are already assigned positive integers 10 and 40

k = 10  

m = 40  

i = 1  # i is initialized to 1

q = 0  # q stores the number of perfect squares between k and m

while i*i < m:  # loop computes perfect square of numbers between k and m

   if i*i > k:          

       q = q + 1  # counts the number of perfect squares

   i = i + 1  #increments the value of i at each iteration

print (q) # prints the no of perfect squares between k and m

If you want to display the perfect squares too use , add print(i*i) statement after if statement  if i*i > k:        

Output:

3

def Fibonacci(n):  # method Fibonacci which has an integer parameter n

   if n <= 1:  # if the value of n is less than or equals to 1

       result = n  # stores the value of n into result

   elif n > 1:  # else if the value of n is greater than 1

       result = Fibonacci(n-1) + Fibonacci(n-2)  

# calls Fibonacci method recursively to associate nth value of Fibonacci #sequence with variable result

   return result     #returns the final value stored in result variable

print(Fibonacci(8))

#calls Fibonacci function and passes value 8 to it

Output:

21