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Readme [11.4K]
3 years ago
8

Given positive integer n, write a for loop that outputs the even numbers from n down to 0. If n is odd, start with the next lowe

r even number. Hint: Use an if statement and the % operator to detect if n is odd, decrementing n if so. Enter an integer: 7 Sequence: 64 20 (2) If n is negative, output 0, Hint: Use an if statement to check if n is negative. If so, just set n = 0. Enter an integer: -1 Sequence: 0 Show transcribed image text (1) Given positive integer n, write a for loop that outputs the even numbers from n down to 0. If n is odd, start with the next lower even number. Hint: Use an if statement and the % operator to detect if n is odd, decrementing n if so. Enter an integer: 7 Sequence: 64 20 (2) If n is negative, output 0, Hint: Use an if statement to check if n is negative. If so, just set n = 0. Enter an integer: -1 Sequence: 0
Computers and Technology
1 answer:
seropon [69]3 years ago
7 0

Answer:

The program in Python is as follows:

n = int(input("Enter an integer: "))

if n < 0:

   n = 0

print("Sequence:",end=" ")

for i in range(n,-1,-1):

   if i%2 == 0:

       print(i,end = " ")

Explanation:

This gets input for n

n = int(input("Enter an integer: "))

This sets n to 0 if n is negative

<em>if n < 0:</em>

<em>    n = 0</em>

This prints the string "Sequence"

print("Sequence:",end=" ")

This iterates from n to 0

for i in range(n,-1,-1):

This check if current iteration value is even

   if i%2 == 0:

If yes, the number is printed

       print(i,end = " ")

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Explanation:

This defines the function

int chkInd(string str1, string str2){    

First, the length of str1 is initialized to 0

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Then exit the loop

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Increase index by 1

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This returns the calculated returned index; if no matching is found, it returns -1

  if (retIndex == 0){   return i;   }else{return -1;}}

}

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