PROGRAMMING/SOFTWARE > Alexander Hamilton High School SCIENCE 101 Lists and Procedures Psuedocode P

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Situation A: Takes the list lotsOfNumbers and uses a loop to find the sum of all of the

odd numbers in the list. Displays the sum.

Sum ← 0 FOR EACH item IN lotsOFNumbers { <IF (INDEX mod 2 = 1) { <sum + list [i]> } > } DISPLAY (sum) Situation B: Write a procedure that takes a positive integer as a parameter. If the

number given to the procedure is no more than 30, the procedure should return the absolute difference between that number and 30. If the number is greater than 30, the procedure should return the number doubled. n ← input IF (n ≥ 0) { <IF (n ≤ 30) { <n← 30 - n DISPLAY n> } ELSE { <n← n * 2 DISPLAY n> } }

Situation C: Write a procedure ConvertToBinary that takes an input as a number from 0

to 16 (including 0 but not 16) and converts it to a binary number. The binary number should be returned as a list.

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Procedure: 1. Store the remainder when the number is divided by 2 in an array.

2. Divide the number by 2 3. Repeat the above two steps until the number is greater than zero.

4. Print the array in reverse order now.

Example: Binary number = 10.

Step 1: Remainder when 10 is divided by 2 is zero. Therefore, arr[0] = 0.

Step 2: Divide 10 by 2. New number is 10/2 = 5.

Step 3: Remainder when 5 is divided by 2 is 1. Therefore, arr[1] = 1.

Step 4: Divide 5 by 2. New number is 5/2 = 2.

Step 5: Remainder when 2 is divided by 2 is zero. Therefore, arr[2] = 0.

Step 6: Divide 2 by 2. New number is 2/2 = 1.

Step 7: Remainder when 1 is divided by 2 is 1. Therefore, arr[3] = 1.

Step 8: Divide 1 by 2. New number is 1/2 = 0.

Step 9: Since number becomes = 0. Print the array in reverse order. Therefore the equivalent binary number is 1010.

List 0

0000

1

0001

2

0010

3

0011

4

0100

5

0101

6

0110

7

0111

8

1000

9

1001

10

1010

11

1011

12

1100

13

1101

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14

1110

15

1111

16

00010000

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