To excel in questions like this, you should first understand the problem.

▎✅ Algorithm Representation

An algorithm can be represented using both flowcharts and pseudocode.

• Flowchart: This represents the algorithm using various shapes:
  – Oval for Start/End
  – Rectangle for Operations
  – Parallelogram for Input/Output
  – Diamond for Decision-making

• Pseudocode: This represents the algorithm using text-based instructions, similar to English words.

▎Hint

To compare three numbers, we should first accept input from the user.

▎Pseudocode

Step 1: Start
Step 2: Input three numbers X, Y, and Z
Step 3: If X > Y then go to Step 5.
Step 4: If Y > Z then 
            Print "Y is the largest"
        Else
            Print "Z is the largest"
        ENDIF
        Go to Step 6
Step 5: If X > Z then
            Print "X is the largest"
        Else
            Print "Z is the largest"
        ENDIF
Step 6: Stop

Flowchart

1. Start: Use an oval shape to represent the start of the process.
2. Input: Use a parallelogram to indicate where the user inputs three numbers (X, Y, Z).
3. Decision (Diamond): Use a diamond shape for the first comparison (Is X > Y?).
   – Yes: If true, go to another decision diamond (Is X > Z?).
     ▪ Yes: Print "X is the largest" (parallelogram).
     ▪ No: Print "Z is the largest" ( parallelogram).
   – No: Go to the next decision diamond (Is Y > Z?).
     ▪ Yes: Print "Y is the largest" (parallelogram).
     ▪ No: Print "Z is the largest" (parallelogram).
4. End: Use an oval shape to represent the end of the process.

@ProgrammingWithFaith2018

✅Exercise: Increment and Decrement

‎

‎1. int a=5;
‎  int  x=++a + ++a;
‎  cout<<x;
‎
‎
‎2. int a=5;
‎   int b=6;
‎   Cout<< ++a + b++;
‎   Cout<<b;
‎
‎3. int x, a, b, c;
‎a = 2;
‎b = 4;
‎c = 5;
‎x = a--+ b++ -++c;
‎cout<<x;
‎
‎4. int x, a;
‎a = 2;
‎x = ++a *--a *++a;
‎cout<<x;
‎
‎5. int x, a = 4;
‎x = ++a + ++a + ++a;
‎
‎cout<<x;
‎

@ProgrammingWithFaith2018
‎

✅ ‎Hint:‎One variable can't store two value at the same time.

‎
‎1.int a=5;
‎int x=++a + ++a;
‎(6+ ++a)
‎(7+7) because one variable can't hold two different value , (6 and 7 ) at the same time so it become 7.
‎
‎7+7=14
‎
‎

Answer 14

‎
‎2.int a=5;
‎   int b=6;
‎   
‎   Cout<<++a +b++// post and pre increment .
‎  6+6 // after this line b will be 7
‎Cout<<++a +b++;  (6+6)
‎  =12
‎Cout<<b;
‎=7
‎
‎

answer=12 and 7

‎
‎3.int x,a,b,c;
‎a=2;
‎b=4;
‎c=5;
‎x=a-- + b++ - ++c;
‎
‎(2+4)-6
‎=6-6=0
‎cout<<x;  
‎
‎

Answer   0

‎
‎4.int x,a;
‎a=2
‎x=++a *--a* ++a
‎(3*2) * ++a , but one variable can't hold two values at the same time.
‎
‎So, (2*2)*++a
‎    (4*3)=12
‎
‎cout<<x;// 12
‎
‎

Answer 12

‎
‎
‎5.int x,a=4;
‎x=++a + ++a + ++a;
‎
‎(++a + ++a) + ++a
‎(5+6) + ++a but we can't store two value at one variable at the same time.
‎
‎So,
‎(6+6)+ ++a
‎(12)+ 7
‎
‎

Answer 19

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