Answer:

Yes. Each of the three sums is calculated to be zero. The flowchart also works for N==1. It is essential to check the logic of a program for simple "boundary cases" like these.

Adding Up Even and Odd Integers

flowchart of program

Here is a skeleton for the program. Compare the skeleton to the flowchart at right.

import  java.util.Scanner;

// User enters a value N
// Add up odd integers,  
// even  integers, and all integers 0 to N
//
class addUpIntegers
{
  public static void main (String[] args ) 
  {
    Scanner scan = new Scanner( System.in );
    int N, sumAll = 0, sumEven = 0, sumOdd = 0;

    System.out.print( "Enter limit value:" );
    N = scan.nextInt();

    int count =  ;
    
    while (   )    
    {
      (more statements will go here later.)

      
    }

    System.out.print  ( "Sum of all : " + sumAll  );
    System.out.print  ( "\tSum of even: " + sumEven );
    System.out.println( "\tSum of odd : " + sumOdd  );
  }
}

First, we should get the counting loop correct. The loop should count from zero up to (and including) the limit. It is often a good idea to write a program in stages (just as a house is built in stages). First, get the foundation correct. The foundation of this program is the counting loop.

QUESTION 3:

Fill in the three blanks.