仅添加奇数

Question

So the question I'm trying to solve the user is supposed to enter any positive number. Then I'm trying to write a program that adds only the odd numbers up to the number the user enters and displays the total. So for example if the user enters 4 my program should add four odd numbers. 1 + 3 + 5 + 7 = 16.
The only tools I have available are for statement, if, if/else if,while loop and println.

I can only figure out how to print out the odd numbers. I know I want to create a variable named total to store the value of adding up all the odd numbers but I don't know how that fits into the program.

import acm.program.*;

public class AddingOddNumbers extends ConsoleProgram {
    public void run() {
        int n = readInt("enter a positive nunber: ");
        int total = 0;
        for (int i = 0; i < n; i++) {
            if (n == 1) {
                println(1);
            } else {
                println((i * 2) + 1);
            }
        }
    }
}

Answer 1

import acm.program.*;

public class AddingOddNumbers extends ConsoleProgram {
    public void run() {
        int n = readInt("enter a positive nunber: ");
        int total = 0;
        for (int i = 0; i < n; i++) {
            if (n == 1) {
                println(1);
            } else {
                println((i * 2) + 1);
                total += (i * 2) + 1;
            }
        }
        println("total : " + total);
    }
}

Answer 2

sum = 0;

for (i = 1; i < n*2; i=i+2)
    sum = sum + i;

Answer 3

This will give you the odd number sum.

  if (n>0) 
   {
    total=0;
      for (int i = 1; i < n; i ++){
        if (i%2 == 1)
            total+=i;
      }
   }

If you want to inclusive of n, then change the condition to i<=n .

Answer 4

Maybe you know how to compute the sum of all numbers up to a given number n ? The formula is quite simple: (n * (n+1))/2 . Now getting the sum of only the odd numbers is a bit trickier but - no worries you can make use only of the previous formula for that. First notice that the sum of all even numbers up to a given number n is:

• (((n/2)* (n/2+1))/2) * 2 if N is even(ie the sum of all numbers up to n/2 times two that is because you have 2+4+6+8+...N = 2*(1+2+3+...n/2) )
• ((((n-1)/2)* ((n-1)/2+1))/2) * 2 if N is odd

In fact if you have integer division the formula is always: (((n/2)* (n/2+1))/2) * 2 = (n/2)* (n/2+1)

So to compute the sum of all the odd numbers up to n you simply subtract the sum of the even numbers from the sum of the all numbers:

(n * (n+1))/2 - (n/2)*(n/2+1)

In fact if you observe closely you will notice that the sum 1+3+...(2*n-1) always equals to n^2 .

This answer should help you solve your problem in all languages and I am leaving the code to you. It is literally one line.

Answer 5

I would use a loop for the odd numbers as well.

for (int i = 0, j = 1; i < n; i++, j += 2) {
    println(j);
    total += j;
}
println(total);

Answer 6

int oddSum = 0;
for (int i = 0; i < n; i++){
  oddSum = oddSum + (i*2) + 1;
}