A for loop puts the start, the stop and the step on one line, which makes off by one errors easier to see and easier to make. Three loops here, each with a different bound.
What to write
Read two integers, n and step.
Using a for loop, print every number from 1 to n on its own line.
Then print the sum of the squares of 1 through n on one line.
Then, counting DOWN from n to 1 with a step of the given size, print each value you land on, one per line.
Assume n is at least 1 and step is at least 1.
Your program reads
Two integers on separate lines: n, then step.
Your program prints
n lines counting up, then the sum of squares, then the values counted down by step.
Worked examples
These are the cases you can see. There are more you cannot, and they use different values, so an answer that prints these numbers as constants will fail.
Example 1 input
5
2
Example 1 output
1
2
3
4
5
55
5
3
1
Example 2 input
1
1
Example 2 output
1
1
1
Your answer
Main.java
Input for the Run button
Run sends whatever is in the input box below. Submitting runs your program against every test case, including hidden ones with different input.
Stuck?
Hint 1
i <= n runs n times when i starts at 1. i < n runs n - 1 times, which is the off by one this exercise is looking for.
Hint 2
The countdown starts at n, tests i >= 1, and subtracts step. It will usually not land exactly on 1, and that is correct.
Hint 3
The counter declared inside the for header disappears at the closing brace, so all three loops can reuse the name i.
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