You start out by finding a suitable problem to solve. Then you write code to solve the problem. After this, you submit the code to us for review. We will then compile your code and run it on some secret input. After some careful deliberation, you will get a judgement informing you whether your code behaved as expected or not.
Your program should read its input from standard input and produce output on standard output.
This can for instance be done using
stderr.write will be ignored. This can be used for debugging your program during development (i.e., you do not have to remove debug output before submitting if you use standard error for debug output). Of course, writing to standard error will take some runtime.
Input will always follow the input specification (so you do not need to validate the input). Your output must follow the output specification.
You are allowed to use all standard libraries included with Dart.
We are currently using Dell PowerEdge R230 servers for judging. These are equipped with an Intel Xeon E3-1220V6 CPU running at 3.0 GHz and 8 GB RAM. A 64-bit Linux kernel is used.
We will inspect the exit code of your program. If it is non-zero, we will judge your submission as Run Time Error.
Solving a problem
Now lets get down to business and write some code. The short tutorial below goes through the solution of A Different Problem.
Step 1: The problem
You are tasked with writing a program that computes the difference between integers. Sounds simple, doesn't it? Well, as we will see, the problem still holds some small difficulties.
Step 2: Reading the input
One thing to note is that the integers can be
fairly large, as large as 1015. Luckily, there is a
64 bit integer type in Dart,
Now that we have determined a suitable type, we just
have to read the data. Reading is done from standard
input. In this problem, we should read until the end
of the file (in other problems, there might be an
integer at the beginning of the input, specifying how
much to read, or there might be a special indicator
denoting that there is nothing more to read). Using
stdin.readLineSync, this can be done as below:
Step 3: Computing the answer
Now that we've read the input, it's time to actually solve the problem. Since 0 ≤ a, b ≤ 1015, we have that −(1015) ≤ a−b ≤ 1015, which means that there is no danger of overflow involved in just subtracting the two numbers a and b. Then, we can just take the absolute value by using the builtin abs method.
Finally, it's time to print the result. Using print (assuming the int variable r holds the result):
Step 4: The solution
Now we are basically done, all that remains is to combine the above parts.
Here is a version of the complete solution.