Hello people…. Today, we are going to make our first non-trivial program submission, and I am going to guide you through reading the problem statement through making a successful submission. I will also be touching the issue of “time limit” and “source limit” in this tutorial. However, I won’t be revealing too much in terms of code, so you will still have to do the hard work!
I’ll be using the Small Factorials problem as the program driving this tutorial.
Here is the problem statement:
You are asked to calculate factorials of some small positive integers.
An integer t, 1<=t<=100, denoting the number of test cases, followed by t lines, each containing a single integer n, 1<=n<=100.
For each integer n given at input, display a line with the value of n!
Time limit: 1s
The problem statement is fairly simple. We all recall that the factorial of a natural number n is the product of all numbers from 1 through n. Recursively, we can define the factorial of a natural number n as the product of n and factorial of (n-1). So, factorial(n) = n X factorial(n-1) and factorial(1) = 1.
You may ask why we haven’t considered factorial(0) = 1? Well, if you read the problem statement carefully, the least number that you will ever be asked to compute the factorial for is 1, so we don’t bother about any number less than 1.
As a first attempt, we would try and think that this program can be trivially coded up using integers or even long long integers, but a quick glance at the upper bound on the input number suggests otherwise. How do you figure out the number of digits in the factorial of 100? We can surely try and find a pessimistic approximation. Do you remember that the number of digits in the product x*y is either digits_in(x) + digits_in(y) or digits_in(x) + digits_in(y) – 1. So let’s try and find out the number of digits for 100!.
Pessimistically, it must be:
digits_in(100) + digits_in(99!), which is:
3 + digits_in(99) + digits_in(98!)
…. and so on….
which is 3+2*90+1*9 = 192
so, in the worst case, we have 192 digits in the result.
This itself should prompt us to not use integers because it will overflow even the biggest integral data type(64-bit) that we have.
The next thing we may think of is to use floating point numbers, but that too can be ignored for the same reasons.
Now that we find ourselves in a fix, we will try and resort to creating our own big integer class. This is where selection of programming language comes into play. Usually, C/C++ is the programming language of choice because of it’s speed and wide adoption. However, now is a good time to look at languages like Java, Python, Caml, etc…. which have built-in support for indefinitely long integral values. So, if you know any of those languages, this problem should be a breeze to solve! And CodeChef allows you to choose one of 30+ different programming languages for coding your solution in!
Once you have coded up a solution in your language of choice and submit it, you find that it says “Wrong Answer”. You wonder “How could that happen” and try to figure out where to start debugging. You enter integers 4, 5, 6 and get the answers as 24, 120, and 720 respectively which are correct. Then you enter the number 50 and you get a really huge number. However, you wonder to yourself “How do I know if this is the correct value of 50!?”. You think for a while and then let it go. Here is where you should think some more.
As a first level of sanity checking, the output should have at least 5 zeros at the end because 50! involves multiplying 10, 20, 30, 40 and 50. If not, you can start debugging. However, even better would be if you knew the exact number of zeros to expect at the end of the output.
Here is how you can find that out. You need to know that you can get a zero at the end of the product of 2 numbers if and only if 10 is a factor of the number. This is possible only if 2 and 5 both are factors of the product.
You can write a simple program to count the number of times a factor of 5 or 2 occurs in all the numbers from 1 to 50, and the number of zeros to expect is the least of the two values. This should validate(to a great extent) your solution’s output.(This co-incidentally is the solution to another problem in the easy set).
What does the “Time limit” mean? It means that the program that you write should be able to compute the factorial of 100 numbers, each of which are <=100 in 1 second.
If I was solving this problem and I saw the constraints mentioned, I would just pre-compute all the values for factorials of numbers from 1 to 100 and store it as strings in my program’s code, and the display the required string depending upon the input. However, to thwart lazy coders like me, the problem setter has set a Code size limit which is “Source limit”. So your program’s source code must be at most 2000 Bytes.
We have already computed the largest result as having at most 192 digits. So, on the average, you can expect each result as having at most 96 digits. If I wanted to store all the values, I would need at least 96*100 bytes of storage, which is 9600 bytes, which exceeds the source limit. This makes it fairly impossible for me to hard-code the answers in the program.