Posted by **anup** on July 11th, 2015
Filed in __Announcement, Features, Practice Problems, Schools__
View Comments

We have done a small change to our practice section. We have gotten rid of our “School” section and added a new “Beginner” section.

We had launched CodeChef for Schools a couple of years ago. Since then, we have visited many schools and one common concern that we have identified is that we do not have a proper place to start for the beginners. And when our intern, Vikesh Tiwari, joined us, this is something he took it upon himself to fix.

This change is a first small step that we took towards that. The “School” section used to contain all the problems from the LunchTime contests. The idea was to have newly added partially graded problems of LunchTime contests under one new tab. However, over a period of time we introduced partial grading in our Long Challenges too, and we realised that the old way of classifying problems based on difficulty levels and not on scoring type was a better one. Also, our Easy section started having problems which ranged from Cakewalk to Simple to Easy and also some Medium level problems! This was undesirable.

So we have made a couple of changes:

- We have introduced a new “Beginner” section where we will keep all the Simple and Cakewalk problems to help a newcomer find the easiest problems on CodeChef and get started.
- We moved all the LunchTime problems from the “School” section into Easy, Medium, Hard and the newly introduced Beginner sections.

As an ongoing endeavour, we will try and classify our problems appropriately so that they are easy to find and attempt. We will also be adding a bunch of very easy problems on CodeChef in the “Beginner” section.

A few other changes that we have done are:

- We have brought back changes that we had undone on the contest pages to reduce the load on our servers during the start of our short contests.
- You will also see an auto reload of contest pages along with the “Announcement” section at the top.
- A more localised notification of the Announcements on the contest pages.
- The most exciting change is that now you can make submissions to a problem of a running contest through the CodeChef IDE! Yeah, thats true. No need to use the submit button of each problem page to send us your solutions. Just use our awesome IDE to debug and submit from the same place.

You can see many more changes coming up on CodeChef to make problem classification better in the coming days. We are working hard to make your learning easier and better. Do keep giving us your feedback.

Cheers!

anup

Posted by **Tojo** on December 1st, 2010
Filed in __Announcement, Features, Practice Problems__
View Comments

Hi everyone,

We have introduced a new challenge tab in the practice section wherein all the challenge problems will be listed. This enables you to find all the challenge problems in a single place without browsing through the other categories. All the challenge problems that belonged to the previous contests have also been added to this area.

Cheers,

Tojo

Hello everyone,

This is a small update to let you know that we have now added some problems from contests being hosted on Codechef in the ‘Peer’ area of the practice section. These problems are available at www.codechef.com/problems/extcontest . As of now, approximately 150 problems from various contests hosted in the past one year have been added to this section. We hope you all have a fun time cracking these

Regards,

Aniruddha.

Posted by **The Chef** on October 27th, 2009
Filed in __Announcement, Features, Practice Problems__
View Comments

Candy Corns,

By popular demand you can now view all submissions for practice problems which you have successfully solved. You’ve solved a problem in Java, you can now see how someone else solved it in C. Hopefully you will use this power for good, and learn from other members as well as assist those who are having trouble.

How else can we improve the site? What can we do, to help you become better programmers?

Let us know,

Amit

Cupcakes,

You’ve asked for it, we’ve done it. A big complaint people have had was the time limits were too strict for programming languages that didn’t start with a C. There are now variable time limits depending on the programming language you have used to submit your solution. The current time limits are as follows (as a multiplier to the time limit listed on the problem page):

- Java – 2X
- Python (PYTH and PYTH 3.6) – 5X
- JS- 2X
- PyPy 2.6.0 – 2X
- Kotlin – 2X
- Ruby – 3X
- PHP – 3X
- Lisp clisp – 3X
- Lisp sbcl – 3X
- Scala – 2X
- C# – 2X
- SCM chicken – 3X
- SCM guile – 3X
- SCM qobi – 3X
- All other languages – 1X

We will continue to tweak the exact multipliers and will let you know if there are any other modifications. Solutions will not be rejudged, you will be required to submit new solutions to see if your program now satisfies the time limit conditions.

Cheers,

The Chef

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

You are asked to calculate factorials of some small positive integers.

**Input:**

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.

**Output:**

For each integer n given at input, display a line with the value of n!

**Example**

**Sample input:**

4

1

2

5

3

**Sample output:**

1

2

120

6

Time limit: 1s

Source limit:2000B

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.

CodeCheffers,

By popular demand we have added a trivial sample question: Life, the Universe, and Everything. Solutions in 25 programming languages can be found in our forums.

Additionally we’ve added a few new, easy practice problems:

Enormous Input Test

Sums in a Triangle

Turbo Sort

The Way to a Friends House Is Never Too Long

Finally we moved some of the more difficult of the easy problems to medium. As always your feedback appreciated.

There are now 23 practice problems available! The problems range in degree of difficulty, we’ve bucketed them in easy, medium and hard. The easiest problem is the ATM. If you’re new to online coding competitions, why don’t you give that a shot first? We will continue to add problems regularly and send out updates through the blog. If you have any questions, check out our forums.

- API < br/ >
- CodeChef DSA Student Offer
- [Live Updates] ACM ICPC Kanpur Regional 2019-20
- Chef’s New Year Present
- [Live Updates] ACM ICPC Amritapuri Regional 2019-20

- About (14)
- ACM ICPC (25)
- Announcement (308)
- API (3)
- Campus Chapters (7)
- CCDSAP (7)
- CCDSAP Stories (5)
- Certification (9)
- College Contests (9)
- Contests (281)
- Events (57)
- FAQ (1)
- Features (53)
- Interviews (25)
- Languages (1)
- Meetup (4)
- Open Source (1)
- Practice Problems (8)
- Prizes (20)
- Problems (15)
- ProblemSetting (1)
- Programmer of the Month (34)
- Schools (13)
- SnackDown (2)
- Tech Talks (23)
- Tutorials (35)
- Uncategorized (1)
- Volunteers (4)
- Winners (123)

- Manraj Singh on In conversation with Taranpreet Singh – an accounting student turned 6-star coder!
- Bhagya Kamal Jain on CodeChef DSA Student Offer
- Avi Srivastava on Tutorial For Small Factorials
- Shloki Jha on Tutorial For Small Factorials
- Shloki Jha on Tutorial For Small Factorials

- January 2020
- December 2019
- September 2019
- July 2019
- June 2019
- May 2019
- March 2019
- February 2019
- January 2019
- December 2018
- November 2018
- October 2018
- September 2018
- August 2018
- July 2018
- June 2018
- April 2018
- March 2018
- February 2018
- January 2018
- December 2017
- November 2017
- September 2017
- August 2017
- July 2017
- June 2017
- March 2017
- February 2017
- January 2017
- December 2016
- November 2016
- October 2016
- September 2016
- August 2016
- July 2016
- June 2016
- May 2016
- April 2016
- March 2016
- February 2016
- January 2016
- December 2015
- November 2015
- October 2015
- September 2015
- August 2015
- July 2015
- June 2015
- May 2015
- April 2015
- March 2015
- February 2015
- January 2015
- December 2014
- November 2014
- September 2014
- August 2014
- July 2014
- June 2014
- May 2014
- April 2014
- March 2014
- January 2014
- December 2013
- November 2013
- October 2013
- September 2013
- August 2013
- July 2013
- June 2013
- May 2013
- April 2013
- March 2013
- February 2013
- January 2013
- December 2012
- October 2012
- September 2012
- August 2012
- July 2012
- June 2012
- May 2012
- April 2012
- March 2012
- February 2012
- January 2012
- December 2011
- November 2011
- October 2011
- September 2011
- August 2011
- July 2011
- June 2011
- May 2011
- April 2011
- March 2011
- February 2011
- January 2011
- December 2010
- November 2010
- October 2010
- September 2010
- August 2010
- July 2010
- June 2010
- May 2010
- April 2010
- March 2010
- February 2010
- January 2010
- December 2009
- November 2009
- October 2009
- September 2009
- August 2009
- July 2009
- June 2009
- May 2009
- April 2009
- March 2009
- February 2009
- January 2009