Rating changes for last rounds are temporarily rolled back. They will be returned soon. ×

PrinceOfPersia's blog

Valentine Algorithm cUp 2015
By PrinceOfPersia, 10 years ago, In English

Are you in love with algorithms? If you are don't miss this round and be algorithm's valentine ;)

On the night before valentine's day (exact time), Valentine Algorithm cUp 2015 is going to take place.

There will be 6 problems and 3 hours to solve them. All of them are written by me(PrinceOfPersia).

This competition is like surprise languages but too different. 3 days before the contest, I will publish tutorial of 5 new algorithm languages(not programming languages) with their compiler codes. Each language is going to be really simple to learn and have at most 6 or 7 commands (they're not like programming languages).

Your program should print the code written in the language the problem says and checker will run it (pay attention that inputs are encoded, don't try to read form input, just print the code). Don't use unnecessary wjitespaces.

Each command of a language, will have a number, shown by w(x) called order. There will be a counter cnt = 0 (initially), every time your code calls this command, cnt increases by x. The order of your code equals the final value of cnt.

Each problem has a limit for your code's order.

If your code gets CE or OLE (order limited exceeded) or WA, you will receive Wrong answer.

Example :

Language (named EXM) : This language is built to process on an integer. There are two valid commands :

  1. M x : multiply the number by x. (Order : w(x) ).
  2. I x : increase the number by x. (Order : w(1) ).

Your program returns the final value of that number.

Your task is given number n, the return value of your code must be number 2n + 1 .

Code in EXM:

M 2
I 1

C++ code (the code you will submit in submit page):

printf("M 2\nI 1\n");

Your code's order will be 2 + 1 = 3 .

UPD: You can download languages tutorial and their compilers here.

Attention: Compile the code for each language, in Windows : g++ -o lang lang.cpp -O2 -std=c++0x and in Mac and Linux g++ -o lang.out lang.cpp -O2 -std=c++0x.

Using in Windows:

For running your code that is saved in a file like x.txt, use exactly lang x.txt in cmd and if you want debug mode enabled, use lang x.txt -d (not lang -d x.txt).

Using in Mac OS X or Linux:

For running your code that is saved in a file like x.txt, use exactly ./lang.out x.txt in terminal and if you want debug mode enabled, use ./lang.out x.txt -d (not ./lang.out -d x.txt).

Sample problem:

Write a program in Prolan that given a string s containing only a and b deletes all bs and converts all as to d. Order shouldn't exceed 106 (1 ≤ |s| ≤ 100).

My code:

(b,)
(a,d)

If you have any question, feel free to ask.

UPD2: Scoring will be 500-1000-1500-2000-2500-3000.

By the way, the main characters of each problem is different from other problems. Actually, for each problem the main characters are an imaginary(or not!) couple :)

UPD3: IDXT's compiler had some bug, you can download the correct source here.

UPD4: For people who want to practice, the new archive(including languages tutorials and bugless compilers) can be found here.

  • Vote: I like it
  • +185
  • Vote: I do not like it

Comments Comments (63)
| Write comment?
»
10 years ago, # |
← Rev. 2   Vote: I like it +5 Vote: I do not like it

The moment I realised when I don't have anyone to spend Valentine's Day with.
D:

But then I realised my country doesn't celebrate it.
:D

But then I realised I don't have a holiday because of it either.
D:

But then I realised that I'll be able to participate in this contest to improve my programming skills.
:D

But then I realised that I'm putting way too many "But then I realised"s.
D:

→ Reply
  • »
    »
    10 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    You're totally wrong. We celebrate valentine's day in Iran.

    → Reply
    • »
      »
      »
      10 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      he's talking about his country which is Kazakhstan looking forward to the contest :D

      → Reply
      • »
        »
        »
        »
        10 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        No, he's not. Read it again :

        But then I realised that I'll be able to write this contest to improve my programming skills.

        So, he's talking about me.

        → Reply
        • »
          »
          »
          »
          »
          10 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          oh my bad you're right :D

          → Reply
        • »
          »
          »
          »
          »
          10 years ago, # ^ |
            Vote: I like it +39 Vote: I do not like it

          I've seen many people use "write" as an (incorrect) form of "take part in" a contest. Plus it actually makes sense if you interpret it that way.

          → Reply
          • »
            »
            »
            »
            »
            »
            10 years ago, # ^ |
              Vote: I like it +7 Vote: I do not like it

            Yeah you're right, I suck at English

            → Reply
»
10 years ago, # |
  Vote: I like it -9 Vote: I do not like it

could't you just read the file instead of a program that output that file?

→ Reply
  • »
    »
    10 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    No, I need to write a compiler for each language and I'm not sure admins add my languages.

    → Reply
»
10 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Sounds interesting! It will be held here in CF? And what about scoring? It will depend only on order or time also makes sense?

→ Reply
  • »
    »
    10 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    In Gym. ACM-ICPC rules.

    → Reply
    • »
      »
      »
      10 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      But UPD2: Scoring will be 500-1000-1500-2000-2500-3000. it seems to be a Codeforces-rules~~~

      → Reply
      • »
        »
        »
        »
        10 years ago, # ^ |
          Vote: I like it +6 Vote: I do not like it

        It's possible to use weighted ACM-ICPC rules: solved problems give 500/1000/something points, and solved problems add to the time penalty, just like Rockethon 2015.

        → Reply
    • »
      »
      »
      10 years ago, # ^ |
        Vote: I like it +5 Vote: I do not like it

      ACM-ICPC rules is ranked by time and the solved number... What about the score?

      → Reply
»
10 years ago, # |
  Vote: I like it +79 Vote: I do not like it

"Are you in love with algorithms? If you are don't miss this round and be algorithm's valentine ;)
On the night before valentine's day (exact time), Valentine Algorithm cUp 2015 is going the take place."

→ Reply
  • »
    »
    10 years ago, # ^ |
      Vote: I like it +41 Vote: I do not like it

    When I saw title of the competition then one thing instantly came to my mind: "Somebody has posted Forever Alone guy in comments". :D

    → Reply
    • »
      »
      »
      10 years ago, # ^ |
        Vote: I like it +25 Vote: I do not like it

      Of course, so many upvotes are waiting for being first with posting it, so why not take them xD?

      → Reply
»
10 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I don't know if this is a bug or not but I can see the contest problems while I'm in coach mode..

→ Reply
  • »
    »
    10 years ago, # ^ |
      Vote: I like it +22 Vote: I do not like it

    It's a feature. Any coach-mode user can view every contest, even if it's pending. Just don't read the problems :)

    → Reply
    • »
      »
      »
      10 years ago, # ^ |
        Vote: I like it +1 Vote: I do not like it

      Okay. I wasn't going to participate in the contest anyway.

      But IMO I think it's better to add problems to such contests maybe about 30 minutes before the contest to seek maximum fairness.

      → Reply
»
10 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Can someone explain solution of sample problem? In documentation, it said that (x, y) replaces the "first occurrences" of x by y. So I don't understand how to use this to deletes all bs and converts all as to d. Thanks in advance.

→ Reply
  • »
    »
    10 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    (b,) replaces the first occurrences of b by an empty string, that means it deletes it.

    → Reply
    • »
      »
      »
      10 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      The question is why is (b,) deleting all strings and not just the first b.

      → Reply
    • »
      »
      »
      10 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      It just replaces "first" occurrence.

      → Reply
      • »
        »
        »
        »
        10 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Read its tutorial again, after deleting the first one, we return to the first line and start over.

        → Reply
»
10 years ago, # |
  Vote: I like it +3 Vote: I do not like it

It would be nice sample problem with answers in every (algorithm) language, is it possible??

→ Reply
  • »
    »
    10 years ago, # ^ |
      Vote: I like it +14 Vote: I do not like it

    Clearly a single problem that has a solution in all five languages is impossible (the input in Autolan, Cursle, and Prolan is a string, while the input in DIT and IDXT is a sequence of four integers). However, here are some easy problems and solutions:

    In Autolan, accept a string iff it has an even number of the letter a. (0 is even.)

    N 2
I 1
A 1 1
E 1 a 2
E 2 a 1

    In Cursle, replace the first occurrence of the letter a with the letter e. It is guaranteed that the string has at least one letter a.

    C a
T 2
T 4
A e
X
T 3
N
T -7

    In DIT/IDXT, move a number in storage 1 to storage 2. It is guaranteed that storage 2 is initially 0. You may leave anything in storage 1/3/4.

    D 1
T 3
I 2
T -3

    Prolan sample problem is given in the blog post.

    → Reply
»
10 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

This is different from Round #291, right?

→ Reply
  • »
    »
    10 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Yes, Valentine Algorithm cUp 2015 is not Codeforces Round #291 (Div. 2) :)

    → Reply
»
10 years ago, # |
  Vote: I like it +65 Vote: I do not like it

sometimes only u want to be a laptop

→ Reply
»
10 years ago, # |
← Rev. 3   Vote: I like it 0 Vote: I do not like it

As it is in Gym section, I guess it will not be rated. Am I ture?

→ Reply
»
10 years ago, # |
  Vote: I like it +3 Vote: I do not like it

Very unusual tasks! So maybe you have some chance to beat tourist in this contest? :P

Btw why we have 5 languages but have 6 tasks?

→ Reply
  • »
    »
    10 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    A language is used for two tasks, of course.

    Now, which languages makes double appearance? I'm betting on Cursle.

    → Reply
    • »
      »
      »
      10 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Your assumption was wrong, so give me a treat. Thanks :)

      → Reply
»
10 years ago, # |
  Vote: I like it 0 Vote: I do not like it

"T d : go to the line Current_line + d (d could be negative). Order : w(1)" In DIT and IDXT why is this command necessary when it is possible to change any storage directly?

→ Reply
  • »
    »
    10 years ago, # ^ |
    ← Rev. 3   Vote: I like it 0 Vote: I do not like it

    "In DIT, write a program to move a number from storage 1 to storage 2."

    D 1
T 3
I 2
T -3

    Now you see why T is necessary.

    → Reply
    • »
      »
      »
      10 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Oh! Damn. I confused line with storage. My bad! Thanks. :)

      → Reply
»
10 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Is this contest d be rated ?

this might be my chance to hit div 1 lol

→ Reply
»
10 years ago, # |
← Rev. 2   Vote: I like it +15 Vote: I do not like it

A bunch of questions. Italics are what I got from inspecting compiler, which means I ask for confirmation.

Autolan:

  • Documentation is mistaken; for the A instruction, it should be third line instead of second line, right? Actually compiler doesn't care.
  • The compiler (or actually interpreter) doesn't seem to enforce N,I,A instructions on the first, second, third instructions respectively.

Cursle:

  • After an A instruction, does the pointer point to the old character or the new character (the one added)? Old character.
  • When the pointer is at last character and N is called, it's a no-op right? Similar with pointer at first character and P is called. Yes.
  • What is the order of an A instruction? w(1).

Cursle, DIT, IDXT:

  • What happens if a T instruction sends the instruction pointer before the first line or after the last line of the program? Program ends.
  • What happens if a C instruction (Cursle), a D instruction (DIT and IDXT), or an X instruction (IDXT) sends the instruction pointer two lines down but it's apparently after the last line of the program? Program ends.

There might be more questions coming...

→ Reply
  • »
    »
    10 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    More question.

    In DIXT, what happens if in an X instruction, the value of b is a storage that points to the number 0? Compiler crashes, it seems.

    → Reply
»
10 years ago, # |
  Vote: I like it +3 Vote: I do not like it

For Cursle doesn't command A(x) have an order?

→ Reply
»
10 years ago, # |
  Vote: I like it 0 Vote: I do not like it

it seems that a "whitespaces" is wrote as "wjitespaces" :D

→ Reply
»
10 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Awesome idea for a contest. Looking forward to take part!

→ Reply
»
10 years ago, # |
  Vote: I like it +4 Vote: I do not like it

Can I participate in a team with my girlfriend?

→ Reply
»
10 years ago, # |
← Rev. 4   Vote: I like it 0 Vote: I do not like it

There seems to be a bug in the IDXT interpreter. When I compile it I get uninitialized variable warnings for a and b on lines 119 and 123. Additionally, the X operation does not work. To fix this, I added the lines a = cm[ln].y; b = cm[ln].z; on line 119. This should appropriately initialize a and b for the X operation.

For the Cursle interpreter, a better debugging output is: cout << "Curren string: "; for (int i = 0; i < a.size(); i++) cout << a[i]; for (int i = 0; i < b.size(); i++) cout << b[i]; cout << " ,Pointer position (0-indexed): " << a.size(); cout << " ,Current line: " << cur + 1 << endl;

→ Reply
»
10 years ago, # |
  Vote: I like it +21 Vote: I do not like it

Does Codeforces server's time utterly lag? It is 18.35 MSK and I still see "8 minutes to contest". (Also check the time left to CF 291; you can see the lag as well.)

→ Reply
»
10 years ago, # |
  Vote: I like it 0 Vote: I do not like it

This was fun.. Thanks :)

→ Reply
»
10 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I have a question in problem F.. It's to be solved using Prolan language and the max. order is 10^7. The order of the single valid command in Prolan is w(|s| + |x| + |y|)... If |s| = 2*10^100+1... How can we pass the given order or did I understand something wrong?

→ Reply
  • »
    »
    10 years ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    |s| = 2 * 100 + 1.

    → Reply
    • »
      »
      »
      10 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      oh... I can't believe I thought about it like that... Feeling very stupid now. Thanks :)

      → Reply
»
10 years ago, # |
← Rev. 2   Vote: I like it +14 Vote: I do not like it

the most dedicated candidate of today to get his valentine: ;)

→ Reply
»
10 years ago, # |
  Vote: I like it +15 Vote: I do not like it

Just because I'm bored, here's an incomplete, rough editorial, based on my solutions. I didn't solve C and F, and thus I don't have the solutions. Resemblance to the Codeforces platform is purely coincidental.

Problem A

The states are the number processed mod 31, plus 1 (because states begin with 1). For example, while processing 1234, you will come to state 2 (), 13 (), 31 (), and 26 (). Accept if you end up on state 1. From state i, make an edge for each j = 0, 1, ..., 9. I can score a 2-minute submission here just because I remembered how to do one with modulo 3.

Problem B

Note that we're basically counting the number of brackets at the outermost level. One approach to eliminate all inner brackets is ([[],[); remove the leftmost leaf at each time. Now we're left with several [], we can assume each [] increments the digit right before it:

(0[],1)
(1[],2)
(2[],3)
(3[],4)
(4[],5)
(5[],6)
(6[],7)
(7[],8)
(8[],9)
(9[],[]0)
([],1)

This is very similar to a former Codeforces Round problem.

Problem D

This is just finding GCD, which can be done by Euclidean algorithm. From a, b, 0, 0, we compute a - b, 0, b, 0 if b < a, otherwise 0, b - a, a, 0. (Possible by simply incrementing storage 3 and decrementing storage 1 and 2; whichever reaches zero faster is the smaller value.) Afterwards, move storage 3 to the empty location, giving a - b, b, 0, 0 or a, b - a, 0, 0. Repeating enough times, we will be left with a single number that is the GCD; move it to storage 1.

Problem E

We need to know how to divide two numbers. Given a and b, with an empty storage c, we can compute a / b and store it to c: decrement a until b|a, then increment c; rinse and repeat until a = 0.

Now label the storage a, b, c, d; we start with n on a and 0 on b, c, d. We will use d as the counter for final answer. Begin with taking b = 2, and test b|a; if that's the case, divide a by b to c, move c back to a, and increment d. Repeat until b doesn't divide a, in which we increment b. Repeat until a = 1 (can be tested by decrementing a twice), where we move d to a.

To see that this is correct, simply note that if b|a, b must be prime; if it's composite, then b = pq for some p, q ≥ 2, and we must have already divided a by p and q before as 2 ≤ p, q < b.

My solution is not that efficient, given that I actually tested whether b is prime first before trying to divide. (The above has omitted this step.) But hey, still works.

→ Reply
  • »
    »
    10 years ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    Aww, I read "are multiplies of" as "are multiples of" and didn't notice "greatest". So I was trying to compute LCM, and I'm not sure that's possible with just 4 registers. :(

    My A and E are exactly the same (up to permutations).

    Here is my B (yeah, I know, it's ugly): ([],X) (X],]) (XXXXXXXXXXX,YX) (XXXXXXXXXX,Y0) (XXXXXXXXX,9) (XXXXXXXX,8) (XXXXXXX,7) (XXXXXX,6) (XXXXX,5) (XXXX,4) (XXX,3) (XX,2) (X,1) (YYYYYYYYYYY,ZY) (YYYYYYYYYY,Z0) (YYYYYYYYY,9) (YYYYYYYY,8) (YYYYYYY,7) (YYYYYY,6) (YYYYY,5) (YYYY,4) (YYY,3) (YY,2) (Y,1) (ZZZZZZZZZ,9) (ZZZZZZZZ,8) (ZZZZZZZ,7) (ZZZZZZ,6) (ZZZZZ,5) (ZZZZ,4) (ZZZ,3) (ZZ,2) (Z,1)

    → Reply
    • »
      »
      »
      10 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Well, there's the sample tests. I also initially thought of LCM until I read the samples.

      → Reply
  • »
    »
    10 years ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    I'll go ahead and post my solution for C.

    Problem C

    This problem is just addition using Cursle. You can solve it by long addition. Keep track of a carry field at the end of the string, along with the answer you are building (Baaa+bbbEcCoooX, c: carry bit, o: out) Look at the carry bit, then check the last digit of b (call it i). Go to "state" b[i+c]. In each state b[i], check the last digit of a (call it j) and go to state a[i+j]. In state a[i], prepend i%10 to out, and set c = i/10. You also have to be aware when a and b become empty. Once both a and b are empty, delete the extra characters. Rather than keeping track of numerical offsets to hop around my states, I used assembly style labels, and wrote a short resolver to calculate the offsets.
    → Reply
    • »
      »
      »
      10 years ago, # ^ |
        Vote: I like it +5 Vote: I do not like it

      I also used long addition. Instead of putting the output after everything, I modify the value of a to be the output. Something like BaaaaC?oooo+bbbbE, where o is the output, b is what remains of the original after consuming the last few digits for output, and C is the carry. For some reason I failed in test 5; probably OLE but I'm not sure.

      → Reply
»
10 years ago, # |
  Vote: I like it +15 Vote: I do not like it

Thank you for very interesting problems! :) I participated in your previous round too and I have to say that I am looking forward to seeing more contests from you.

→ Reply