All problems were created and prepared by me and magnus.hegdahl. Thanks to BucketPotato, Monogon, and Priyam2k for valuable feedback and testing. Special thanks to MikeMirzayanov for coming up with the idea and preparing the unique checker for 1663H - Cross-Language Program. I hope you enjoyed the contest, and found it to be an interesting series of puzzles.

Example solution code written by Agnimandur is provided for each problem.

### 1663A - Who Tested?

**Clue 1**

The problem statement is the title of the problem.

**Clue 2**

Re-read the original announcement blog.

**Solution**

If you carefully read the announcement blog, you will notice that I especially thanked BucketPotato for testing. Just print "BucketPotato".

**Solution Code (Python)**

```
print("bucketpotato")
```

### 1663B - Mike's Sequence

**Clue 1**

Did you notice the peculiar constraints on the variable $$$r$$$?

**Clue 2**

"Mike" is of course MikeMirzayanov. "Mike's sequence" has to do with Codeforces ratings.

**Clue 3**

3000 is the cutoff for the highest rating: Legendary Grandmaster.

**Solution**

Mike's sequence are the Codeforces rating thresholds: [1200,1400,1600,1900,2100,2300,2400,2600,3000]. Your task is simply to print the threshold immediately above the provided rating $$$r$$$.

**Solution Code (Python)**

```
R = int(input())
ratings = [1200,1400,1600,1900,2100,2300,2400,2600,3000]
higher = [r for r in ratings if r > R]
print(higher[0])
```

### 1663C - Pōja Verdon

**Clue 1**

Aenar Targaryen was a dragonlord of Old Valyria.

**Clue 2**

The main clue is the title: "Pōja Verdon".

**Solution**

Aenar Targaryen was a dragonlord of Valyria (an empire from a *A Song of Ice and Fire*) who moved his family to Westeros right before Valyria was destroyed.

`pōja`

means `their`

and `verdon`

means `sum`

(or `amount`

based on context) in High Valyrian (you can easily translate this on Duolingo).

In a nutshell: Print the sum of the array.

**Solution Code (Python)**

```
n = int(input())
nums = list(map(int,input().strip().split()))
print(sum(nums))
```

### 1663D - Is it rated - 3

**Clue 1**

Did you notice that the problem had a special "constraints" section? That's highly unusual for a *Codeforces* problem.

**Clue 2**

The length of S is 3. Are there any famous three letter programming contests?

**Solution**

The capital variable letters and the "constraints" section are designed to mimic how atcoder.jp formats their problem statements.

The problem is asking whether the Atcoder contest (ABC,ARC,AGC) is rated for a rating $$$X$$$. (Note that the judge ignores Atcoder heuristic contests for simplicity).

- ABC's are rated for $$$[0,1999]$$$
- ARC's are rated for $$$[0,2799]$$$
- AGC's are rated for $$$[1200,\infty]$$$

**Solution Code (Python)**

```
s = input().strip()
x = int(input().strip())
if (s=="ABC" and x<2000) or (s=="ARC" and x<2800) or (s=="AGC" and x>=1200):
print('yes')
else:
print('no')
```

### 1663E - Are You Safe?

**Clue 1**

Do you recognize the story of the Minotaur?

**Clue 2**

The problem is a word search, but not for the word "minotaur".

**Solution**

Perhaps you recognize the story of the minotaur? Less known is the hero who saved the kids from the minotaur: his name was Theseus [thee-see-us].

The problem is just a word search (horizontal, vertical, diagonal) for `theseus`

. You should only go in the forward direction when you search for a word.

**Bonus Clues (for other problems)**

The grid contains other clues embedded in it. Perhaps you can find `esoteric`

, `java`

, and `script`

(hints for problem G). You might be able to find `atcoder`

(hint for problem D). Also `hero`

(hinting at Theseus for this problem).

There are also some red herring words out there.

**Solution Code (C++)**

```
#include <bits/stdc++.h>
using namespace std;
void good() {
cout << "YES";
exit(0);
}
int main() {
int N; cin >> N;
char grid[N][N];
for (int i = 0; i < N; i++) {
string row; cin >> row;
for (int j = 0; j < N; j++) grid[i][j] = row[j];
}
string word = "theseus";
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (i < N-7) {
string w = "";
for (int k = 0; k < 7; k++) w += grid[i+k][j];
if (w == word) good();
}
if (j < N-7) {
string w = "";
for (int k = 0; k < 7; k++) w += grid[i][j+k];
if (w == word) good();
}
if (i < N-7 && j < N-7) {
string w = "";
for (int k = 0; k < 7; k++) w += grid[i+k][j+k];
if (w == word) good();
}
}
}
cout << "NO";
}
```

### 1663F - In Every Generation...

**Clue 1**

Where is the quote from?

**Clue 2**

"buffy", "the", "vampire", "slayer"

**Clue 3**

"buffy" (length 5), "the" (length 3), "vampire" (length 7), "slayer" (length 6). Remember how $$$3 \le |s| \le 7$$$? Now go back to the sample and see what you can do.

**Solution**

The quote is from the well known TV series Buffy the Vampire Slayer.

Perhaps you notice the lengths of the words `buffy`

, `the`

, `vampire`

, `slayer`

are unique integers in the set {3,5,6,7}.

Given the string $$$s$$$, simply "add" that string with the target word of the same length.

Here's how adding two letters work:

- Convert both letters to numbers in $$$[0,25]$$$.
- Add the two numbers up modulo 26.
- Convert the number back into a letter.

You can verify your addition algorithm is correct by checking whether "tourist" + "vampire" = "ooggqjx" (as provided in the sample).

Finally, if the length of $$$s$$$ is $$$4$$$, the answer is "none", because there is no word of length 4 in Buffy the Vampire Slayer.

**Solution Code (Python)**

```
s = input().strip()
words = ["","","","the","","buffy","slayer","vampire"]
if len(s)==4:
print("none")
else:
target = words[len(s)]
ans = ""
for i in range(len(s)):
a = ord(s[i])-ord('a')
b = ord(target[i])-ord('a')
ans += chr((a+b)%26 + ord('a'))
print(ans)
```

### 1663G - Six Characters

**Clue 1**

Input is a normal string of letters, but output consists of non-letter *characters*?

**Clue 2**

Think of an esoteric programming language.

**Clue 3**

JSF (an esoteric programming language equivalent to Javascript that only uses 6 characters).

**Solution**

The April Fool's Contests love of esoteric programming languages, combined with the reference to "six characters", will eventually lead you to JSFuck.

The six characters is a pun: the input is a string of length 6 like "abcdef", but the output is a string \textit{consisting} of only 6 characters: `[ ] ( ) ! +`

.

Brief research will demonstrate a mapping from each english letter to a JSFuck string. Two letters can be concatenated with "+".Note that the optimal mapping will make your solution file approximately 12KB in size, well within the 64KB requirement of Codeforces.

However, the mapping from each letter to javascript is not unique! However, the legend mentions that Aenar needs to go to the string's "home". JSFuck's home website is here, which is where the jury's mapping is created from.

Alternatively, instead of hardcoding the generated strings, you can simply use the encoding algorithm provided in the Github.

**Solution Code (Python and Javascript)**

### 1663H - Cross-Language Program

**Solution**

Google research into how Pascal and C++ work will eventually lead you to a solution like this. Many people used the fact that `\`

can be used to continue single line comments to the next line in C++ but not in Pascal.

```
// \
begin end.
int main() {}
```

memeI did all possible word searches (normal, normal with diagonal, normal with diagonal with reverse, straight up grid BFS) for "MINOTAUR" and then for both "MINOTAUR" and "THESEUS" but didn't consider doing for just "THESEUS", fml

I did all possible word searches for

`theseus`

, got WA. Then I did only the forward-diagonal (i.e.`[i0+k][j0+k]`

) search and got AC. Meanwhile the editorial says you had to do horizontal and vertical searches too. Who's in the right?Test cases were a bit weak, you had to do horizontal, vertical, and diagonal.

IMO it is quite mean, that testcases had backwards-going "theseus" in them but were not considered safe. :D I checked for theseus first (but was also checking theseus backwards ord the other diagonal) but it only passed test 1. Then I checked for minotaur && !theseus as the unsafe condition and got to case 5.

same, smh

I printed "red panda" for problem A and got accepted???

Ha ha that's an easter egg :)

Problem A accepts either "bucketpotato" or "red panda" (or various similar combinations).

Is it normal that i wrote program writing "Red panda" for A?

what is meant by "AGC" in problem D?

See https://atcoder.jp/contests/

AGC -> AtCoder Grand Contest

Ah yes I totally forgot thanks!

## is it rated?

why did my rating change ._.

Bro it is rated! No way lol

What the actual f-k?

https://codeforces.com/contest/1663/submission/152390392

This gets AC in E.

At first I thought that the generator used the same rng, but surely it doesn't work like this, right?

Maybe there are like only ~6 tests so it's not too hard for random to pass.

E has 9 tests, so in theory there's a 1/512 chance of this happening lol.

Rejudge and see how lucky they really are lmao

Actually, if I recall correctly, CF re-runs your submission two more times in case of getting wrong answer. So the probability of failing all three attempts is $$$\frac{1}{8}$$$, probability of guessing answer in one test case is $$$\frac{7}{8}$$$, while you get AC with $$$\left(\frac{7}{8}\right)^9 \approx 0.3$$$ probability. Really worth a try!

UPD: It seems that I confused WA and TLE solutions. While it is rather a well-known fact that CF re-runs TLE solutions, it is very strange to do it also with WA ones. Does CF really do it? Is it right to behave so?

CF doesn't seem to do it, because if codeforces actually did it and the chance of bypassing all 9 test with a random solution was 0.3, then with 10 tries i should get AC with 97.8% probability, but i don't. That only can mean that CF make only one attempt on failure or that i'm very unlucky :)

You should try playing poker.

You have chance to pass any problem if you are lucky. Just pring random strings.

can anyone tell me why I am getting WA in E? I did just like in editorial :(. My submission

wtf bro

The problem was that you accessed out-of-bounds memory; I have corrected your solution and it now gets AC: 152390393.

It is RATED! WTF???

I bet the rating changes will be reverted tomorrow :)

It's obvious, but still very unexpected

problem G

`Hard-coding strings`

solution code isn't availableI guess the rating changes are only for today .

Another Solution of HAre the rating updates an April Fool's joke?

Submission Any particular reason why this code for C wasn't accepted?

Because you wrote ans=+x Instead of ans+=x

T_T

Did it work?

yes T_T

WOW

Well, it is rated :D

I write my user name in A get wrong Why

The answer is either "BucketPotato" or "Red Panda"

During the contest, after studying Pascal for an hour, I found a clumsy solution for H.

codeI wonder what I was doing :<

How can I become a expert? It is so hard!!!!!!!!!! T_T

Now you are a expert lol

Huchung huchung func sunc