https://r.recruit-jinji.jp/code_fes/us/index.html

Details will be announced in early August. In Japanese website, it says there will be 20 slots for international students.

According to Japanese website, the finals will be 25th-26th in November.

# | User | Rating |
---|---|---|

1 | tourist | 3602 |

2 | LHiC | 3287 |

3 | Radewoosh | 3266 |

4 | W4yneb0t | 3181 |

5 | TakanashiRikka | 3178 |

6 | moejy0viiiiiv | 3122 |

7 | izrak | 3109 |

8 | anta | 3105 |

8 | Um_nik | 3105 |

8 | ershov.stanislav | 3105 |

# | User | Contrib. |
---|---|---|

1 | rng_58 | 178 |

2 | csacademy | 167 |

3 | tourist | 165 |

4 | Petr | 164 |

4 | Errichto | 164 |

6 | Swistakk | 158 |

7 | matthew99 | 152 |

8 | Zlobober | 151 |

9 | zscoder | 143 |

10 | GlebsHP | 137 |

https://r.recruit-jinji.jp/code_fes/us/index.html

Details will be announced in early August. In Japanese website, it says there will be 20 slots for international students.

According to Japanese website, the finals will be 25th-26th in November.

AtCoder Grand Contest 018 will be held on Sunday (time). The writer is maroonrk.

The point values will be 300 — 700 — 800 — 1100 — 1600 — 1700.

Let's discuss problems after the contest.

Recently I solved an interesting constructive task. Can you solve it? I'll reveal the source a bit later.

*N*(*N* + 1) soccer players stand in a row. The height of the *i*-th player is *h*_{i}. No two of them are of the same height.

Sir Alex wants to remove *N*(*N*−1) players from this row leaving a new row of 2*N* players in which the following *N* conditions hold:

- No one stands between the two tallest players,
- No one stands between the third and fourth tallest players,
- No one stands between the two shortest players.

Find one possible way to achieve this.

Constraints: 2 ≤ *N* ≤ 500, 1 ≤ *h*_{i} ≤ 10^{9}, *h*_{i} are pairwise distinct.

I couldn't find a place to discuss this.

When will we get results? Is anyone confident with B or F?

AtCoder Grand Contest 017 will be held on Sunday (time). The writer is semiexp.

The point values will be 200 — 400 — 1000 (500) — 1100 — 1200 — 1600.

Let's discuss problems after the contest.

AtCoder Grand Contest 016 will be held on Sunday (time). The writer is sugim48.

The point values will be 300 — 700 — 700 — 1000 — 1400 — 1600.

Let's discuss problems after the contest.

**UPD**: Now the editorial (check page 5) is ready.

I'll add some quick comments here.

A.

Try all characters from 'a' to 'z', and assume that we want to get a string that consists only of this character. What greedy strategy should we follow?

B.

Suppose that there are *C* colors in total. Then, each *a*_{i} must be either *C* or *C* - 1, and it means that the difference between the maximum and the minimum is at most one. Do some case-analysis using this fact.

C.

Suppose that *H* is not a multiple of *h*. Put a positive number in a cell if its row-number (0-based) is a multiple of *h*, otherwise put a negative number. Can you choose the "positive number" and the "negative number" properly?

D.

Create a new cell that keeps the xor of all elements. Each operation can be considered as a swap between this cell and another element in the sequence. Reduce it to a graph problem.

E.

Can you get YES/NO for a given pair of vertices (or more generally, a given set of vertices) in *O*(*M*)? Restate it in terms of graphs, then find an *O*(*NM* + *N*^{3}) solution.

F.

The task is about grundy numbers. Can you get *O*(*bell*_{number}(*N*)) solution by trying all possible grundy-number sequences? Can you reduce it to *O*(3^{N}) by DP?

```
x = 0;
while(x < 1){
y = x;
x += rand(); // returns a real number between 0 and 1, uniformly at random
}
```

What is the expected value of y?

It's not very hard, but the result is surprising.

CF Handle | Country |
---|---|

1. kevinsogo | Philippines |

2. tourist | Belarus |

3. vepifanov (7 years in a row!) | Russia |

4. marcin_smu | Poland |

5. simonlindholm | Sweden |

6. mnbvmar | Poland |

7. Endagorion | Russia |

8. drinkless | Russia |

9. -XraY- | Russia |

10. zemen | Russia |

11. semiexp | Japan |

12. Snuke | Japan |

13. SpyCheese | Russia |

14. TeaPot | Russia |

15. SnapDragon | Canada |

16. Zlobober | Russia |

17. moejy0viiiiiv | China |

18. s-quark | China |

19. hogloid | Japan |

20. Egor | Russia |

21. KAN | Russia |

22. Ahyangyi | China |

23. rng_58 | Japan |

24. LHiC | Russia |

25. ifsmirnov | Russia |

26. ivan.popelyshev | Russia |

27. Lewin | United States |

28. PavelKunyavskiy | Russia |

29. aitch | Sweden |

This year I haven't decided whether I'll participate (the flight won't be paid). Anyway, here is the top 25:

CF Handle | Trip to Moscow |
---|---|

1. tourist | Easy |

2. Um_nik | Easy |

3. rng_58 | Hard |

4. krijgertje | Medium |

5. W4yneb0t | Medium |

6. Milanin | Easy |

7. Petr | Easy |

8. dreamoon | Hard |

9. Eryx | Medium |

10. LHiC | Easy |

11. Alex_2oo8 | Easy |

12. moejy0viiiiiv | Hard |

13. aid | Easy |

14. I_love_Tanya_Romanova | Easy |

15. jcccccccccccccccccccccsb | Hard |

16. fqw | Hard |

17. shik | Hard |

18. izban | Easy |

19. maksay | Easy |

20. ilyakor | Easy |

21. Swistakk | Medium |

22. snuke | Hard |

23. Miceren | Hard |

24. mnbvmar | Medium |

25. KAN | Easy |

AtCoder Grand Contest 015 will be held on Saturday (time). The writer is DEGwer.

The point values will be announced later.

Let's discuss problems after the contest.

By the way, the writer is currently flying.

**1. Tsinghua** (Win: 40%, Gold: 90%, Medal: 100% after some rounding)

- CF max: 3441, 3203, 2967 (agg = 3516)
- Open Cup: 1, 1, 11, 9 (median = 5)
- China Final: 1
- MIPT: 1, 1, 1, 2, 4, 1, 4

**2. Warsaw** (Win: 15%, Gold: 75%, Medal: 95%)

- CF max: 3122, 2962, 2826 (agg = 3237)
- Open Cup: 2, 3 (median = 2.5)
- Petrozavodsk: 4, 2, 4, 3, 2, 2, 1, 2

**3. SPb ITMO** (Win: 15%, Gold: 50%, Medal: 90%)

- CF max: 2880, 2870, 2860 (agg = 3104)
- Open Cup: 7, 3, 2, 1, 13, 1, 2, 1, 3, 7, 13, 8, 8, 12, 3, 2, 1, 12 (median = 3)
- Petrozavodsk: 4, 1, 3, 1, 6, 5, 3, 7, 5
- MIPT: 6, 2, 3, 1, 5, 2, 1

**4. SPb SU** (Win: 10%, Gold: 40%, Medal: 85%)

- CF max: 3170, 2951, 2936 (agg = 3281)
- Open Cup: 9, 21, 18, 5, 3, 4, 9, 9, 6, 1, 2, 2, 7, 1, 4, 23, 20, 4 (median = 5.5)
- Petrozavodsk: 7, 3, 6, 5, 8, 1, 4, 5, 1
- MIPT: 2, 4, 2, 3, 1, 10, 8

**5. Seoul** (Win: 5%, Gold: 40%, Medal: 85%)

- CF max: 3174, 2919, 2760 (agg = 3250)
- Open Cup: 4, failed, 4 (median = 4)

**6. MIPT** (Win: 5%, Gold: 30%, Medal: 75%)

- CF max: 2928, 2757, 2756 (agg = 3063)
- Open Cup: 16, 29, 27, 28, 7, 24, 17, 15, 9, 15, 17, 3, 9, 4, 10 (median = 15)
- Petrozavodsk: 1, 5, 5, 5, 1, 6, 6, 1
- MIPT: 3, 6, 5, 4, 6, 3

**7. MIT** (Win: 5%, Gold: 30%, Medal: 75%)

- CF max: 2900, 2637, 2540 (agg = 2981)
- Open Cup: Equivalent to 2.5 in NAIPC

**8. KTH** (Gold: 10%, Medal: 60%)

- CF max: 2761, 2637, 2273 (agg = 2870)
- Open Cup: 2, 17, failed, 7 (median = 12)
- MIPT: 5, 5, 16, 7, 7, 2

**9. Fudan** (Gold: 5%, Medal: 55%)

- CF max: 2543, 2367, 2327 (agg = 2667)
- Open Cup: 6, 13, 9, 6, 13, 18 (median = 11)
- China Final: 5
- MIPT: 12, 3, 7, 15, 13, 4, 5

**10. New South Wales** (Gold: 5%, Medal: 50%)

- CF max: 3073, 2429, 2006 (agg = 3082)
- Open Cup: 6, 8, 16 (median = 8)
- MIPT: 14, 9, 10, 16, 2, 5, 7

**11. Tokyo** (Medal: 50%)

- CF max: 2762, 2694, 2424 (agg = 2903)
- Open Cup: 29, 5, 10, 12, 14, 29 (median = 13)
- MIPT: 11, 10, 6, 8, 15, 6, 6
- Barcelona: 1, 1, 1, 2, 3, 1, 1

**12. SJTU** (Medal: 50%)

- CF max: 2707, 2609, 2582 (agg = 2874)
- Open Cup: 26, 3, 18, failed, 21, 12, 10, 20, 26, 14 (median = 19)
- China Final: 2
- MIPT: 8, 7, 4, 10, 3, 9, 10

I won't be surprised if the following teams get a medal:

**13. NTU** (Medal: 45%)

- CF max: 2772, 2341, 2220 (agg = 2808)
- Open Cup: 11, 15, 21, 10, 2 (median = 11)

**14. Waterloo** (Medal: 45%)

- CF max: 2925, 2908, 2703 (agg = 3100)
- Open Cup: Equivalent to 12.5 in NAIPC
- Barcelona: 3, 2, 4, 1, 1, 3, 2

**15. Perm SU** (Medal: 40%)

- CF max: 3037, 2850, 2578 (agg = 3127)
- Open Cup: 6, 18, failed, 3, failed, 9, 18, 7, 24, 15, 14, 16, 16, 7, 21, 16, 21, 17 (median = 16)
- Petrozavodsk: 8, 12, 10, 14, 10, 10, 2, 6, 6
- MIPT: 4, 8, 11, 5, 20, 7, 9

**16. Peking** (Medal: 40%)

- CF max: 2920, 2566, 2338 (agg = 2965)
- Open Cup: 21, 3 (median = 12)
- China Final: 6

**17. Helsinki** (Medal: 35%)

- CF max: 2721, 2280, 1516 (agg = 2745)
- Open Cup: 15, failed, 9, failed, 16, 4, 17, 11, 30, 18, 2, failed, 8 (median = 16)
- Petrozavodsk: 3, 26, 25, 11, 11, 7, 10, 2, 9
- MIPT: 10, 16, 14, 6, 23, 17, 3

**18. BSUIR** (Medal: 30%)

- CF max: 2825, 2625, 2466 (agg = 2943)
- MIPT: 7, 13, 8, 11, 9, 13, 12

===

- Expected win of top 7 = 0.95
- Expected gold of top 10 = 3.75
- Expected medal of top 18 = 11.05

And I guess Tokyo Institute of Technology will be around 25th, Keio will be around 40th.

AtCoder Grand Contest 014 will be held on Saturday (time). The writer is yutaka1999.

The point values are 300 — 500 — 700 — 900 — 1400 — 2400. Note that the contest duration is unusual (130 minutes).

Let's discuss problems after the contest.

Let's discuss problems.

B is very nice. Until the end of the contest, I thought it was a marathon task. Then yutaka1999 told me the following: for an odd prime *p*, *a*[*p*] = 2 - *p*%4.

And I like geometry tasks like J.

There are lots of important tournaments in this season, and we moved some AtCoder contests to avoid collision. Please check https://atcoder.jp/.

We'll add more ARC/ABCs in May.

AtCoder Grand Contest 013 will be held on Saturday (time). The writer is maroonrk.

The point values are 300 — 500 — 700 — 900 — 1600 — 2000. Note that the contest duration is unusual (150 minutes).

Let's discuss problems after the contest.

AtCoder Grand Contest 012 will be held on Saturday (time). The writer is camypaper.

The point values are 300 — 700(200) — 1000 — 1000 — 1000 — 2000.

Let's discuss problems after the contest.

Let's discuss problems.

Does anyone have an idea for D?

There will be ARC 070 this Saturday, and interactive problems may be used there.

You can try an example of interactive problem on AtCoder here: https://practice.contest.atcoder.jp/tasks/practice_2

Here is my solution for 100 points:

```
#include <fstream>
#include <string>
using namespace std;
int main(void){
int N,Q,i,j;
scanf("%d%d", &N, &Q);
string s;
for(i=0;i<N;i++) s += (char)('A' + i);
for(i=0;i<N;i++) for(j=0;j<N-1;j++){
printf("? %c %c\n", s[j], s[j+1]);
fflush(stdout);
char ans;
scanf(" %c", &ans);
if(ans == '>') swap(s[j], s[j+1]);
}
printf("! %s\n", s.c_str());
fflush(stdout);
return 0;
}
```

AtCoder Grand Contest 011 will be held on Sunday (time). The writer is semiexp.

The point values are 300 — 400 — 800 — 900 — 1300 — 1700 (?).

Let's discuss problems after the contest.

Mujin Programming Challenge 2017 will be held on February 25th, 21:00 — 23:00 JST.

This is an online contest, and the top 30 people will be awarded prizes. The prize for the first place is $2000.

Please check the official site for details.

You will be asked to fill a questionnaire when you register, so please register at least several minutes before the contest.

UPD: Point values are 900(500) — 1300(300) — 1300 — 1800. The full scores correspond to C, D, E, F in AGC, and partial scores correspond to B, A.

If you are a contestant, you can be relaxed and you can do anything (except for cheating). It's perfectly fine if you just guess the solution and submit it without knowing why (though personally I don't find it very beautiful). However, if you are a writer, you need to prove your solution. Here is the list of things you have to prove:

Does your solution always return correct answers for all possible valid inputs?

- GOOD: Strict proof.
- BAD:
*My intuition tells that this is correct!* - BAD:
*I tried really hard to come up with counterexamples, but I couldn't. It must be correct!*

Does your solution always run in time for all possible valid inputs?

- GOOD: It's
*O*(*n*^{2}) and the constraints say*n*≤ 1000. It should work. - GOOD: For this problem we can prove that the slowest case is xxx. Experimentally, my solution works for the input xxx under the given TL.
- BAD:
*I tried really hard to generate various testcases, and my solution passed all cases!*

Randomized algorithms are not hackish ways of solving problems. The writers should prove that for any valid input, the intended solution works correctly with very high probability. Please check here for an example of such proof. Another example is Rolling Hash: please check here. Note that, for example when we compute 10^{5} hashes for strings of lengths 10^{5}, you need four hashes of prime modulo around 10^{9}, not two. (Practically two works but we can't prove that).

Especially in geometry problems, we sometimes use epsilons. However writers should be careful about the use of epsilons.

- GOOD: We know |
*b*| and |*d*| are up to 10^{4}. Let's use*eps*= 10^{ - 9}for comparing two fractions*a*/*b*and*c*/*d*. - BAD:
*Let's use**eps*= 10^{ - 9}! (without reasons)

When the intended solution uses complicated double operations (like sqrt, trigonometry, log, lots of fractions, etc.) such analysis may be hard. In this case, one possible way is to add constraints like "even if we move a point by up to 10^{ - 3}, the answer doesn't change".

AtCoder Grand Contest 010 will be held on Saturday (time). The writer is yutaka1999.

The point values are 300 — 500 — 700 — 1000 — 1600 — 1600.

Let's discuss problems after the contest.

My thoughts on yesterday's problem D and hashing.

http://rng-58.blogspot.jp/2017/02/hashing-and-probability-of-collision.html

Codeforces (c) Copyright 2010-2017 Mike Mirzayanov

The only programming contests Web 2.0 platform

Server time: Aug/21/2017 07:28:31 (c2).

Desktop version, switch to mobile version.

User lists

Name |
---|