CSES - Datatähti 2025 alku - Results
Submission details
Task:Kortit II
Sender:rottis
Submission time:2024-11-01 23:36:50 +0200
Language:Ruby
Status:READY
Result:0
Feedback
groupverdictscore
#10
#20
#30
#40
#50
Test results
testverdicttimegroup
#10.07 s1, 2, 3, 4, 5details
#20.45 s2, 3, 4, 5details
#3--3, 4, 5details
#4--4, 5details
#5--5details
#6--5details

Code

mod_by = 10**9 + 7

def factorial(n)
  (1..n).reduce(1, :*)
end

def choose(a, b)
  # a choose b
  (factorial(a) / (factorial(b) * factorial(a-b))).to_i
end

def recursive_get_ways(chosen_nums, choices, p1_wins)
  if choices.empty?
    p1_wins_calc = 0
    chosen_nums.each_with_index do |num, ind|
      if num > ind + 1
        p1_wins_calc += 1
      end
    end
    if p1_wins == p1_wins_calc
#      p "found!"
#      p chosen_nums
      return 1
    else
      return 0
    end

  else
    total = 0
    choices.each do |choice|
      # optimize by avoiding draws
      if choice == chosen_nums.length + 1
        next
      end
      total += recursive_get_ways(chosen_nums + [choice], choices - [choice], p1_wins)
    end
    return total
  end
end

row_count = gets.chomp.to_i
rows = []
row_count.times do
  rows.append(gets.chomp.split(" ").map(&:to_i))
end

rows.each do |row|
  n = row[0]
  original_n = n
  p1_wins = row[1]
  p2_wins = row[2]
  p1_moves = []
  p2_moves = []

  if p1_wins + p2_wins > n
    puts(0)
    next
  end

  draws = n - p1_wins - p2_wins
  
  # we need to choose "draws" from n
  # ways_to_draw = n choose draws
  # this gives us the number of possible choices of numbers to draw with
  # in the end you should multiply this with ways_to_choose_other_numbers and multiply it by its factorial (unique orderings of the solution)
  ways_to_choose_drawing_numbers = choose(n, draws)
  
  n -= draws
  
  if (p1_wins + p2_wins > 0) && (p1_wins == 0 || p2_wins == 0)
    puts(0)
    next
  end

  ways_to_choose_other_numbers = recursive_get_ways([], (0...n).to_a, p1_wins)

  # code here
  # note: we can fix p1 moves to be [0...n] since the permutations of the final moves are already accounted for
  # brute-force: n! (too slow. 24000 for first part and others are no easier)
  # too many....
  # brute force not enough
  # we can (maybe) assume the first moves[p2_wins] are p2 wins and the last moves[p1_wins] are p1 wins
  # 
  
  puts(
    (ways_to_choose_drawing_numbers * ways_to_choose_other_numbers * factorial(original_n)) % (mod_by)
  )
end

Test details

Test 1

Group: 1, 2, 3, 4, 5

Verdict:

input
54
4 4 0
3 1 3
3 2 2
4 0 4
...

correct output
0
0
0
0
0
...

user output
0
0
0
0
0
...

Test 2

Group: 2, 3, 4, 5

Verdict:

input
284
6 1 0
5 0 2
7 1 5
7 7 5
...

correct output
0
0
35280
0
36720
...

user output
0
0
1834560
0
131040
...

Test 3

Group: 3, 4, 5

Verdict:

input
841
19 3 12
19 19 13
19 7 13
20 11 15
...

correct output
40291066
0
0
0
0
...

user output
(empty)

Test 4

Group: 4, 5

Verdict:

input
1000
15 12 6
7 1 6
44 4 26
6 6 5
...

correct output
0
5040
494558320
0
340694548
...

user output
(empty)

Test 5

Group: 5

Verdict:

input
1000
892 638 599
966 429 655
1353 576 1140
1403 381 910
...

correct output
0
0
0
249098285
0
...

user output
(empty)

Test 6

Group: 5

Verdict:

input
1000
2000 1107 508
2000 1372 249
2000 588 65
2000 1739 78
...

correct output
750840601
678722180
744501884
159164549
868115056
...

user output
(empty)