https://www.codingame.com/training/easy/count-as-i-count
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Created by @Magicien-d-oz,validated by @bbb000bbbyyy.
If you have any issues, feel free to ping them.
They don’t have forum accounts.
TBF, they barely have CG accounts at all: they only solve community puzzles in the exact same language as the puzzle author.
CG’s been notified last time already, but have apparently chosen not to act on it.
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Thank you for the gentle reminder.
Indeed, those two accounts are fake accounts. I’ve tried to investigate and understand from which regular account they could come, without result for now.
I’ve removed their moderation rights. The approvals they’ve done can’t be easily undone though.
4 Likes
It seems I misunderstand the calculation logic behind this puzzle as I fail all test below 47.
For me the (bad) solutions are
45 : 38
46 : 17
47 : 7
48 : 3
49 : 1
50 : 1
What is the right solution for 45 and 46? Maybe that would help to see where do I go wrong…
Thanks.
45 -> 40, the rest is correct.
Thanks, this helped a lot. It turned out that my implementation of the “max-4-rounds” rule was faulty.
Hello,
I’ve been stuck on the puzzle Count as I count for two days. I get the good result for tests #1 (47) and #3 (49) but I get wrong results (way too high) as soon as the score to reach is a little high. Can anyone please help me ? I’ve tried to mix loops and recursion and I think I’m on the right path, but I’m not going anywhere with my code. Thank you.
Hi everyone,
my code solve problem till to 39, but when i try to solve 38 I found an incorrect result.
These my permutations:
(follow this rule:
X|Y|Z are the pins fallen (permutations) –
[X,Y+40] is the Y pins fallen altogether (permutations of X,Y+40),
[X+40,Y] is the X pins fallen (permutations of X+40,Y) and so on.
==> (sum of permutations)
1|11 (2) – [1,51] (2) ==> (4)
1|1|10 (3) – [1,1,50] (3) ==> (6)
1|2|9 (6) – [1,42,9] (6),[1,42,49] (6),[1,2,49] (6) ==> (24)
1|3|8 (6) – [1,43,8] (6),[1,43,48] (6),[1,3,48] (6) ==> (24)
1|4|7 (6) – [1,44,7] (6),[1,44,47] (6),[1,4,47] (6) ==> (24)
1|5|6 (6) – [1,45,6] (6),[1,45,46] (6),[1,5,46] (6) ==> (24)
1|1|1|9 (4) – [1,1,1,49] (4) ==> (8)
1|1|2|8 (12) – [1,1,42,8] (12),[1,1,42,48] (12),[1,1,2,48] (12) ==> (48)
1|1|3|7 (12) – [1,1,43,7] (12),[1,1,43,47] (12),[1,1,3,47] (12) ==> (48)
1|1|4|6 (12) – [1,1,44,6] (12),[1,1,44,46] (12),[1,1,4,46] (12) ==> (48)
1|1|5|5 (6) – [1,1,45,5] (12),[1,1,45,45] (6) ==> (24)
1|2|2|7 (12) – [1,42,2,7] (24),[1,42,42,7] (12),[1,42,42,47] (12),[1,2,42,47] (24),[1,2,2,47] (12) ==> (96)
1|2|3|6 (24) – [1,42,3,6] (24),[1,42,43,6] (24),[1,42,43,46] (24),[1,2,43,6] (24),[1,2,43,46] (24),[1,2,3,46] (24) ==> (168)
1|2|4|5 (24) – [1,42,4,5] (24),[1,42,44,5] (24),[1,42,44,45] (24),[1,2,44,5] (24),[1,2,44,45] (24),[1,2,4,45] (24) ==> (168)
1|3|3|5 (12) – [1,43,3,5] (24),[1,43,43,5] (12),[1,43,43,45] (12),[1,3,43,45] (24),[1,3,3,45] (12) ==> (96)
1|3|4|4 (12) – [1,43,4,4] (12),[1,43,44,4] (24),[1,43,44,44] (12),[1,3,44,4] (24),[1,3,44,44] (12) ==> (96)
2|10 (2) – [42,10] (2),[42,50] (2),[2,50] (2) ==> (8)
2|2|8 (3) – [42,2,8] (6),[42,42,8] (3),[42,42,48] (3),[2,42,48] (6),[2,2,48] (3) ==> (24)
2|3|7 (6) – [42,3,7] (6),[42,43,7] (6),[42,43,47] (6),[2,43,7] (6),[2,43,47] (6),[2,3,47] (6) ==> (42)
2|4|6 (6) – [42,4,6] (6),[42,44,6] (6),[42,44,46] (6),[2,44,6] (6),[2,44,46] (6),[2,4,46] (6) ==> (42)
2|5|5 (3) – [42,5,5] (3),[42,45,5] (6),[42,45,45] (3),[2,45,5] (6),[2,45,45] (3) ==> (24)
2|2|2|6 (4) – [42,2,2,6] (12),[42,42,2,6] (12),[42,42,42,6] (4),[42,42,42,46] (4),[2,42,42,46] (12),[2,2,42,46] (12),[2,2,2,46] (4) ==> (64)
2|2|3|5 (12) – [42,2,3,5] (24),[42,42,3,5] (12),[42,42,43,5] (12),[42,42,43,45] (12),[2,42,43,5] (24),[2,42,43,45] (24),[2,2,43,5] (12),[2,2,43,45] (12),[2,2,3,45] (12) ==> (156)
2|2|4|4 (6) – [42,2,4,4] (12),[42,42,4,4] (6),[42,42,44,4] (12),[42,42,44,44] (6),[2,42,44,4] (24),[2,42,44,44] (12),[2,2,44,4] (12),[2,2,44,44] (6) ==> (96)
2|3|3|4 (12) – [42,3,3,4] (12),[42,43,3,4] (24),[42,43,43,4] (12),[42,43,43,44] (12),[2,43,3,4] (24),[2,43,43,4] (12),[2,43,43,44] (12),[2,3,43,44] (24),[2,3,3,44] (12) ==> (156)
3|9 (2) – [43,9] (2),[43,49] (2),[3,49] (2) ==> (8)
3|3|6 (3) – [43,3,6] (6),[43,43,6] (3),[43,43,46] (3),[3,43,46] (6),[3,3,46] (3) ==> (24)
3|4|5 (6) – [43,4,5] (6),[43,44,5] (6),[43,44,45] (6),[3,44,5] (6),[3,44,45] (6),[3,4,45] (6) ==> (42)
3|3|3|3 (1) – [43,3,3,3] (4),[43,43,3,3] (6),[43,43,43,3] (4),[43,43,43,43] (1) ==> (16)
4|8 (2) – [44,8] (2),[44,48] (2),[4,48] (2) ==> (8)
4|4|4 (1) – [44,4,4] (3),[44,44,4] (3),[44,44,44] (1) ==> (8)
5|7 (2) – [45,7] (2),[45,47] (2),[5,47] (2) ==> (8)
6|6 (1) – [46,6] (2),[46,46] (1) ==> (4)
12 (1) – [52] (1) ==> (2)
The total sum is 1638 instead of 1776.
i cannot understand what miss in my solution.
I don’t understand the logic of the exercice. Why in the exemple, it is possible to have “P1 P1 P1” and not “1 1 1”?
3 Likes
Because if you have 1 it meen you dropped one pin but if you dropped one pin you need to take the number on it. So you can’t have 1.
I haven’t the slightest idea how to do this one 
1 Like
Is there a maximum number of pins? For example if the initial score is 0 can I knock down 50 pins?
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There is an impossible test case for this one, I didn’t see it.
Unclear description. What does it mean number without P? “when the number is not accompanied it means the number of keels dropped” - and what is the sum of their numbers?
The sum of their numbers doesn’t matter, if you hit 3 of them you get 3 points.
The value of a keel only matters when it is the only one being hit.
1 Like
I don’t know how I didn’t see 
“-If a player knocks down more than one pin, he marks the number of pins knocked down”
Ok, thank you
Just curious if anyone can do it with equation other than while for or recursive
I doubt it is possible to find a closed-form solution for the very general case.
However, we can find some formulas. 4 months ago, I published my solution (in python3) where, too bored to write a boring recursive solution, I tried to find such a formula. I almost succeeded. (but this is ugly and uncommented).
First, I ignore the different counting with one or more pins to get a more elegant exercise. So there is exactly one possibility to score each value between 0 and N (I start from 0 because it simplify a lot the calculus)
The function Combin(N, k, n) gives the number of possibilities to score n within k rounds with possible scoring ranging from 0 to N values. As you can see, for k fixed, it is possible to find a closed-form formula, which can be generalized quite easily for all k via sums and products (but this is no more closed-form)
The case disjunction in the print is to catch up with the different scoring with 1 or more pins.
To find out how I came up with these formulas, take N infinite (the problem become much more simpler). Then this is the same method with N finite, but this is a bit more involved.
4 Likes
yea i was trying to come up with a single formula based on “n pick r” formulas but in the end using loops/recursion didn’t timeout so took the lazy way out. then again this is a programming site and not a discrete math course.
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