Cubes

Cubes for Masha\href{https://codeforces.com/contest/887/problem/B}{\text{Cubes for Masha}}

Can anyone clarify it to me what the problem is actually asking?

#ComputerScience

Note by Real Sakib25
7 months, 2 weeks ago

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Comments

Look at the input given:

1
2
3
4
3
0 1 2 3 4 5
6 7 8 9 0 1
2 3 4 5 6 7

This means you got 3 cubes, each of these cubes whose faces are given to be the 3 sextuplets of integers.

Now's let's form all the non-negative integers starting 0 by using each of the die at most once to form the digits.

Can we form the number 0 from these 3 dice? Yes.
Can we form the number 1 from these 3 dice? Yes.
Can we form the number 2 from these 3 dice? Yes.
Can we form the number 3 from these 3 dice? Yes.
\vdots
Can we form the number 85 from these 3 dice? Yes.
Can we form the number 86 from these 3 dice? Yes.
Can we form the number 87 from these 3 dice? Yes.
Can we form the number 88 from these 3 dice? No. (Because there's no 2 dice that has a face of 8)

So the answer is 87.

Pi Han Goh - 7 months, 2 weeks ago

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Shouldn't it be like this?

  • Can we form the number 87 from these 3 dice? Yes.

  • Can we form the number 88 from these 3 dice? No. (No two cubes with digit 8)

So the answer is 87.

Real Sakib25 - 7 months, 2 weeks ago

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You're right. I've updated my text.

Pi Han Goh - 7 months, 2 weeks ago

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@Pi Han Goh Thank you

Real Sakib25 - 7 months, 2 weeks ago

@Nazmus Sakib, That is, there are n dice, with a number of 0-9 on each face. Find the longest consecutive number in the group, you can not use all the dice. For example, there is only one 8 in example 1, so 88 cannot be combined, and 1 -87 can be combined, all answers are 87, this is a water problem, n range is 1-3, direct violence is enough, save all combinations into an array, and then traverse to find out the number that cannot be combined.

Half pass3 - 7 months, 2 weeks ago
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