ABC006C

https://gyazo.com/2f6b59dcf7787876c9b5f7e08d21cb47

Thoughts.

Assuming they are all 2, the problem is to make up a maximum of N combinations of +1 or +2 for the extra number.

A naive double loop will not make it in time.

If the remainder is less than or equal to N, just give that number of 1's.

If the remainder is more than N, we can make them all 3 and assign the remainder to 4.

Constant Order

Official Explanation

It's kind of a totally different story.

If you decide on one, you can do a Tsurugame count, so you do a full search for one, or you can find out that the old man is 0/1, so you do a full search for the rest.

AC with the above solution without searching.

code:python

def solve(N, M):

IMPOSSIBLE = (-1, -1, -1)

x = M - 2 * N

if x < 0:

return IMPOSSIBLE

if x <= N:

return (N - x, x, 0)

x -= N

if x <= N:

return (0, N - x, x)

return IMPOSSIBLE

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