HHKB2020

I made a calculation error in D, which looks easy at first glance, and got into it and solved time was 3 questions. I should have skipped it and done E.

https://gyazo.com/96fdb60be9b5c7f4c36b92c975424089

https://gyazo.com/9d6d93ac3d485534cc5a6bed6d31b23f

https://gyazo.com/d25766b2c359add23bb67af11630b668

Counting.

The code to read the map in one dimension with a guard was created and registered in a snippet.

code:python

def solve(data):

ret = 0

for i in allPosition():

ret += 1

ret += 1

return ret

https://gyazo.com/3bd559ee36a1dd8e504a027b0b98bb28

code:python

def solve(N, XS):

cur = 0

for i in range(N):

cur += 1

print(cur)

It's a double loop, but the inside is only called 200,000 times at the highest, so it's okay.

The size of "used" is N or N+1 and RE.

https://gyazo.com/4e292857475488888efb440a7ce310a6

The problem that seems simple at first glance but gets bogged down with "Oh, there are times when it overflows" and "Oh, the number of reductions depends on the width of the overflow".

Maybe I could have solved it if I'd calmed down and sorted it out, but I'd melted too much time away, and I was too impatient to think straight.

In addition, the naive solution is like this

code:python

def blute(N, A, B):

ret = 0

for ax in range(0, N - A + 1):

for ay in range(0, N - A + 1):

for bx in range(0, N - B + 1):

for by in range(0, N - B + 1):

if ax <= bx < ax + A or ax < bx + B <= ax + A:

if ay <= by < ay + A or ay < by + B <= ay + A:

continue

ret += 1

return ret

I confirmed that dividing by (1-x)^5 yields a quadratic equation.

code:python

array([ 36300, -151980, 238728, -166632, 43560, 0, 0,

0, 0, 0])

After I saw the big coefficient, I was like, "Oh, this was three variables, what am I supposed to do?

I'm not good at math, so computer.

Official Explanation

I'd like to count the non-overlapping cases, but it's hard, so I'll subtract the overlapping cases from the total.

https://gyazo.com/28c8b91789fd402d10da52759be21205

I ran out of time in D. I read it later and this one seems easier for me to solve. Maybe I should have read all the problems first or used the Pomodoro timer to avoid getting bogged down.

Thoughts.

For example, if a square is illuminated by a square where two lights can be placed, the square will be illuminated by three of the four possible light placement combinations

Find K for all squares

For example, to see how many squares continue upward, just look at the square above you and add 1.

Do this for all four directions.

The sum of the number of illuminated tiles per lighting pattern is the sum of the number of illuminated lighting patterns per tile

https://gyazo.com/4ece1b83003aa351486285e44b87a67c

Expected Maximum Value] of random variable

max(a,b,c) <= x ⇔ a<=x and b<=x and c <= x

like that, and inequality is easier to handle.

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