I used my own hash map with modulo 1e5, and buckets with size equal to one

it's the case where count of '(' is greater than ')'.

For example:
(( or (()

its wa is something like a Ivan's Car problem

[code deleted]

Edited by moderator 27.07.2023 17:22

There is no checker in this problem. In my solution I
1. Print new line after query even if there is no strings which satisfy it
2. Don't print new line after last query

Pls, explain me why i'm wrong
Let T is answer.
t in [0,T]: x(t)=vx*t, y(t)=vy*t-g*t^2/2
t > T: x(t) = x(T) + (vx+ux)(t-T) = (vx+ux)t - ux*T
       y(t) = y(T) + (vy - g*T + uy)*(t-T)-g * (t-T)^2/2 = (-g/2)*t^2 + (vy+uy)*t-y*T

t1, t2: x(t1)=L, x(t2)=L+l
t1 = (L+ux*T)/(vx+ux)
t2 = (L+l+ux*T)/(vx+ux)
tv: y'(tv)=0, tv = (vy+uy)/g

Then i use conditions:
0<T<L/vx, T<2*vy/g
h<y(t1)<h+d, h<y(t2)<h+d
if tv in (T-eps, T+eps) then y(tv)<h+d

My algo:
find all candidates from conditions and then check that candidate is good

My code:
[code deleted]


Edited by moderator 27.07.2023 17:23

play with precision

5
0 0
4 0
4 2
2 4
0 2

answer:
3.640719

u visit end vertex more than one time, idk why we need only 1 time it visit, but by that i got ac

a0a*
F

My WA output N (infinite)
I interpret this as (a)concat(0)concat(a*) = {a, aa, aaa, aaaa, ...}

If you are solving through comparing Min and Max distance from some border point to center of figure, then check for circle in following way: Max - Min <= 3.

why greedy with catching two mins dont work
pseydotests:
#1:
test

test, universal

base
#2:

universal, base

base, universal

#3:
universal

base

test

universal

Try the test
8
1 7 4 6 3 5 2 8
and
8
8 2 5 3 6 4 7 1
.
Both tests have solutions.

double

1000000000 1000000000 1000000000
0 0 0
answer:
It is a kind of magic

Please give me some tests, my program passed all the tests on the forum

Edited by author 22.07.2020 04:11

you don't know how to write bfs, try watching videos for beginners

precision

Try this test:
9 6
2 3 2 5 3 4 1 6 1

Answer:
8
1 6 1 2 2 5 3 4 3