I precalculated digits for segment from 1e5 to 2e5, then i computed all digits from 1 to 1e5(if n >= 1e5), then i stepped from 1e5 to n with step size = 1e5. And after all this I computed digits from last step to n. It does about 1e9/1e5 steps and works 0.015 sec.

I think that it's dumb solution, so I want to know, how to solve this task in O(log10(n)) or smth similar.

I wrote two programs, one AC, one WA.
I run these two programs for many times but they always make
the same answer.
Is it strange?

a= input().split()

l = int(a[0])
h = int(a[1])
ω = int(a[2])

g = 9.81
ω = ω * 2 * 3.14159 / 60
t = (h / 100) / (g * 0.5)
x = ω * t * l

if x % l <= l/2:
    print("butter")
else:
    print("bread")

why do I get runtime error?
n = int(input())
for i in range (n) :
    a=list (map (int,input().split()))
a.sort()
c = a[::-1]
d = c[0] + c[1] + c[2]
e = (len(c) // 2)
print(d, e)

Time limit. Where is my mistake?
cnt1 = int(input())
lst1 = input()
lst1 = lst1.split()
cnt2 = int(input())
lst2 = input()
lst2 = lst2.split()
cnt3 = int(input())
lst3 = input()
lst3 = lst3.split()
cnt = 0
for i in range(cnt1):
    for j in range(cnt2):
        if lst1[i] == lst2[j]:
             for k in range(cnt3):
                    if lst1[i] == lst3[k]:
                        cnt += 1
        else:
             continue
print(cnt)

Edited by author 31.10.2017 14:42

Vectors calculations uses square roots. Could this lead to not very precise floats and wrong results?

Edited by author 05.10.2023 13:50

Edited by author 05.10.2023 13:50

I just don't know what is wrong.I need some test in order to pass test 8

Check  that you indeed turn after reaching point B, i.e. that intersection point is not between A and B or you don't even reach B.

it is said that the testcase 9 is as follow：
4 4 1 11.1
1 2 2.0 0.5 0.5 0.5
2 3 1.05 1 0.01 1.0
3 4 1.05 1 0.01 1.0
4 2 1.05 1 0.01 1.0

the right answer is YES.

but i think there is no possibility to increase his money, so the answer should be NO.
could someone explain this test case to me.? thanks a lot..

Edited by author 19.09.2010 18:45

At the beginning of the turn, does player choose only among the parallelepipeds generated during the last turn, or among all of them, including the previous turns?

For example, K=1, n1=11

First player dissects it at x=2. We have now two parallelepipeds, of size 1 and 9.
Second player chooses the one of size 9, and dissects it at x=3, generating parallelepipeds of size 2 and 6.
Does first player have choice between 1, 2 and 6? Or only 2 and 6?

Thanks to authors :)

Try this test:
10 100 50 10 10 10 99
1 1 1 1 1 1 1 1 1 1

give me test,pleasee

It's kinda tedious since you need most operations too

tell me the hint, i want to solve this problem.

First off in your text editor switch from utf8 to cp437 (for example I use vscode and there's a tab in the bottom left that I can click to change the encoding). Only then, go to another comment on this problem (template for msvs 2010) and copy-paste the table from there.

For anyone struggling with this problem, here's how I did it:
segment trees with lazy updates for the dp
O(sqrtN) updates for the string itself (the logN retrieval with the segment trees instead of the O(1) with the sqrt blocks retrieval was too big of a hog it seems, maybe I messed something up)
Manacher for the edge cases
and dp[i] = dpEven[i] + dpOdd[i] from Manacher
So I used the segment trees to do most of the work, and then for the edge cases I used Manacher, so on average I'd get O(sqrt(N) + log(N) + K) for updates and O(log(N) + K) for queries.
Does anyone have a less ugly solution? My source code is aroung 15kb (although I used long names for clarity) and around 550 lines long. Also I've seen people's code run a lot faster, under a second (mine ran in 2.7 s) and I was wondering how they did it. Thanks!

rt