I have WA4. Can someone give me that test ?

Wow. This problem is really cool. Try using the naive method and you're sure to get TLE#26.
Well, my solution(AC 0.234) goes this way:

if the first term is not 1 then the ans is 1.
if the first i terms are 1<<0, 1<<1, 1<<2, ... 1<<(i-1) then all numbers from 1 to (1<<i) - 1 can be expressed as a sum of some of the above terms. Thus if a number k is present all numbers from k to k+((1<<i) - 1) can be skipped. Proceed using this approach.

By the way 1<<i represents pow(2, i).

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

Answer: 0

Use int64_t :)

lines = []
for line in stdin:
    for i in range(0, len(line.split())):
        lines.append(line.split()[i])
lines = [int(i) for i in lines]
lines_rev = list(reversed(lines))
numbers = []
for i in range(0, len(lines_rev)):
    numbers.append(lines_rev[i]**0.5)
for i in range(len(numbers)):
    print(f"{numbers[i]:.4f}")

fix this plz

I've got many RE16 with using python 3.4. Did anybody else get it? How did you solve it?

a = int(input())
f = []
for i in range(a):
    s = int(input())
    f.append(s)
if f.count(3) >=1:
    print('None')
else:
    if f.count(5) == len(f):
        print('Named')
    elif sum(f) / len(f) >=4.5:
        print('High')
    else:
        print('Common')

Can anyone give tests data or some discription?


Edited by author 01.11.2023 14:24

1000000000 1000000000 300
answer: 97395891

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

Hello,
I am trying this question, however, I am unable to understand how to construct the solution. My Dp is weak so I thought to practice here, however I am having a tough time.
Can someone please help in describing how to start thinking about such problems?

we have 12 hours on clocks

If you have WA10, make sure you don't have 'I' as a card value instead of 'T'. For some reason it was very easy for me to mix these letters in the statement...

u can search for intersections instead of looking for cutouts, just invert the segments of the second function
use min max instead of a lot of conditions and you get a very simple code :)

My idea is to reverse the operations of adding characters and doubling a string. I have an iterative algorithm in O(n * log(n)). At each iteration, I count dp - the length of the maximum substring ending with index i, obtained by the doubling operation. Then I look for the maximum among the maximum substrings and truncate the current string to leave this substring. But this is a greedy solution. Is it possible to come up with a counterexample to it?

admins, could you pls fix this?

My first attempt use string hashing to check the palindrome with Segment Tree lazy prop.
O(Q*K*logK*logN), esp. for the update query and it's TLE 11

However, using Manacher algorithm with Segment Tree you can achieve O(Q*(K+lgN))

Little help:
- WA#3 : You're likely answering too much. (K involved)
- WA#9 : N=10^5, overflow somewhere

Help!
What is wrong?


[code deleted]

Edited by moderator 05.11.2023 02:38

var N: integer;
begin
readln(n);
if (abs(n)>10000) then writeln('Îøèáêà ââîäà') else
if N>=0 then writeln(((1+n)*n)/2)
else writeln(((1+n)*(abs(n)+2))/2);
end.