-lm -O2 -std=c++11
Morning
Before
T1
As if half the answer?
Or $ dp $
T2
Unclear.
T3
rectangle?
During
T1
Dichotomous answer, greedy judge?
It must be because of all the values are a function monotonous?
Seems like if the pan than the maximum, then the ......
Consider three points (on-site $ YY $)
Large play table ...... $ gedit $ scene of the explosion.
We function divided into four categories:
$$
k>0\, b>0\\
k<0\, b>0\\
k>0\, b<0\\
k<0\, b<0
$$
Found last category waste =. =
Then the first three categories, the $ 1 \, 3 $, when $ t $ tends to positive infinity they must be selected
$ 2 $ beginning may be selected.
If all of the $ 2 $ does not have a single option is growing. Direct-half
If $ 2 $ has been selected, that is, a single valley ~
Began $ YY $-thirds?
If you can guarantee $ \ log $ it is great!
If just a valley and then in the back half?
(Front row directly if $ 0 $)
$ Er $ ...
First hit half of it, how much can lie dotted
Think about how the $ check $ labeled $ \ Theta (N) $
More than a $ \ log $ to $ T $ the $ QAQ $
Think $ ing $
Half +-thirds (funny
In this way, the large sample and make life difficult for ......
Hell miles ...... $ WA0 $ !!!!!
T2
$ 3 $ minute hand
(Tl $ $ third $ T2 \, 3 $ min) =. =
Vigorously code Gaussian elimination ......
That can be represented by another $ x_1 $ of ......
T3
The violence code.
After
|
16
|
Miemeng | 66
03:05:47
|
21
03:05:47
|
13
03:05:47
|
100
03:05:47
|
Evening
I took the test? ?
Drink Grass $! $
Tribute to $ 0x223 $!
Before
T1
greedy? ? ? ? ? ?
T2
$tarjan$？？？？？
T3
$ Exgcd $? ? ? ? ? mathematics.
During
T1
Directly engaged in it ......
3 4 4 0 2 1 2 5 6 1 2 7 2 3 3 3 2 2 2
T2
~~~ giant magic ......
Single 举子 collection?
$3^{17}=129140163$
Not experts to work out a sample $ emm $ ......
The devil tickets ......
Find an edge set it acyclic ......
violence. the complexity:
$\Theta(2^M \times M)$
Not large, about funny :(
$$287452830644856679165810939428862445578063655160866980680588178644246189273244291058106368$$
Play out, almost forgot to clear $ V $ ......
Think of the last $ 10 $ points when ...... $ emm $
Consider inclusion and exclusion.
The number of minus 1 ring
The number of rings plus 2
The number of minus 3 rings
Plus the number of rings 4
The number 5 minus the ring
Plus the number of rings 6
...
Now we must consider the maximum number of rings
With a ring and can maintain a pressure level bitset.
$\Theta(2^{C_n^3})$
:( very small if bigger ~~
$$5016456510113118655434598811035278955030765345404790744303017523831112055108147451509157692220295382716162651878526895249385292291816524375083746691371804094271873160484737966720260389217684476157468082176$$
Sides must find a way to replace point $ QoQ $
$dp$？
Consider like pressure!
T3
I knew it:
$lcm(a,b)=a \times b \div gcd(a,b)$
(Number theory only $ gcd $)
$$ NYY \, NYY \, \ NNYY, YNN $$
Failed exam $ QnQ $
After
|
4
|
Miemeng | 100
03:14:52
|
40
03:14:53
|
20
03:14:54
|
160
03:14:54
|