A
Obviously, the answer is
\ [C_n ^ m \]
Anyway, I did not see the obvious
Papers for the ball, the date of the box, then we put the problem into \ (n \) balls into the \ (m \) boxes in
And so on, is really the m box? ?
From the meaning of the questions, we have chosen, in fact, is m ~ n balls. Thereby, it is possible to increase a number m + 1 box, used to hold the rest of the ball is not selected.
Here, 1 ~ m number box is not empty, but m + 1 box is possible, in order to eliminate the impact of this difference, we have added a number n + 1 ball, placed at the end to ensure that in any case m not empty box number +1
Flapper law
In \ ((n + 1) -1 \) placed in the spaces \ ((m + 1) -1 \) plates, which spacer is \ (m + 1 \) one set is not empty, so the answer is \ (C_n ^ m \)
Let Kankan data
id=1：
Do not say, hit the table number combination
id=2：
Think about it, make a linear inversion table + factorial table, fucks Do It
id=3：
Modulus is small, Lucas water through Theorem
id=7：
n, m and large, modulus and large, ah Zezheng
Segmented play table
wtf？？？
With id = 2 a similar situation, but also hit the factorial table, but \ (10 ^ 9 \) watch must fight for it, taking into account the \ (T \) is very small, we can play the segment table, in \ (1 \) ~ \ (10 ^ 9 \) in every one million even a factorial (of course% p)
For example: b [1] = 1, b [2] = 1000000, b [3] = 2000000, ......!!!
Inverse direct use of Fermat's Little Theorem (quick power)
Then continue fucks
B
Tree Mo team
deaf practices: DFS necklace + HH sequence (ojbk I had forgotten yellow necklace What the hell, just remember this name as well as the practice of Luo Gu block \ (O_2 \) would T80）
dfs order to convert the tree into a linear problem
Offline, sorted by the right end section (due to a point corresponding to the period interval dfs sequence). Then the pointer moves left shift shift
For each number we record a
C
Since the subject of the request is from a certain point and connected to the starting point, then back to the starting point and another point connected
There are three approaches:
1. propose to build a new random map and point connected to the starting point
2. Binary Packet
3. Figure regular Jianxin
Specific solution to a problem boss:
Easy to find violence \ (N ^ 2logN \) enumeration ring 1 and two adjacent points, then you just need to find without No. 1 point fucks shortest path between two points can be
considered for more than one enumeration , one point will split into two, one point, one exit point, and then all the points one set of two adjacent points into
one of which is connected to the set point, it is connected to another set point, then run the shortest time point that can be calculated from two points each selected from a set of all points of the ring
two treatments:
- Direct randomly divided set of points, then the error probability \ (75% \) , more than a few random
- In binary packet, enumerated order bits do this a collection of bits, in order to make another set, the complexity of the \ (Nlog ^ 2N \)
to sum up
A : No, promiscuity, cheat points, the portion T of the results Lucas
B: No, promiscuity, violence
C: wow, immortal question, did not see the most basic out
I hate bundle test! ! ! !
Even if I did not bundled test points
Expect Score: \ (60 + 80 + 0 = 60 ~ \?)
Actual Score: \ (45 + 20 + 0 = 65 \)
New knowledge learned today:
Piecewise play table
2. Binary Packet
3. tree Mo team
Digression
% Out of question people OBlack seniors
The immortal gods immortal question to do
10 relates to three questions method, TQL .......
A title is what I heard last year are stuck, involving a total of seven practices (exam when I say why only id = 1/2/3 / 7,4 5 6 where to go, and now should be glad that there have been only four methods QwQ ....)