Title Link: HDOJ-5159There are too many ways to do this problem. BC's second question can also be very water.Algorithm OneThe algorithm I wrote at the time of the game was this:Preprocess out all the answers, and then output for each query directly.Enquiries (A, b) are recorded as (A, b).The answer to (a, b) is from the answer (A, b-1).The answer to (a, 1) is the average of 1 to a, which is quite obvious.If B > 1, then we consider in the first B, we draw each kind of card probability is 1/a, and
Probability methodDemands out and expectations, the basic theorem of expectation, and the expectation of each part of the desired and.E (sum) = e (1) + E (2) + ... + e (x);The probability P =1-(1-1/x) ^b, which is selected in B-time, is only selected and not selected in each of the numbers in B.So the expectation of each number is also i* (1-1/x) ^b)Get the sum expectation.#include Combination methodAnd all the results may appear and divide by the total kind.#include Hdu5159--
started writing. In order to ensure that I could push it backwards, I would use the stack. Before the result is written, I would have to go out of the submit time ,,, rating just ran Rz.
After going out for a while in the evening, I came back and wrote the question. The first time I did it again, I used the Fast Power naively. In fact, I didn't need the fast power to be faster, then the answer is not set to long and Wa again, and then after several times of messing around, the AC will become
DZY Loves Topological Sorting (BC #35 hdu 5195 topsort + priority queue ),
DZY Loves Topological Sorting
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission (s): 264 Accepted Submission (s): 63
Problem DescriptionA topological sort or topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge
(U → v)
From vertex
U
To vertex
V
,
U
Comes befo
Bash cannot handle floating-point operations and lacks specific operations. These operations are important computing functions. Fortunately,BCThis problem can be solved.BCNot only is it a flexible and precise tool with multiple functions, but it also provides some convenient functions that many programming languages have. because it is a complete UNIX tool, it can be used in pipelines,BCIt is also very common in scripts.
BC uses scale to solve decima
Since the first four days have been looking at Bird's Linux books computer some basic knowledge, today only touch the basic commands, from today on a daily record of their own Linux learning process.Cal: CalendarCalCal 2015: List all calendars for 2015Cal 5 2015: List calendar for May 2015Date: DayDate: Displays the current dateBC: CalculatorsBC: Count into Calculator functionecho $LANG: Displays the display language supported by the current terminalExampleChange current LANG:LANG=ZH_CN. UTF-8Ma
is to judge whether the two sets of integers are equal.
Because the elements in the collection are heterosexual, the elements must be different before you can compare the two sets
So the main problem is to remove the weight
After sorting the elements in the collection, it's OK to go heavy ...
This question is the most simple one in the BC ... b question to now or WA is also drunk ... Thinking of the same Satan. Think about it tomorrow.
_ooooo_//o8888
The game is still not done, BC is to do a problem, it is estimated to lose points, slowly improve, always end the fate of a problem ...
Ideas:
Dynamic planning, but I just started to learn, and will not do ah ... First find the law for a long day, the results of nothing to find out, and began to deep search,
Results sample can be over, hand over the time, the depth is too big ah, not correctly estimated ... After the game to see the puzzle, dynamic
Victor Serial port Control detailed description
Class/function
Header file
Description
Tybcommdevice
Vcl.ybcommdevice.hFmx.ybcommdevice.h
Serial control
Tvictorcomm
Vcl.victorcomm.hFmx.victorcomm.h
Multi-threaded serial port class
Tcommqueue
Vcl.victorcomm.hFmx.victorcomm.h
Serial data Queue (serial FIFO cache)
King asked the famous doctor Flat Magpie said: "Your family brothers three people, are skilled in medicine, in the end which is the best." ”
The Flat Magpie answer: "The elder brother is best, elder brothers second, I worst." ”
The king asked again, "then why are you most famous?" ”
The Flat Magpie answer: "Elder brother Cure, is cure before the illness attack." Since the general people do not know that he can eradicate the cause of the disease beforehand, so his fame can not be spread out
expires is a Web server response message header field that, in response to an HTTP request, tells the browser that the browser can cache data directly from the browser before the expiration time, without having to request it again.
Cache-control is consistent with expires, which indicates the validity of the current resource, whether the browser caches data directly from the browser or re-sends the request to the server. But Cache-
Control | Upload This is an example of selecting a province and then uploading the click event to the top level of the control in the page handling event. The bold part is the entire upload event process. Upload events (exposure events), exposure attributes, complex property management, style management, and so on are advanced topics for ASP.net server custom controls.
Using System;
Using System.Web;
Using
There is a widely spread story: Wei wenwang asked Bian Que: "which of the three of your brothers is the best in medicine ?" Bian Que answer: "Long Brother is the best, middle brother is the second, and I am the worst ." Wen Wang asked: "Why are you the most famous ?" Bian Que answer: "Long Brother is treating diseases before and after the onset of illness, and has no feeling. The average person knows that he has exceeded the cause in advance, so he is well known, at the beginning of his illness,
Preface:
Third, it is easy to name, because according to the suggestions of netizens. So I changed it to Asp.net control development. Well, let's get started. I talked about the basic concepts of composite controls in the previous section. Today I am going to learn more about how the composite control view tracks the status of sub-controls. How does one save the subcontrol status? How does one load the subc
Card
Time limit:10000/5000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total Submission (s): 975 Accepted Submission (s): 425
Special Judge
Problem Description There is x cards on the desk, they is numbered from 1 to X. The
The Mook Jongaccepts:506submissions:1281Time limit:2000/1000 MS (java/others)Memory limit:65536/65536 K (java/others)Problem DescriptionZjiaq want to become a strong mans, so he decided to play the Mook Jong. Zjiaq want to put some Mook jongs in his
This is a created
article in which the information may have evolved or changed.
A simple simulation
#include #include #include using namespace Std;int a[105], b[105], f[ 105];bool CMP (int a, int b) {return a 0 && flag) {F[i] = (N/a[i] + 1) *
This is Ali's latest test, and the following code prints the required string. If you have other questions, you can modify the following code to provide additional functionality.
As you can see from the title, the character growth is 4 times times
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