graylog collector

Expected n cards. Card collector Time Limit: 2000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others) Total submission (s): 1700 accepted submission (s): 776 Special Judge Problem descriptionin your childhood, do you crazy for collecting the beautiful cards in the snacks? They said that, for example, if you collect all the 108 people in the famous novel Water Margin, you will win an amazing award. As a smart boy, you no

Php collector. This article is an article collection function implemented by php. it is only a simple function, but the source code is very rare. it is suitable for small collection functions. This article is an article collection function implemented by php. it is just a simple function, but the source code is very rare, suitable for small collection functions. This article is an article collection function implemented by php. it is only a simple fun

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=2602The bare ...1#include 2#include 3#include 4#include 5#include 6#include 7#include 8#include 9#include Ten#include One#include A#include -#include -#include the#include -#include Set> -#include -#include + - using namespacestd; + A Const intMAXN =555555; at intN, M; - intV[MAXN]; - intW[MAXN]; - intDP[MAXN]; - - intMain () { in //freopen ("in", "R", stdin); - intT; toscanf"%d", T); + while(t--) { -Memset (DP,0,sizeo

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=2639Test instructions: Finding the K-Solution for 01 backpacksDP (I, j) represents the I-optimal solution with a capacity of JFor the second-dimensional operation and the 01 backpack almost the same, but we only need to pay attention to each time to take the first k large.1#include 2#include 3#include 4#include 5#include 6#include 7#include 8#include 9#include Ten#include One#include A#include -#include -#include the#include -#include S

. All the input numbers is nonnegative and no more than 100.Outputfor Each test case, display a single line containing the case number and the number of different patterns Mr. B can Make with these stones, modulo 1,000,000,007,Which is a prime number.Sample Input31 1 121 2Sample OutputCase 1:15case 2:8Hintin the first case, suppose the colors of the Stones Mr. B have is B, G and M, the Differen T patterns Mr. B can form are:b; G M BG; BM; GM; GB; MB; MG; BGM; BMG; GBM; GMB; MBG; MGB.SourceField=

http://acm.hdu.edu.cn/showproblem.php?pid=2639The main idea is, to the backpack race bones, for the K-optimal solution, on the basis of the ordinary 01 backpack, add one-dimensional space, then f[i,v,k] can be understood as the first I items, put in the capacity v backpack, the value of the K-optimal solution. The time complexity is O (NVK).Talk is cheap.Look at the code.ImportJava.util.Scanner; Public classBonecollector { Public Static voidMain (string[] sure) {intT; Scanner SC=NewScanner (syst

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=2602The title requirement is as much as possible, so the initialization time is 0.If the requirement is exactly full, the initialization time except dp[0] = 0, the other must be set to-inf, indicating the illegal situation.1#include 2 using namespacestd;3 intT;4 intdp[1010][1010];5 intv[1010], w[1010];6 intMain ()7 {8 //freopen ("In.txt", "R", stdin);9 //freopen ("OUT.txt", "w", stdout);Tenscanf"%d", T); One while(t--) A { -

Problem DescriptionThe Title of this problem be familiar,isn ' t it?yeah,if you had took part in the ' Rookie Cup ' competitio N,you must has seem this title. If you haven ' t seen it before,it doesn ' t matter,i would give you a link:The link:http://acm.hdu.edu.cn/showproblem.php?pid=2602Today We is not desiring the maximum value of bones,but the k-th maximum value of the bones. NOTICE That,we considerate and ways that get the same value of bones is the same. That means,it'll be a strictly decr

dp+ State Compression1#include 2 using namespacestd;3 Doublep[ at];4 Doubledp[1 at];5 intMain ()6 {7 intN;8 while(SCANF ("%d", n)! =EOF)9 {Ten for(intI=0; i) Onescanf"%LF",p[i]); Adp[(11]=0; - for(intI= (12; i>=0; i--) - { thedp[i]=1.0; - Doublesum=0.0; - for(intj=0; j) - { + if(i (1Continue; -sum+=P[j]; +Dp[i]+=dp[i| (1P[j]; A } at //printf ("%lf%lf==\n", dp[i],sum); -dp[i]=dp[i]/

 Test instructions: There are n kinds of cards, eating snacks when you will eat some cards, tell you in a bag of snacks to eat each card of the probability, to collect all the cards need to buy a snack bag number of expectations.Idea: First state compression, then probability DPThe state transition equation is d[i] = P ' * (d[i]+1) + sigma (d[i |, using d[i] to indicate how many packets need to be bought by the state I to the target. (1 #include Copyright NOTICE: This article for Bo Master or

, time and effort, inventory is often very long. Especially the book wholesale market. Often need daily inventory, heavy workload. Overtime is often required. Time consuming and laborious.Inventory machine: The use of Chinese-code inventory machine for book inventory, equipment to replace human operation, to the book industry to bring efficient and high-speed inventory. The process of scanning completed the background inventory order generation, no manual manual entry process. No data import exp

[Hacker Collector Edition] ends with a jump $ Ing = imagecreatetruecolor (700,410); $ red = imagecolorallocate ($ ing, 255,255, 0); $ yellw = imagecolorallocate ($ ing, 45); imagefill ($ ing, 119,102,135,152, $ red); $ a = array (93,123, 30, 52,152, 66,102, 90, 30); imagefilledpolygon ($ ing, $, 10, $ yellw); $ a1 = array (, 25, 25); round ($, $ a1, 10, $ yellw); $ a2 = array (227,108,227,127,245,134,228,139,227,157,215,143,196,149,208,132,196,117,

This article describes how to create a keyword query collector for Baidu Dictionary using PHP. For more information, see Samples collected by Baidu dict The collection of all the result data after Baidu dict dictionary translation, of course, comes with the 13.5w word library and simple collection cases. here I will write the main class dict. class. php is released. the Project address is http://github.com/widuu/baidu_dict. if you have a specific dir

Topic Links:http://acm.hdu.edu.cn/showproblem.php?pid=4336Main topic:Each pack of snacks has a card, a total of n different cards, the probability of obtaining the N cards is p[i] (1 What is the expectation of collecting all the cards.Ideas:PI indicates the probability of getting the first card and EI indicates the expectation of the card.Suppose there are two cards now, by test instructions:E1 = 1/p1,e2 = 1/p2,e12 (indicates the expectation of buying 1 or 2 of one packet) = 1/(P1+P2).When we ca

Test instructionsThere are n kinds of cards, inside each package, there may be a card or not, the probability that each card is known to appear in the face, to obtain n kinds of cards, the need to eat the face of the package number of expectationsAnalysis:n is very small, with the shape of pressure, before doing the pressure of this problem, but the probability of how to push the unclear, now seems to be the basic probability DPDp[s] means the number of packets that are expected when the number

1. Title Description: Click to open the link2. Problem-Solving ideas: The topic is the classic 01 knapsack problems, I reverse enumeration of the writing is not introduced, mainly about the classic error. If the input n items are set to 0~n-1, the correct wording and the wrong wording are given.3. Code:(the correct wording)#pragma COMMENT (linker, "/stack:1024000000,1024000000") #include (Error one)for (int i=0;i(Error two)for (int i=0;i(Three wrong wording)for (int i=0;iThe above three kinds of

= -; - DoubleP[MAXN]; $ Doubledp[1MAXN]; $ intN; - - voidsolve () { the intMST =1N; - intmask = MST-1;WuyiDp[mask] =0; the -Per (I,0, Mask) { Wu DoubleTMP =0.0; - AboutDp[i] =1.0; $Rep (J,0, N) { - if(I (1j)) - Continue; -Dp[i] + = p[j] * dp[i| (1j)]; ATMP + =P[j]; + } theDp[i]/=tmp; - } $ the DoubleAns = dp[0]; theprintf"%.06lf\n", ans); the } the - intMain () { inIos::sync_with_stdio (false); the #ifndef Onl

Probability dp/mathematical expectation/pressure dp/tolerance principleThe 14th question in Kuangbin summaryWhat a wonderful way to do it ... The kuangbin of the code to see the good ...1 //hdoj 43362#include 3 #defineRep (i,n) for (int i=0;i4 #defineF (i,j,n) for (int i=j;i5 #defineD (i,j,n) for (int i=j;i>=n;--i)6 Const intn= A;7 8 Doublep[n],f[1N];9 intMain () {Ten intN; One while(SCANF ("%d", n)! =EOF) { A Doublett=0.0; - Rep (i,n) { -scanf"%LF",p[i]); thett+=P[i]; -

Topic Link: Click to open the linkTest instructionsThere are n kinds of cards, every packet of instant noodles have a certain probability to get one of the cards (may also not get cards)Ask about the expectations of the number of instant noodles that the dragon needs to eat.Ideas:Dp[i] means that you already have a card that has a status of I, how many packages are needed to have all the cards,Apparently dp[(1And the answer is dp[0];Using example two example, the following DP equation is directl

String path = Request.getcontextpath ();String basepath = request.getscheme () + "://" +request.getservername () + ":" +request.getserverport () +path+ "/";%>-String dbip= "192.168.1.20:3306";String dbname= "diancai360_haiwai20150330";String dbuser= "root";String dbpwd= "root12";String dbsql= "jdbc:mysql://" +dbip+ "/" +dbname+ "? user=" +dbuser+ "password=" +dbpwd+ "useUnicode=true Characterencoding=utf-8 ";Class.forName ("Com.mysql.jdbc.Driver"). newinstance ();Connection conn= drivermanager.g