amd 3450

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Hdoj 3450 HDU 3450 counting sequences ACM 3450 in HDU

Miyu original, post Please note: Reprinted from __________ White House Question address: Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 3450 Description: Counting Sequences Time Limit: 2000/1000 MS (Java/others) memory limit: 32768/65536 K (Java/Others)Total submission (s): 312 accepted submission (s): 105 Problem descriptionfor a set of sequences of integers {A1, A2, A3 ,... an}, we define a sequence {aI1, ai2, ai3... AIK} in which 1 Inpu

"BZOJ-3450" tyvj1952easy probability and expectation DP

3450:tyvj1952 Easy Time limit:10 Sec Memory limit:128 MBsubmit:468 solved:353[Submit] [Status] [Discuss] DescriptionOne day WJMZBMR is playing osu~~~ but he is too weak to push, some places entirely by luck: (Let's simplify the rules of the game.Have n click to do, success is O, failure is x, score is calculated according to comb, continuous a comb there is a*a points, comb is a great continuous o.Like Ooxxxxooooxxx, the score is 2*2+4*4=4+16

HDU 3450 (tree array + dp)

Question link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 3450 Counting Sequences Time Limit: 2000/1000 MS (Java/others) memory limit: 32768/65536 K (Java/Others)Total submission (s): 1815 accepted submission (s): 618 Problem descriptionfor a set of sequences of integers {A1, A2, A3 ,... an}, we define a sequence {aI1, ai2, ai3... AIK} in which 1 Inputmultiple test cases the first line will contain 2 integers n, D (2 Outputthe Number of perfect

Bzoj 3450:tyvj1952 Easy expectation/probability, dynamic planning

3450:tyvj1952 Easy Time limit:10 Sec Memory limit:128 MBsubmit:431 solved:325[Submit] [Status] [Discuss] DescriptionOne day WJMZBMR is playing osu~~~ but he is too weak to push, some places entirely by luck: (Let's simplify the rules of the game.Have n click to do, success is O, failure is x, score is calculated according to comb, continuous a comb there is a*a points, comb is a great continuous o.Like Ooxxxxooooxxx, the score is 2*2+4*4=4+16

Hdu 3450 tree array optimized dp

Hdu 3450 tree array optimized dp I won't talk about the meaning of the question; hdu2227 is almost just that this question is not incremental, but the adjacent difference is not greater than d. In fact, they are the same and there are also differences; This question does not contain a range and the size relationship between trees has an impact on the results. Therefore, the original id is used in the tree array. Due to the large number of values, the

Poj 3450 merge ate identity (suffix array)

Http://poj.org/problem? Id = 3450 I did this during my vacation. At that time, I used KMP. This time I used a suffix array. Concatenate n strings and separate them with different characters that do not appear in the input string. Separate each character in The New String according to the region of the atomic string and mark it with the LOC array. The length of a binary common substring. Check whether the height values of N consecutive Suffixes in the

Hdu 3450 tree array optimized dp and hdu3450

Hdu 3450 tree array optimized dp and hdu3450 I won't talk about the meaning of the question; hdu2227 is almost just that this question is not incremental, but the adjacent difference is not greater than d. In fact, they are the same and there are also differences; This question does not contain a range and the size relationship between trees has an impact on the results. Therefore, the original id is used in the tree array. Due to the large number of

POJ-3450 impersonate identity

Label: style blog Io ar OS for SP on 2014 Enumerate all substrings of a string and match other strings. # Include POJ-3450 impersonate identity

Hdu 3450 tree-like array optimized DP

;} int Searchl (int A ) { int Left=1,Right=Tt; while (LeftRight ) { int Mid=(Left+Right)/2; if (Mark[Mid]A)Left=Mid+1; else if (Mark[Mid]==A ) return Mid; Else Right=Mid; } return Left;} int Main () { int I,J,D; while (~scanf("%d%d",N,D) { for( I=1;IN;I++) {scanf("%d",Num1[I]);Num2[I]=Num1[I]; }Sort(Num2+1,Num2+1+N);Tt=0;Num2[0]=-1; for (I=1;IN;I + +) { if( Num2[I]!=Num2[I-1]) {Tt++;Mark[Tt]=Num2[I]; } }Memset(Coun,0 ,sizeof( Coun)); int Sum=0; int X=Searchl(Num1[1])

Bzoj 3450 Easy

Attention to detail ah ...Same as the last.#include #include#include#include#defineMAXN 300500using namespacestd;intN;CharS[MAXN];Long DoubleDP1[MAXN],DP2[MAXN];intMain () {scanf ("%d\n%s",n,s); for(intI=1; i) { if(s[i-1]=='o') {dp1[i]=dp1[i-1]+1;DP 2[i]=dp2[i-1]+2*dp1[i-1]+1;} Else if(s[i-1]=='x') {dp1[i]=0;DP 2[i]=dp2[i-1];} Else{dp1[i]= (dp1[i-1]+1)*0.5;DP 2[i]=dp2[i-1]+(2*dp1[i-1]+1)*0.5;} } printf ("%.4lf\n", Dp2[n]); return 0;}Bzoj 3450

HDU 3450 & HDU 2836 tree Array

Source: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 3450 Question: give some numbers and ask how many sets are there. The absolute value of the difference between adjacent numbers is less than D. Idea: Tree array. After finding the upper and lower bounds, it is the same as HDU 2227. The method for finding the upper and lower bounds is binary. Code: #include

BZOJ-3450 Easy probability dp "mathematical expectation"

contribution to the expectation, it is assumed that in the processing of the position I, I have a continuous o, which can be considered a special case.    Suppose this bit is o then? i = (l + 1) ^ 2-l ^ 2 = 2 * l + 1, while taking expected E (? i) = 2 * E (L) + 1;Suppose this bit is x then? i = 0;Suppose this one is? Then E (? i) = P1 * X1 + p0 * X0 (where P1 is the probability of taking one of the bits, P0 is the probability of taking 0, X1 for this bit to take 1 of the score changes, X0 the s

HDU 3450 Segment Tree + two points

Click to open linkTest instructions: Give a sequence of numbers, and ask you that the absolute value of the difference between the two adjacent number is greater than 2, and the 9901Train of thought: see not at all, did not think is the topic of line tree, weak cry ~ ~ ~, see the solution of Daniel, is to know how the matter, for the current number A, then it is the last element can be composed of the situation is a-d to a+d and, can also think, A-d has formed the M kind of situation, Then in th

ZOJ 3450 Doraemon ' s Railgun (DP Group backpack)

the city. Group I is located at different (X i , Y i ) and contains W i soldiers. After days T0 , all the enemy would begin to attack the city. Before it, the railgun can fire artillery shells to them.The Railgun is located at ( X0 , Y0 ), which can fire one group at one time, the artillery shell would fly straightly to The enemy. But in case there is several groups in a straight line, the railgun can only eliminate the nearest one first if Doraemon Wants to attack further one. It took T days t

POJ 3450 Corporate Identity to find the longest common substring of all characters

each trademark would be at least 1 and at most C Haracters.After the last trademark, the next task begins. The last task was followed by a line containing zero.OutputFor each task, output a single line containing the longest string contained as a substring in all trademarks. If there is several strings of the same length, print the one and that is lexicographically smallest. If There is no such non-empty string, output the words "IDENTITY LOST" instead.Sample Input3aabbaabbabbababbbbbbbabb2xyza

BZOJ 3450: Tyvj1952 Easy, bzojtyvj1952

BZOJ 3450: Tyvj1952 Easy, bzojtyvj1952 Time Limit: 10 Sec Memory Limit: 128 MB Submit: 874 Solved: 646 [Submit] [Status] [Discuss] Description One day, WJMZBMR is playing osu ~~~ But he is too weak, and in some places he depends entirely on luck :(Let's simplify the rules of this game.There are n clicks to do. If the operation succeeds, it is o. If the operation fails, it is x. The score is calculated by comb. For a consecutive comb, there is a *

POJ-3450 Corporate Identity (kmp+ suffix array)

each trademark would be at least 1 and at most C Haracters.After the last trademark, the next task begins. The last task was followed by a line containing zero.OutputFor each task, output a single line containing the longest string contained as a substring in all trademarks. If there is several strings of the same length, print the one and that is lexicographically smallest. If There is no such non-empty string, output the words "IDENTITY LOST" instead.Sample Input3aabbaabbabbababbbbbbbabb2xyza

Asynchronous modules define AMD and asynchronous modules amd

Asynchronous modules define AMD and asynchronous modules amd It is called Asynchronous Module Definition, and is defined by Asynchronous components (or modules. AMD is a mechanism in which components and their dependencies can be asynchronously loaded.Define Method define(id?, dependencies?, factory); Component ID The component ID is the unique identifier of a c

AMD specs and how to turn AMD into COMMONJS

Original: http://villadora.me/2014/05/23/amd-define-and-how-to-translate-amd-to-commonjs/There has been a lot of controversy between Commonjs and AMD, and both have evolved and converged on the project. Personally, Commonjs is more developer-oriented, and for developers, it requires a clear version and management, less code and more interference, and less configu

What about AMD FX-8370? AMD FX-8370 Evaluation

Although AMD's day-to-day task is to zen the momentum, but after all Zen the fastest also have to be the end of the year or the beginning of the IPO. In such a gap, the addition of the upgrade is the pain before the dawn. As a representative of AMD High-performance CPU--FX series, although its historical evaluation has long been conclude, but the market always has a place for the FX series. This year, AMD s

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