precalculus inequalities

Topic Links:Intervals Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 24379 Accepted: 9274 DescriptionYou are given n closed, integer intervals [AI, bi] and n integers c1, ..., CN.Write a program:Reads the number of intervals, their end points and integers C1, ..., CN from the standard input,Computes the minimal size of a set Z of integers which have at least CI common elements with interval [AI, bi], for each i= ,..., N,

arbitrarily far apart, output-2. Otherwise output The greatest possible distance between person 1 and N.Sample Input 14 2 11 3 82 4 152 3 4Sample Output 19Test Instructions:n individual stand into a row, there are two people between the minimum distance, there are two of the maximum distance between people, ask the first person to nth person distance between the maximum is how much;Ideas:Such inequalities can be obtained by test instructions:x Line:d

equation: initial conditions are L( x, y; 0) = F( x, y) Multi-scale edge detection and blob detection: Gradient operator for edge detection Over 0 points detection: two differential invariance equations Three differential invariance inequalities are satisfied: Blob detection: Laplascause equation or determinant of Hessian matrix Autom

Test instructions: There are n intervals [a, b], and each interval has a value of C. Find the elements in a set so that each interval has at least C elements in this set, asking the smallest set size.Idea: Set d[i+1] means 0 to i how many number in this set, obviously for each interval, d[b+1]-d[a]>=c, in order to meet the requirements of the topic. But this does not make all the collections connected. Continue to excavate the condition, according to d[] 's definition can get, 0Thus three

See http://www.cnblogs.com/proverbs/archive/2013/02/01/2713109.html for details (I think this is enough detail, I will not repeat it)Let's recall the course of doing this problem! At first it was a bit of a mistake, but it came back immediately and clearly test instructions. Wrote a n^2 algorithm. Obviously, for the data of N In the beginning there are doubts, why can directly in the queue after the first maintenance to take the first place? The reason is that I Slope optimization is the fact th

-confidence, there is no need to proveBut sometimes it's not so sure, it's good to just prove it in the head.51Nod Greedy Starter Tutorial _ Perfect stringJohn thinks that the perfect degree of a string is equal to the perfection of all the letters in it.The perfect degree of each letter can be assigned by you, different letters of the perfect degree, respectively, corresponding to an integer between 1-26John doesn't care about the letter case. (That is, the letters F and F) are of the same degr

optimization, monotone queue optimization and so on. Each is a special, enough for you to study for a period of time, but also the knowledge is very practical, there are many topics if you do not take the initiative to find it difficult to meet, generally I look for the way to look at other people's blog by looking at other people's blog tags or blog mention of the term, And then I'm going to search and see what it is.To lift a few chestnuts, slope optimization DP recommends this article:Http:/

The basic thought of deep learningSuppose we have a system s, which has n layers (S1,... SN), its input is I, the output is O, the image is expressed as: I =>S1=>S2=>.....=>SN = o, if the output o equals input I, that is, input I after this system changes without any information loss (hehe, Daniel said, it is impossible.) In the information theory, there is a "message-by-layer-loss" statement (processing inequalities), the processing of a information

Topic Links:It's all on the mindTime limit:2000/1000 MS (java/others)Memory limit:65536/65536 K (java/others)problem DescriptionProfessor Zhang has a number sequencea1,a2,.. . ,an . However, the sequence isn't complete and some elements is missing. Fortunately, Professor Zhang remembers some properties of the sequence:1. For everyi∈{1,2,.. . ,N} ,0≤ai≤ .2. The sequence is non-increasing, i.e.a1≥a2≥. . ≥an .3. The sum of all elements in the sequence are not zero

Let $A =\{a_1$$\overline{\varepsilon} (A) =\limsup\limits_{n\to \infty}\frac{\log| a\cap\{1,2,\cdots,n\}|} {\log n}.$$We can similarly define the lower exponential density.We have the following result claimed in gaps and the exponent of convergence for an integer sequence.Main Result: $$\overline{\varepsilon} (a) =\tau (a) =\limsup\limits_{n\to \infty}\frac{\log n}{\log a_n}.$$Proof. Let $\overline{\varepsilon} (A) $$\limsup\limits_{n\to \infty}\frac{\log n}{\log a_n}\ge \limsup\limits_{n\to \in

ChannelTest instructions: A point on a two-dimensional coordinate that builds a length and a minimum tree containing all pointsIdeas:Define state Dp[i,j] represents the minimum cost (length of the branch) that the point I points to J merges togetherState transition equation: dp[i,j]= min (dp[i,k]+dp[k+1,j]+cost (i,j)) iCost (I,J) =py[k]-py[j]+px[k+1]-px[i];Cost (I,J) monotonically decreasing function when J is fixedWe guess cost (i,j) satisfies the quadrilateral inequalitiesProof: F (i) =cost (i

This article transferred from: http://www.cnblogs.com/void/archive/2011/08/26/2153928.htmlAlways do not know what type of differential constraint is the problem, and recently in writing the shortest way to look at the next, the original is to give some form as x-yThe magic is that this kind of problem can be converted into the shortest path problem in graph theory, the following begins in detail:(1) For example, given three inequalities, B-ABy the que

Poj_1755 At first, I thought it was necessary to enumerate three routes, but after I thought about it, I only needed to enumerate two routes, because I could regard the total distance as a fixed value and set it to INF. In this way, for contestant I, if the first item In x length and the second item in Y length can run faster than any contestant J, then there will be X/V [I] + Y/U [I] + (INF-x-y) /W [I] But there are several questions to consider: The first one is whether A1 and A2 are

, there is a big gap between what graduate students learn in college and practical applications. For example, many teachers in the computer department of a university do not know how software products are actually developed. The programming and project management mentioned above are both in books and theoretically, therefore, the enthusiasm of graduate students is naturally poor. What is the status of graduate education?In summary, if the tutor earns money from the tutor, the student must

Interval DP, cut DP[I][J] is not cost-independent (no validity)DP[I][J] represents the cost of interval i,j, so as long as the enumeration of the cutting method is done, the interval is divided into smaller intervals. O (n^3)Looked at the quadrilateral inequalities, proved too long.#include //variables don't take ignore left or something like that.using namespacestd;Const intMAXN =Wuyi;intCUT[MAXN];intDP[MAXN][MAXN];Const intINF =0x3fffffff;intMain ()

"Dichotomy + Computational Geometry" HDU 4033 Regular polygon Topic Link: hdu 4033 Regular Polygon TopicThe edge length of the polygon is known by the distance from one of the inner points in the regular polygon to all vertices.Binary problems generally exist with an unknown amount of equation (equation relationship), through the scope of the two unknown amount of search, the geometric relationship of this topic is: Know one side, the internal angle and (around the inner point) equals 360 degree

Title Link: http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=34651IdeasDifferential constraint system.Set the operation on node U and Sum[u], then the Edge (u,v) weight is d-sum[v]+sum[u]. For the biggest problem with the minimum value we think of the two-point answer, set two points to X, then the problem becomes the question whether there is a solution to determine the minimum value of x. For weights we have inequalities d-sum[v]+sum[u]>=

value, that is, |x (n) |(4) theory has proved that in the program design of autocorrelation Linear Predictive Coding (LPC), the reflection coefficient Ki satisfies the following inequalities: |ki|2. Statistical analysis methodFor variables that cannot be defined in theory, statistical analysis is used to determine its dynamic range. The so-called statistical analysis, is to use enough input signal sample value to determine the dynamic range of variab

Title: http://www.lydsy.com:808/JudgeOnline/problem.php?id=3143Analysis:Easily if you know the mathematical expectations of each side, then you can be greedy to the desired size of each side, so the question is how to ask for the expectation of each side.If you can't find a way to ask, you can first find the expectations of each point.Easy to get f[i]=∑f[j]/d[j] j->i with sideSpecial, for the starting point, because it is in the beginning, so it should be f[1]=1+∑f[j]/d[j], for the end, to the f

."Problem-solving ideas"This is a comparison of water noi of the topic, mainly and check the application of the set, first sort, put the equation in one and check the concentration, and then to check the inequalities, if there is not satisfied, then output ' no ', all meet the output ' YES '.Simple and check set can get 90 points, after discretization can get full marks (obviously, I will not write)1 typeeqq=Record2 L,r:longint;3 End;4 var5EQ,NEQ:Arra