precalculus inequalities

optimize the state of the transfer of purpose 2823 Left-leaning tree can merge heap Suffix tree a very useful data structure is also a hotspot for the contest questions 3415 3294 Search More troublesome search topics training 1069 3322 1475 1924 2049 3426 Wide-search state optimization using m-binary number storage state, conversion to string hash table weight, bitwise compression storage state, bidirectional wide search, A * algorithm 1768 1184 1872 1324 2046 1482

Description a row of n piles of stones in a playground. It is now necessary to merge the stones into a pile in order.　　It is stipulated that each of the adjacent 2 stones can be merged into a new pile each time, and a new pile of stones is counted as the score of the merger.Try to design an algorithm that calculates the minimum score to combine n heap stones into a pile.　　The first line of Input is a number n.The following n rows are a number a per line, representing the number of stones. Output

the right-hand If the equal sign is the same as the slope, then the two lines parallel the vector, A, a, and a collinear line.But this is the case where the slope of both lines exists. Let's discuss the situation where the slope does not exist:(1). The slope of the line where a is located does not exist as \ (a.x=0\).Wakahara inequality is still established as \[0>a.y*b.x\], and then discuss \ (a.y\) positive and negative, you can prove that the line B is still located in a straight line of the

used to easily prove some inequalities, such as (f (x) =\ln{x},\,p_k=\dfrac{1}{n}\), can prove \ (\prod x_k\leqslant \dfrac{1}{n}\sum x_k\).\[f (X_k) =f (x) +f ' (x) (X_k-x) +\frac{1}{2}f "(\xi) (x-x) ^2\geqslant f (x) +f ' (x) (x_k-x) \tag{5}\]\[\SUM\LIMITS_{K=1}^N{P_KF (X_k)}\geqslant f\left (\sum\limits_{k=1}^n{p_kx_k}\right) \tag{6}\]1.2 Differential 1.2.1 tangent and normal plane of multi-element functionNow, using the method of differential to

This is a half-plane of the naked problem, the first half-plane to write, just say I half-plane cross understanding it.The so-called half-plane intersection, is to find a large pile of two of yuan an inequality of the intersections, and each two yuan of the solution set of an inequality can be seen in a straight line above or below the standard equation of the contact line can be obtained. So these inequalities can be converted to some half-plane, whi

Title Link: http://poj.org/problem?id=1201Test instructions: given n (1Idea: S[i] represents the number of less than equals I, so that you can directly follow the input to establish an inequality after the conversion to a network can be;It is important to note that the original two inequalities s[i-1]-s[i] In fact, only after processing the input edge, linear change monotonicity can be;#pragma COMMENT (linker, "/stack:1024000000,1024000000") #include

1. The concept of linear programmingLinear programming is to study the extremum problem that makes a linear objective function take maximum (or minimum) under a set of linear inequalities or equality constraints .2. Standard form of linear programmingfeatures : The objective function is great , the equality constraint , and the variable is non-negative .MakeThe matrix expression of the linear programming standard form is:Conventions:How to make a stan

link according to the above information, and output this quantity if there is an infinite number of possible output-1.AnalysisTo see the information of a bunch of inequalities, the first thought is the differential constraint,When the total length s is determined, we can do the difference constraint, judge whether there is a negative ring to determine the legalityConnecting Edge 对于dis(Ai,Bi)>=Di， 如果AiThen with the Bellman_ford, after the n-

Test instructions: Give the positive integer k, find all the integers so that the 1/k=1/x+1/y is satisfiedAccording to the basic inequalities, k1#include 2#include 3#include 4#include 5#include 6#include 7#include 8#include 9#include Ten #definemod=1e9+7; One using namespacestd; A -typedefLong LongLL; - the intMain () { - intk,i,j,a,b,x,y; - while(cin>>k) { - intans=0; + for(y=1; y2*k;y++){ - if(y!=k) x= (k*y)/(

these n integers is m[1], m[2],..., M[n] Then it must is the case thatW[m[1]] andS[M[1]] > s[m[2] [> ... > S[m[n]]In order for the answer to is correct, n should be as large as possible.All inequalities is strict:weights must is strictly increasing, and speeds must be strictly decreasing. There may is many correct outputs for a given input, and your program only needs to find one.Sample Input6008 13006000 2100500 20001000 40001100 30006000 20008000 1

Test instructionsAt the beginning of the skating club there are 1 to n skates each K-double, the X-foot person can wear X-to-x+d skates,;There are m operations, each containing two number ri,xi, representing the person who came with the XI Ri number foot (xi may be negative);For each operation, the output skates are sufficient;nExercisesFirst of all, this is a two-figure matching problem, obviously there is no intersection between shoes and people;Then there is a hall theorem:The set of two vert

Topic Link:https://uva.onlinejudge.org/index.php?option=com_onlinejudgeitemid=8page=show_problem problem=2473ExercisesFirst we can get the constraints of shape such as Xi-xj If there is a negative circle, the original difference system has no solution.Simple proof:We might as well set this ring to X1,X2...XN.That is, there are inequalities x1 All add up to 0 Therefore, the solution must satisfy Sigma (Y) >= 0, if there is a negative circle, there is d

-independent coordinates. 1. Non-integrable differential constraint with time derivative 2. Solvable constraints/single-sided constraints with inequalities To solve a single-sided constraint method: 1. Constraint is not solvable, the constraint equation takes the equation 2. Remove the constraint and add an independent coordinate the "generalized coordinates" still describe the spatial location. The number is the same as the Freedom s . Spanned s -dim

Chapter 1 introduce to VectorsTwo core operations of linear algebra: vector addition and multiplication. Combining these two operations is a linear combination of vectors.1.1 Vector and Linear combinationsTwo basic operations: vector addition and multiplication.two x vector v and W linear combination of : Cv+aw. where C and a are arbitrary scalars.1.2 Lengths and Dot productsdot product : The inner product of a vector. ------------- in relation to length and angle, the "geometry" part of linear

driving plan cannot reach the current petrol station. In other words, to enable the car to continue to move forward, it is necessary to ensure that the sum of C is always greater than 0.If the cost and the sum of gas than gas, obviously the car will not be able to complete a lap, the following proof if cost and less than the gas and the sum, there must be a solution to allow the car to complete a lap.Now c[0]+c[1]+...+c[n-2]+c[n-1]>=0 , we sum the first I of C, assuming that when i=j, this sum

Linear programmingConstraints, a linear function is reached to the extremum. That Both the objective function and the constrained linear programming are called linear programming.Common FormsLinear programming is convex optimizationConvex optimization:convex function planning on convex sets is called convex programming.It can be proved that the linear set is a convex set, which satisfiesA linear function is a convex function, which isBut not strictly convex.Unified formTo facilitate a unified so

Reference: Http://www.cnblogs.com/jackge/p/3231767.html speaks very well.Sentiment: the best situation must be a complete picture, but not allowed, so must be some conform to u->v, but V can not go to u,In this case, maximize the number of edges, eventually forming two graphs, then apply inequalities#include #include#include#include#include#includestring>#include#include#include#include#includeusing namespaceStd;typedefLong LongLL;Const intn=1e5+5;Con

) =e^{-x}-\left (1-\frac{x}{a}\right) ^{a},x\in [0,a]$. Determine the positive or negative of its minimum value. $f (0) =0,f (a) =e^{-a}>0$.$ $f ' (x) =-e^{-x}+\left (1-\frac{x}{a}\right) ^{a-1}$$If $f ' (x) $ no 0 points, then $f (x) $ maximum at the end of the minimum value, that is, $f (x) >0,x \in [0,a]$. If there are 0 points, just take advantage of the 0-point nature.The other side proves similar to the inequality.5. (Jordan inequality) set $0\leq x \leq \frac{\pi}{2}$, then an inequality

/*Two-point answer (note accuracy) for each answer (S1+S2+S3 ...) /(T1+t2+t3 ...) >=ans when the ANS has a larger space for the above inequalities if the enumeration of each road appears too violent to degenerate into: S1-t1*ans+s2-t2*ans+s3-t3*ans ... >=0 differential constraints run the longest way if dis[n]>0 or have a positive ring (beginning this forget) ans is legal*/#include#include#include#includeConst Doublejing=0.0001;using namespacestd;Doub

The core of this section is how to relate the hoeffding inequalities to the feasibility of machine learning.This PAC is very image and accurate, describing the "current possibility is probably right", that is, a probability of the last.Hoeffding's connection to machine learning is:If the number of samples is large enough, the learning effect obtained on the training set can be translated onto the test set. In other cases,Here is only the "training set