precalculus inequalities

#1223: InequalitiesTime Limit:1 SecMemory limit:256 MBTopic Connection http://hihocoder.com/problemset/problem/1223DescriptionGiven n an inequality about X, the maximum number of questions is set.Each inequality is one of the following forms:X X X = CX > CX >= CInputThe first line is an integer n.The following n rows, one inequality per line.Data range:1OutputAn integer line that represents the maximum number of inequalities that can be set at the sam

single positive integer (each one representing a mouse). If 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 Inpu

][J] =Maxnum; - for(intK = I;k ) +Dp[i][j] = min (dp[i][j],dp[i][k] + dp[k+1][J] + sum[j]-sum[i-1]); A } at } -cout1][n]; - return 0; -}View CodeWikioi 3002 Stone Merge 3Title DescriptionDescriptionThere are n heap of stones in a row, each pile of stones have a weight of w[i], each merge can merge adjacent two piles of stones, the cost of a merger is the weight of two piles of stone and w[i]+w[i+1]. Ask what sort of merger sequence it takes to minimize the total merg

Test instructions: Given some intervals, the length of a set requires that there are at least two sets in each interval.Solution: Greedy or differential constraint. Greedy idea is very simple, as long as the interval by the right edge of the order, if the last two elements in the collection is not within the current interval, the number of the last two in the interval is added to the set, if only one element in the interval to add one, if the two elements are not added in the interval.The differ

dp/Quadrilateral Inequalities Naked title Ring Stone Merge ...Remove the ring as a chain1 //hdoj 35062#include 3#include 4#include 5#include 6#include 7#include 8#include 9 #defineRep (i,n) for (int i=0;iTen #defineF (i,j,n) for (int i=j;i One #defineD (i,j,n) for (int i=j;i>=n;--i) A #definePB Push_back - #defineCC (A, B) memset (A,b,sizeof (a)) - using namespacestd; the intGetint () { - intv=0, sign=1;CharCh=GetChar (); - while(!isdig

function of VC dimension H and also the subtraction function of sample number N.The definition of VC dimension is described in detail in svm,slt. E relies on H and N, shut up expects the least risk, and only cares about its upper bound, i.e. e minimization. So, you need to choose the right H and N. This is the structural risk minimization of structure Risk MINIMIZATION,SRM.SVM is the approximate implementation of SRM, the concept of SVM has another big basket. Stop there.Physical meaning of 1 n

dp/Quadrilateral Inequalities After the POJ 1739 Post office that question, it is easy to write out the rules of the equation:DP[I][J]=MIN{DP[I-1][K]+W[K+1][J]} (indicates the minimum cost of dividing the first J points into blocks i)$w (l,r) =\sum_{i=l}^{r}\sum_{j=i+1}^{r}a[i]*a[j]$Then there is $w (l,r+1) =w (l,r) +a[j]*\sum\limits_{i=l}^{r}a[i]$So: W[i][j] clearly satisfies the monotonicity of the interval inclusionsThen we boldly guessed tha

Linear programmingFirst of all, generally all linear programming problems can be converted to the following standard type :But we can see that all of these are inequalities, and we prefer the equation in our calculations, so we introduce this new concept: slack type :It is clear that our final request is that all constraints to the left of the variable are not less than 0. To solve such problems, we have a very convenient model algorithm: simplexBase

Set $a _1, a_2, \cdots, a_n\in\mathbf{n^*}$, and different. Verification: $${a_1\over1^2} + {a_2\over2^2} + \cdot + {A_n\over n^2} \ge {1\over1} + {1\over2} + \cdots + {1\over n}.$$Solution One:Consider using the basic inequalities $a + b\ge 2\sqrt{ab}$ to eliminate the molecules on the left. $$\because {A_k\over 1^2} + {1\over a_k} \ge 2\cdot{1\over k},\ k=1, 2, \cdots n.$$ $$\therefore \sum_{k=1}^{n}{a_k\over k ^2} + \sum_{k=1}^{n}{1\over A_k} \ge 2

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 14006000 12002000 1900Sample Output44597Sort by decreasing the weight increment rate and finding the biggest string is mainly the path problem.#include #include#inclu

bunch of inequalities, I can't testify anyway)#include #include#defineEPS 1e-8#defineN 1000010using namespacestd;intedgenum,n,k;intVet[n],head[n],size[n],son[n],next[n];DoubleF[n];Doubleans=0;voidAddintUintv) {Edgenum++;vet[edgenum]=v;next[edgenum]=head[u];head[u]=Edgenum;}voidDfsintu) { inte=head[u];size[u]=1; son[u]=0; while(e>0){ intv=Vet[e]; DFS (v); Son[u]+=size[v];size[u]+=Size[v]; E=Next[e]; } e=head[u];if(e==0) {F[u]=1.0; if

https://vjudge.net/problem/UVA-11478Given a graph, each edge has a weighted value. Each time you can select a node v and an integer d, reduce the weight of all edges at the end of the V by D, increase the weight of all edges starting at V by D, and finally let the minimum value of all edges be greater than 0 and as large as possible. Damn, the book was wrong. >0 not nonnegative wa several times because of thisConsidering the constraints of each edge, Di represents the Halum amount of IW-dv

3002 Stone Merge 3time limit: 1 sspace limit: 256000 KBtitle level: Diamonds Diamond SolvingTitle DescriptionDescriptionThere are n heap of stones in a row, each pile of stones have a weight of w[i], each merge can merge adjacent two piles of stones, the cost of a merger is the weight of two piles of stone and w[i]+w[i+1]. Ask what sort of merger sequence it takes to minimize the total merger cost.Enter a descriptionInput DescriptionFirst line an integer n (nSecond row n integers w1,w2...wn (WI

characters at the end of the size is returned 0, if there is no comparison to the location of the "\" size is also returned corresponding value. If the current element of the str1 current element is greater than str2 returns an integer, otherwise a negative number is returned.Code implementation: #include #include int my_strcmp (const char *STR1, const Char *str2)/*STR1 and str2 do not need to change */ { assert (str1); assert (STR2); while (*STR1==*STR2) { if (*str1== ')/* can indicate

Test instructions: There are in the kindergartenNa little friend,LXHGWWThe teacher now wants to assign sweets to the children and ask every child to get the sweets. But the children also have jealousy, will always put forward some requirements, such as Xiao Ming does not want to small red points to the candy more than his, so in the allocation of sweets,LXHGWWneed to meet the children'sKa requirement. Kindergarten sweets are always limited,LXHGWWto know how many sweets he needs at least to make

2 Lagrangian duality (Lagrange duality)Put aside the above two planning questions first to look at the existence of equality constraints of the extremum problem, such as the following optimization problem:The objective function is F (w), and the following is the equality constraint. The usual solution is to introduce the LaGrand day operator, which is used here to represent the operator and get the Lagrange formula asL is the number of equality constraints.Then, the partial derivative is equal t

, and only if, x = cy, the equation is set.16) We then prove the triangular inequalities. Theorem is |x + y | 17) We then draw the definition of the vector angle, which is introduced by the analogy of a geometric triangle. The expression of the vector angle is obtained by cosine theorem, cosθ= a b/|a| | b|18) by introducing the concept of vector angle, we draw the method of judging vector perpendicular. If A. b = 0. Then two vectors are intersecting a

smaller towns with distances less than 20km, and try to determine whether the company can make the entire region's residents hear the radio program normally.inputThe first behavior is two integer n,m, which is the number of small towns and the small town pairs that are next less than 20km. The next M-line, 2 integers per line, indicates that the distance between the two towns is less than 20km (numbering starts from 1).OutputIf the requirement is met, Output 1, otherwise output-1.Input Sample4

Tree-related knowledge points:Relative to the Tsinghua University Press this "discrete mathematics" about the tree of this chapter is relatively simple, the concept is not much, about trees, forests, leaves, sub-fulcrum, spanning tree, the minimum spanning tree concepts are very simple, here no longer, the following records several theorems and important algorithm steps.Theorem 1: Set tProof: Set T has k leaf, then there are n-k branch points (the degree of the fulcrum is greater than or equal t

Set out the $x,y,z$ three unknown quantity to represent three kinds of units of combat effectiveness respectively.Then all kinds of inequalities can be expressed as $ax+by+cz\geq 0$ form.Notice the $z>0$, then divide both sides by $z$ to get $ax+by+c\geq 0$.Then after the half-plane intersection to find all vertices, all vertices are brought into the evaluation for each query.#include 　　BZOJ1829: [Usaco2010 mar]starc StarCraft