how to round to nearest whole number

Link Storage Stack API details see my blog: Chained storage for stacks-API implementationsNearest matchAlmost all compilers have the ability to detect if parentheses matchHow do I implement symbolic pair detection in the compiler?#include Algorithm IdeasStart scanning from the first characterIgnored when ordinary characters are met,Pressed into the stack when the left sign is metPOPs the top symbol from the stack when the right symbol is met and matchesMatch succeeded: Continue reading into next

# ALGORITHM:K-NN (nearest neighbor classification algorithm)# AUTHOR:KERMIT.L# time:2016-8-7#==============================================================================From numpy Import *Import operatorImport Matplotlib.pyplot as PltDef creatdataset ():Group = Array ([[1.0, 1.1], [1.0, 1.0], [0, 0], [0, 0.1]])Labels = [' A ', ' a ', ' B ', ' B ']Return Group,labelsGroup,labels = Creatdataset ()Plt.figure (1)Plt.plot (group[:,0],group[:,1], ' or ')P

#-*-coding:utf-8-*-__author__= ' Ghostviper ' "" "K Nearest neighbor Collation Algorithm" "" fromnumpy Import*importoperatordefcreatedataset (): group=array ([ [1.0,1.1], [1.0,1.0],[0,0], [0,0.1]]) labels=[' A ' , ' A ', ' B ', ' B ']returngroup,labelsdefclassify0 (InX, dataset,labels,k): #shape Get the size of the array in the dataset datasetsize=dataset.shape[0]# Copy the matrix based on the input elements and make a difference to the known data se

Algorithms based on nearest neighbor, often used in various situations,For example, 100,000 users, for each user to find the most similar users,When N is particularly large, the efficiency is not very high, such as when the n=10^5, it is not very good, because the violence of the time complexity of the law is O (n^2).This requires special means, there are two commonly used methods, one is the Kdt tree (and ball trees), one is a local sensitive hash (a

Sub-arrays and the closest 0 issues:For array A of length n, the values of the contiguous Subarray and the nearest 0 are obtained.such as: 1,-2,3,10,-4,7,2,-5; the array of neutrons and the value closest to 0 is 0, and the Subarray is -4,7,2,-5.Program implementation:1 /***************************************2 FileName NearZeroSubArray.cpp3 Author:godfrey4 CREATEDTIME:2016/5/35 ****************************************/6#include 7#include 8#include 9#i

Working principle:Given a training dataset, for a new input instance, find the nearest K-instance to the instance in the training dataset (that is, the K-neighbor above), where the majority of the K-instances belong to a class, and the input instance is categorized into this class.code example:knn.py from Import *import operatordef CreateDataSet (): = Array ([[[1.0,1.1],[ 1.0,1.0],[0,0],[0,0.1]]) labels= ['b','b ','b','b'] return Group,lab

Taiwan Big machine skill and cornerstone are finished, but no programming has been, now intends to combine Zhou Zhihua "machine learning", the machine to learn the actual combat, the original book is Python2, but I feel python3 better use some, so plan to use Python3 write it again. Python3 and Python2 different places will be in the program in the bid.Code and data: HTTPS://GITHUB.COM/ZLE1992/MACHINELEARNINGINACTION/TREE/MASTER/CH2K-Nearest Neighbor

Recent point to question:In n points on a two-dimensional plane, the distance of the nearest pair of points is fast.Java implementations:Package P2;import static Java.lang.math.*;import Java.util.arraylist;import Java.util.arrays;import Java.util.collections;import Java.util.comparator;import Java.util.scanner;public class CloestPair {//coordinate point static Class Point {public ' point () {}public point (Double x, double y) {this.x = X;this.y = y;}

The example in this article describes the method that jquery implements to find the nearest parent node. Share to everyone for your reference, specific as follows: This shows the nearest table to find the current control More interested readers of jquery-related content can view the site: A summary of jquery operations Dom node methods, a summary of jquery extension techniques, a summary of c

Import java.util.*; To find the nearest common ancestor public class getlowesttree{//Two fork tree node in a binary tree two nodes are defined as the publicly static class node{public int value; Public Node left; public Node right; public Node (int data) {this.value=data; }///Find the nearest public ancestor of two nodes Getlowestancestor (node Head,node n1,node n2) {//Recursive exit if (head==nu

1. Background In the future, the blogger will update the machine learning algorithm and its Python simple implementation regularly every week. Today's algorithm is the KNN nearest neighbor algorithm. KNN algorithm is a kind of supervised learning classifier class algorithm. What is supervised learning and what is unsupervised learning? Supervised learning is the algorithm used when we know the target vector, unsupervised learning is used when we do

The following three scenarios are required to give a solution and resolve: 1: Two fork tree is the search binary tree. 2: Two fork tree each node has a parent (three fork chain) 3: It is the ordinary two-fork tree. (Achieve time complexity as O (N))For the first case, a binary search tree, we can determine the location of the two nodes according to the size of the data, if the root node is on the 2 side of the root node is the nearest public ancestor,

Paste the Python code belowknnclassify.py1 fromNumPyImport*2 Importoperator3 4 defCreatdataset ():5Group = Array ([[[1.0,1.1],[1.0,1.0],[0,0],[0,0.1]])6Labels = ['A','A','B','B']7 returnGroup, Labels8 9 defclassify (inx,dataset,labels,k):TenNumSamples =Dataset.shape[0] OneDiffmat = Tile (InX, (numsamples,1))-DataSet ASqdiffmat = diffmat**2 -Sqdistances = sqdiffmat.sum (Axis = 1) -distances = sqdistances * * 0.5 theSorteddistindicies =Distances.argsort () -ClassCount = {} - forIinchxran

Http://codeforces.com/problemset/problem/598/CThe main idea is to give you the starting point of a vector vector is the minimum output of the 0,0 from the beginning of the (*) vectorThis is the first contact with the polar angle.The distance from a point to the origin of a function image is the polar diameter, and the angle between the polar diameter and the x-axis is the polar angle.Sorting by polar angle is from the rightmost beginning of the third quadrant to the far right of the second quadr

The K approximation algorithm is an instance-based learning, and when using the algorithm we have to have an example training sample data close to the classification results.Advantages: High accuracy, insensitive to outliersDisadvantages: Time complexity and space complexity are relatively large. (If the training sample dataset is large, it takes a lot of space to hold the data, and it takes time to predict the data and train the sample data set for each piece of data.) ） The infrastruc

); $ }Panax Notoginseng //blacked out this point -VIS[RT] =1; the //dealing with RT-related queries + for(inti =0; I ) { A intQV =Que[rt][i]; the if(Vis[qv]) { +Ans =fin (qv); - return true; $ } $ } - return false; - } the - intMain () {Wuyi intTC; thescanf"%d", TC); - while(tc--) { Wuscanf"%d", n); - init (); About for(inti =0; I 1; i++) { $ intu, v; -scanf"%d%d", u, v); - Tree[u].push_back (v); -deg[v]+

/************************************************************************//* non-recursive method The following is a simple algorithm with an O (n) complexity to solve the LCA problem 1 The path from root to N1 is found and stored in a vector or array. 2) Locate the path from the root to the N2 and store it in a vector or array. 3) Traverse these two paths until a different node is encountered, and the previous one is the lowest common ancestor. *//***********************************************

Test instructions: to seek LCA ... This tree is a one-way side, with an entry level of 0 for the root and only a set of queries.Solution: St For LCA (only this). DP[I][J] indicates who the first 2j ancestor of the first point, the transfer equation dp[i][j] = dp[dp[i][j-1]][j-1]. DFS one side record dp[i][0], and the depth of the point, the time of the LCA to the depth of the point to the same depth, if at this time two points coincident that the LCA is the current point, or else rise at the sam

This problem has a high accuracy requirement. Use atan2 ...#include #include#include#include#includeusing namespacestd;Const intmaxn=100000+Ten;intN;structx{Long Doublex, y; Long DoubleK; intID;} S[MAXN];BOOLcmpConstXa,Constxb) { returna.kB.K;}intMain () {scanf ("%d",N); for(intI=1; i) {cin>>s[i].x>>s[i].y; S[I].K=atan2 (s[i].y,s[i].x); S[i].id=i; } sort (S+1, s+1+n,cmp); s[0]=S[n]; Long Doublemin=99999; intAns1,ans2; for(intI=0; i) { Long Doublejiao=s[i+1].k-S[I].K; if(jiao0) Jia

] = true; For (INT I = head [u]; I! =-1; I = edge [I]. Next) {If (! Vis [edge [I]. to]) {tarjan_lca (edge [I]. to); Fa [edge [I]. to] = u ;}} if (U == Y ! Flag) {If (vis [x]) {printf ("% d \ n", findfa (x); flag = true; return ;}} if (u = x ! Flag) {If (vis [y]) {printf ("% d \ n", findfa (y); flag = true; return ;}} int T,, b; int main () {CIN> T; while (t --) {scanf ("% d", N); Init (); bool T1 [maxn] = {0}; rep (I, 1, n-1) {scanf ("% d", A, B); addedge (, b, 1); T1 [B] = true;} scanf ("% d