round to nearest million calculator

the height of the specified node, in fact, such as the following:Finds the height of the node p. Note with only the height of the tree alone is different int getheight (bitnode* root, Bitnode * p, int h = 1) {if (!root) return 0;if (p = = Root->lchild | | p = = Root->rchild) return H + 1;return getheight (Root->lchild, p, h+1) = = 0? GetHeight (Root->rchild, P, h+1): GetHeight (Root->lchild, p, h+1);}The first sequence of tests used is listed asabc## #de # #fg # # #The corresponding two-fork t

Test instructions: Given a tree, ask the nearest public ancestor of two nodes.Input: The first line T, which indicates the number of test groups.Each set of test data contains an n, which indicates the number of nodes, the n-1 line below is the connected edge, and the last line is the queryOutput: Total T-line, representing the test results for each group/*Multiply LCA Note this is a tree, so the side is one-way, deep search from the root node to star

Solution One:The total number of n in the array, and the 22 difference to find out, you can get the minimum value pair. The time complexity is O (N2). N2 the square of the value nThe code is as follows:Double mindifference (double arr[], int n){if (n return 0;Double Fmindiff = fabs (arr[0]-arr[1]);for (int i = 0; i for (int j = i + 1; j {Double tmp = fabs (Arr[i]-arr[j]);if (Fmindiff > TMP)Fmindiff = tmp;}return Fmindiff;}Solution Two:if the array is ordered, an O (n*log2n) algorithm can be used

3Sum ClosestGiven an array S of n integers, find three integers in S such, the sum was closest to a Given number, target. Return the sum of the three integers. You may assume this each input would has exactly one solution.For example, given array S = {-1 2 1-4}, and target = 1.The sum is closest to the target is 2. (-1 + 2 + 1 = 2).Idea: This question and the previous question 3Sum same strain, just deformed. So the whole solution of this problem refer to the above question, the details are slig

If the segment is parallel to the x-axis (y-axis), the point is too far from the line where the segment is perpendicular, perpendicular is easy to calculate, and then calculates the perpendicular, if perpendicular on the line is returned perpendicular, otherwise the end of the perpendicular near the endpoint; The two ends of the line segment are PT1 and pt2, with a slope of k = (pt2.y-pt1. Y)/(pt2.x-pt1.x); the linear equation is: y = k* (x-pt1.x) + pt1.y. Its perpendicular slope is-1/k, the ver

intN;6 intcmpConst void*a,Const void*b)7 {8 return*(int*) a-* (int*) b;9 }Ten voidSearchintx) One { A intmid,l,r,ans=-1; -mid=n/2; -L=0; theR=N; - while(l1) - { - if(a[mid]>x) r=mid; + ElseL=mid; -Mid= (r+l)/2; + } A if(L +1==n| | x-a[l]1]-x) ans=A[l]; at Elseans=a[l+1]; -printf"%d\n", ans); - } - intMain () - { - intm,f; in inti; -scanf"%d",n); to for(i=0; i) +scanf"%d",a[i]); -Qsort (A,n,sizeof(int), CMP); thescanf"%d",m); * while

Asked yesterday, so I immediately thought of a method. The problem is that given an unordered array, finding the nearest K number to the median requires O (n) complexity. The idea is as follows: First, we need to know that the median of unordered Arrays can be obtained within the O (n) time by using linear time selection. (1) O (n) calculates the median of unordered arrays. (2) subtract the median from all the numbers in the array and tak

the array. I know that there are several training data sets.DiffMat = tile (partition, (dataSetSize, 1)-dataSet # tile: Functions in numpy. Tile expands the original array into four identical arrays. DiffMat obtains the difference between the target and the training value.SqDiffMat = diffMat ** 2 # each element is squareSqDistances = sqDiffMat. sum (axis = 1) # multiply the corresponding column to obtain the square of each distance.Distances = sqDistances ** 0.5 # Start, get the distance.Sorted

distance from the sample) to improve. Another disadvantage of this method is that the calculation workload is large, because the distance from the text to all known samples must be calculated for each text to be classified before K Nearest Neighbor points can be obtained. Currently, the common solution is to edit known sample points in advance and remove samples that do not have much effect on classification. This algorithm is more suitable for autom

() forJinchRange (32): Returnresult[0,32*I+J] =Int (liststr[j])returnReturnresultdefhandwritingclasstest (): Trainingfilelist= Listdir ('trainingdigits') M=Len (trainingfilelist) #Get Training DataTrainingmat = Zeros ((m,1024)) Hwlabels= [] forIinchRange (m): Filenamestr=Trainingfilelist[i] Filestr= Filenamestr.split ('.') [0] Classnumstr= Int (Filestr.split ('_') [0]) hwlabels.append (CLASSNUMSTR) trainingmat[i,:]= Image2vector ('trainingdigits/%s'%filenamestr) Testfilelist= Listdir (

Def classify0 (inx,data_set,labels,k):DATA_SET_SIZE=DATA_SET.SHAPE[0]. shape[0] indicates How many rows to get Data_set,. shape[1] that gets the number of columnsalso note The use of brackets after shapeDiff_mat=tile (InX, (data_size,1))-data_sizeTile can extend an array, the above code is the InX in the array copy data_size , anyway, it is not good to express, you can Baidu. Using this will save you from using a for loop. Sq_diff_mat=dif_mat**2Sq_distances=sq_diff_mat.sum (Axis=1)Axis=1 mean

2. When predicting data classification, the ' Dict ' object has no attribute ' Iteritems 'Such as:The most common workaround is to change the order of environment variablesSuch asNote: Which version is above, and who is the Python version in CMD.Such asAnother example:Then you can do this by predicting the classification of your data:Implementation of KNN classification algorithm based on K-nearest neighbor algorithm in machine learning combat

) - returnSORTEDCLASSCOUNT[0][0]knn-classify03.3 Using KNN in the above data and calculating the rate of miscarriage1 defdatingclasstest ():2HoRatio = 0.50#Hold out 10%3Datingdatamat,datinglabels = File2matrix ('DatingTestSet2.txt')#Load Data setfrom file4Normmat, ranges, minvals =autonorm (Datingdatamat)5m =Normmat.shape[0]6numtestvecs = Int (m*hoRatio)7Errorcount = 0.08 forIinchRange (numtestvecs):9Classifierresult = Classify0 (normmat[i,:],normmat[numtestvecs:m,:],datinglabels[numtes

https://oj.leetcode.com/problems/3sum-closest/Similar to the 3sum. The difference is that we need to approximate a value this time, in fact, similar to equality, with the L and R pointers constantly moving, and then repeatedly take the smallest.classSolution { Public: intn,m; intThreesumclosest (vectorint> num,inttarget) {Vectorint> a=num; N=a.size (); Sort (A.begin (), A.end ()); intms=numeric_limitsint>:: Max (); intres=-1; for(intI=0; i){ intx=A[i]; intl=i+1, r=n-1; whi

"%(Classifierresult, datinglabels[i]))if(Classifierresult! = Datinglabels[i]): Errorcount + = 1.0Print("The total error rate is:%f"% (errorcount/float (numtestvecs))) Print(Errorcount)defimg2vector (filename): Returnvect= Zeros ((1,1024)) Fr=open (filename) forIinchRange (32): Linestr=Fr.readline () forJinchRange (32): Returnvect[0,32*I+J] =Int (linestr[j])returnReturnvectdefhandwritingclasstest (): Hwlabels=[] trainingfilelist= Listdir ('trainingdigits')#load the training setm =Len (trainingfil

1. What is k nearest neighbor Popular Will, if I were a sample, the KNN algorithm would be to find a few recent samples, see what categories they all belong to, and then select the category with the largest percentage of their category. KNN is the full name of K-nearestneighbor,k is the number of samples we are looking for, k=1 is to find the most recent samples, and then their own category is the category of the sample.The KNN also calculates its di

This is Lintcode's title: http://www.lintcode.com/zh-cn/problem/subarray-sum-closest/ Give an array and K, find a sub-array, and nearest k, if there are multiple returns to any one. Of course n squared can be solved, but extra space can be onlogn. The idea is to use the S (i)-S (j) approach. First calculate all the S (i), then deposit Map,key is the position of the last element Sum,value. Then the sum is sorted, and then the sum is traversed once, it

Tags: long search limit Data sel technology share structure latitude and longitude return database table Structure 2. The key is this SQLSELECT Address_, Longitude_, Latitude_, ROUND (6378.138 * 2 * ASIN (SQRT (POW () (? * PI ()/180-latitude_ * PI ()/180)/ 2), 2) + cos (? * PI ()/+) * cos (latitude_ * PI ()/+) * POW (SIN ((? * PI ()/180-longitude_ * PI ()/180)/2), 2)) ) * +) as Distance_um from xf_lfhz_health_roll_addires where business_id =? ORDER by Distance_um ASC LIMIT 0, 1SqlThis SQ

, you can search in a smaller scope, rather than all users. So for this example, I've added 4 where conditions, only for distance calculations for users with longitude and latitude greater than or less than 1 degrees (111 km) of the user, and an index to the longitude and latitude two columns in the data table to optimize the speed at which the where statement executes. The final SQL statement is as follows $sql=SELECT * from users_location where latitude >‘.$lat.‘-1 and Latitude ‘.$lat.‘+1 and

A supervised KNN neighbor algorithm:(1) Calculate the distance between the points in a well-known category dataset and the current point(2) Sorting in ascending order of distance(3) Select K points with a minimum distance from the current point(4)