 关于比赛胜负的预测问题，怎么使用C语言的程序的编写的代码技术实现的呢？

Problem Description
Nils and Mikael are intergalaxial fighters. Now they are competing for the planet Tellus. The size of this small and unimportant planet is 1 < X < 10000 gobs. The problem is that their pockets only have room for one gob, so they have to reduce the size of the planet. They have available 1 <= K <= 6 FACTORweapons characterized by numbers F1, F2, . . . , Fk, all less than 0.9. As is commonly known, a FACTORweapon will blow off part of the planet, thus reducing the planet to a fraction of its size, given by the characteristic. Thus, with e.g. F1 = 0.5 an application of the first weapon will half the size of the planet. The fighter who reduces the size to less than, or equal to, 1 gob can take the planet home with him. They take turns attacking the planet with any weapon. If Nils starts, who will win the planet? Assume that both Nils and Mikael are omniscient and always make a winning move if there is one.
Technical note: To ease the problem of rounding errors, there will be no edge cases where an infinitesimal perturbation of the input values would cause a different answer.Input
The first line of input is N <= 100, the number of test cases. Each of the next N lines consists of X, K and then the K numbers F1, F2, . . . , Fk, having no more than 6 decimalsOutput
For each test case, produce one line of output with the name of the winner (either Nils or Mikael).Sample Input
4
6 2 0.25 0.5
10 2 0.25 0.5
29.29 4 0.3 0.7 0.43 0.54
29.30 4 0.3 0.7 0.43 0.54Sample Output
Mikael
Nils
Nils
Mikael
Go , SuSu 的编写_course
20200107Problem Description When SuSu and his friends visited GanQuan village , something dangerous happened on them . They were trapped in a mysterious cave , what was worse ,there was a terrible monster lived in the cave , so they had to escape from the cave as quickly as possible . But little monsters were also in the cave, as a result ,when SuSu was seen by the little monsters ,they would notice their boss to catch SuSu back. Little monsters' visual field is showed in figure A. The grid colored red represents where the monster stands, the arrow represents which direction the monster walk to ,and the grids colored blue and red represents where the very monster can see at this moment. The monsters often walks in a straight line, and when they faces a wall or the boundary of the maze in front of him, they will cost 1 seconds to turn back, and then walk back along the straight line, until reach a wall again. If SuSu was seen by the monsters ,it's impossible for them to escape from the cave . SuSu needs your help , please tell him the minimum time he needs to walk from the start point to the exit of the cave , be careful not to be seen by the monsters. Every seconds SuSu can move upward downward leftward and rightward , and can also make no movement. Figure A Input The first line of input gives the number of cases, T (at most 90). the first line of each case has four numbers n,m. (2<=n,m<=50) then n lines with m characters describe the maze 'A' represents the init position of SuSu. 'B' represents the exit position of the cave. '.' represents the grids can be walked on. '*' represents the wall which can not be stepped on. Then follows a number k (at most 50). Next k lines with three integers x , y , d (1 <= x <= n,1 <= y <= m,1 <= d <= 4).represents a monster walking to d direction is in (x,y) positon (the topleft grid is (1,1) ) at 0 seconds . The monster walks up when d == 1. walk down when d == 2.walk left when d == 3.walk right when d == 4. Output If SuSu can get to the exit in 1000 seconds ,output the minimum time he need. print "胜败兵家事不期 卷土重来是大侠" otherwise. Following the case number (start with 1). Sample Input 1 3 4 *.*. .A.B ***. 1 3 4 1 Sample Output Case 1: 2
排序，输出出现次数最多的颜色，用C语言怎么实现？_course
20181229Problem Description Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color and find the result. This year, they decide to leave this lovely job to you. Input Input contains multiple test cases. Each test case starts with a number N (0 < N <= 1000)  the total number of balloons distributed. The next N lines contain one color each. The color of a balloon is a string of up to 15 lowercase letters. A test case with N = 0 terminates the input and this test case is not to be processed. Output For each case, print the color of balloon for the most popular problem on a single line. It is guaranteed that there is a unique solution for each test case. Sample Input 5 green red blue red red 3 pink orange pink 0 Sample Output red pink
输出多项式的系数使得给定的方程成立，怎么用C语言的程序的设计的代码编写的思想方法来实现的_course
20190613Problem Description Li Zhixiang have already been in “Friendship” oceangoing freighter for three months. The excitement has gradually disappeared. He stands on the board, holding the railing and watching the dazzling ocean in the sun silently. Day after day, the same scenery is monotonous and tasteless, even the merry seagulls following the freighter cannot arouse his interest. Hearing the footsteps behind, he turns back to see the old captain is coming towards him. The captain has understood his idea, however, he starts a new topic with the young man. “Do you know how far our voyage is?” The captain asks. Li Zhixiang feels ashamed because he can not answer. Then the captain says with a smile, “5050 miles. Do you still remember the story of 5050?” This time the young man really blushes. The old captain continues saying:” You definitely know the story of 5050. When the German mathematician, “the prince of mathematicians”, Gauss was 10 years old …” Young man remembers this story and goes on to tell, “ When Gauss was 10 years old, he could add a list of integers from 1 to 100 in a few seconds, which shocked the teachers.” The old captain adds, “Gauss has many other stories like this. When he entered the university at the age of 17, he was able to construct heptadecagon by compass and straightedge. His university teachers were also impressed by his ability. Not only could college graduate students fail to do it, but also they felt hard to understand Gauss’s constructing process.” At this time, vicecaptain greets the old captain. The old captain says to Li Zhixiang: “Come over to my office tonight, let’s continue the conversation.” It is still calm and tranquil in the evening. The freighter travels smoothly on the sea in the silver moonlight. The captain tells the young man the following words. Among the mathematicians through the ages, there are three greatest mathematicians: Archimedes, Newton and Gauss. Most of Gauss’s mathematical achievements are difficult to understand. Nevertheless, there are some comparatively easy. For instance, when it comes to solving multivariate system of linear equations, there is a solution called “Gauss Elimination”. In the navigation business, many problems can be solved by “Gauss elimination”. If you are interested in it, I will show you a simple question. Try it.” Input There are several test cases. In the first line of each case, a number n indicates that there are n equations. The following n lines, each line has n+1 numbers, ai1,ai2,ai3…..ain, bi(1<= i <=n), these numbers indicate the coefficients of systems of the equations. ai1*x1+ai2*x2+......ain*xn=bi. Input is terminated by the end of file. Output For each given systems of equations, if there are solutions, output n solutions in the order of appearance in the equations（n<=100）, each solution number is in one line. If solution is not integer, show it in fraction. If no solution, output “No solution.” Leave a blank line after each case. Sample Input 2 1000000000000000000000000 1000000000000000000000000 1000000000000000000000000 1000000000000000000000000 1000000000000000000000000 0 1 0 4 Sample Output 1/2 1/2 No solution.
怎么运用 C 语言的程序，Cylinder_course
20190810Problem Description Using a sheet of paper and scissors, you can cut out two faces to form a cylinder in the following way: 1. Cut the paper horizontally (parallel to the shorter side) to get two rectangular parts. 2. From the first part, cut out a circle of maximum radius. The circle will form the bottom of the cylinder. 3. Roll the second part up in such a way that it has a perimeter of equal length with the circle's circumference, and attach one end of the roll to the circle. Note that the roll may have some overlapping parts in order to get the required length of the perimeter. Given the dimensions of the sheet of paper, can you calculate the biggest possible volume of a cylinder which can be constructed using the procedure described above? Input The input consists of several test cases. Each test case consists of two numbers w and h (1 ≤ w ≤ h ≤ 100), which indicate the width and height of the sheet of paper. The last test case is followed by a line containing two zeros. Output For each test case, print one line with the biggest possible volume of the cylinder. Round this number to 3 places after the decimal point. Sample Input 10 10 10 50 10 30 0 0 Sample Output 54.247 785.398 412.095
首字母变大写 应该如何实现呢_course
20200526Problem Description 输入一个英文句子，将每个单词的第一个字母改成大写字母。 Input 输入数据包含多个测试实例，每个测试实例是一个长度不超过100的英文句子，占一行。 Output 请输出按照要求改写后的英文句子。 Sample Input i like acm i want to get an accepted Sample Output I Like Acm I Want To Get An Accepted
C 语言来实现进制转换_course
20190819Problem Description 输入一个十进制数N，将它转换成R进制数输出。 Input 输入数据包含多个测试实例，每个测试实例包含两个整数N(32位整数)和R（2<=R<=16, R<>10）。 Output 为每个测试实例输出转换后的数，每个输出占一行。如果R大于10，则对应的数字规则参考16进制（比如，10用A表示，等等）。 Sample Input 7 2 23 12 4 3 Sample Output 111 1B 11
偶数求和 是怎么实现的呢_course
20200526偶数求和 Problem Description 有一个长度为n(n<=100)的数列，该数列定义为从2开始的递增有序偶数，现在要求你按照顺序每m个数求出一个平均值，如果最后不足m个，则以实际数量求平均值。编程输出该平均值序列。 Input 输入数据有多组，每组占一行，包含两个正整数n和m，n和m的含义如上所述。 Output 对于每组输入数据，输出一个平均值序列，每组输出占一行。 Sample Input 3 2 4 2 Sample Output 3 6 3 7
Range 区间的问题_course
20200416Problem Description Some automobiles display the estimated driving range, that is, the distance you can expect to drive it (without adding fuel) before running out of fuel. Here is how it works: periodically, the vehicle's computer records the odometer reading and the weight of fuel in the fuel tank. From this data, the fuel consumption over a certain distance can be computed. From the fuel consumption and the most recent measurement of fuel tank contents (which we assume is current for all practical purposes), the range can be calculated. Intervals over which the quantity of fuel increased (fuel was added to the tank) will not be used in the computations. For example, in the first problem instance of the sample input, the interval where the fuel weight increased from 29.9 kilograms to 34.2 kilograms will not be used. In this example, 16.3 kilograms of fuel were consumed over a distance of 228.6 kilometers. Therefore, the most recently measured fuel contents of 31.2 kilograms will enable you to drive another 438 kilometers (rounded to the nearest integer). The input will always contain at least one interval (two consecutive lines of input) where no fuel was added to the tank. Input The input contains data for a number of problem instances. Each problem instance consists of three or more (odometer reading, fuel weight) pairs, one pair per line. Distances are measured in kilometers and fuel weight in kilograms. All input numbers will be given to one decimal place. The end of each problem instance will be signaled by a (0.0, 0.0) pair. The last problem instance will be followed by a (1.0, 1.0) pair. Output For each problem instance, print the range, rounded to the nearest integer. Sample Input 18400.5 43.2 18440.4 40.4 18482.7 37.0 18540.2 33.1 18585.3 29.9 18620.8 34.2 18664.6 31.2 0.0 0.0 18400.5 43.2 18440.4 40.4 18482.7 37.0 18540.2 33.1 18585.3 29.9 0.0 0.0 1.0 1.0 Sample Output 438 415
从Golang的stdin读取_course
20150315<div class="posttext" itemprop="text"> <p>I'm trying to read from Stdin in Golang as I'm trying to implement a driver for Erlang. I have the following code:</p> <pre><code>package main import ( "fmt" "os" "bufio" "time" ) func main() { go func() { stdout := bufio.NewWriter(os.Stdin) p := []byte{121,100,125,' '} stdout.Write(p) }() stdin := bufio.NewReader(os.Stdin) values := make([]byte,4,4) for{ fmt.Println("b") if read_exact(stdin) > 0 { stdin.Read(values) fmt.Println("a") give_func_write(values) }else{ continue } } } func read_exact(r *bufio.Reader) int { bits := make([]byte,3,3) a,_ := r.Read(bits) if a > 0 { r.Reset(r) return 1 } return 1 } func give_func_write(a []byte) bool { fmt.Println("Yahu") return true } </code></pre> <p>However it seems that the <code>give_func_write</code> is never reached. I tried to start a goroutine to write to standard input after 2 seconds to test this.</p> <p>What am I missing here? Also the line <code>r.Reset(r)</code>. Is this valid in go? What I tried to achieve is simply restart the reading from the beginning of the file. Is there a better way?</p> <p><strong>EDIT</strong></p> <p>After having played around I was able to find that the code is stuck at <code>a,_ := r.Read(bits)</code> in the <code>read_exact</code> function</p> </div>
模仿通配符的定义实现字符串的模糊查询问题，怎么采用C程序的语言的编写的思路实现_course
20190419Problem Description WisKey downloaded much software in winter vacation, and the disk was in a state of confusion. He wastes many times to find the file everyday. So he wants a tool to help him do this work. The file name consists of lowercase letters. The name pattern is a string of lowercases, '?'s and '*'s. In a pattern, a '?' matches any single lowercase, and a '*' matches none or more lowercases. Let’s do this~ Input The first line of input contains two integers N (0 < N <= 10000) and M (0 < M <=100), representing the number of file names and the number of word patterns. Each of the following N lines contains a file name. After those, each of the last M lines contains a name pattern. You can assume that the length of patterns will not exceed 6, and the length of file names will not exceed 20. There are multiple cases in the data file, process to end of file. Output For each pattern, print a line contains the number of matched file names. If there is no file name that can match the pattern, print "Not match". Sample Input 4 5 this the an is t* ?h*s ??e* *s e Sample Output 2 1 1 2 Not match
数字的多次的迭代的算法的问题的解决的方式，用的是C程序的语言的编写的过程怎么做_course
20190409Problem Description For any positive integer n, we define function F(n) and XEN(n). For a collection S(n)={1,2,...,2n}, we select some numbers from it. For a selection, if each selected number could not be divided exactly by any other number in this selection, we will call the selection good selection. Further, we call a good selection best selection if the selection has more elements than any other good selection from S(n). We define F(n) the number of elements in the best selection from S(n). For example, n=2, F(n)=2. From the collection {1,2,3,4}, we can make good selection just like {2,3} or {3,4}, but we can't make any larger selection. So F(2) = 2. Then we pay attention to XEN(n). For every S(n), there are always some numbers could not be selected to make up any best selection. For instance, when n=2, 1 is always could not be chosen. What's more, for every S(n), there is a number k which satisfies that all the number, from 0 to k, are always could not be chosen. Now we let XEN(n)=k: n=2, F(n)=2, XEN(2)=1; n=4, F(n)=4, XEN(4)=1. You should write a program to calculate the value of F(n) and XEN(n) with a given number n. Input Your program is to read from standard input. There are multiple cases. For each case, one integer n (1 ≤ n ≤ 10^7) in a line. Output Output two integers with one space between them in one line per case. Sample Input 2 4 Sample Output 2 1 4 1
SCV集合的查询和数组操作的问题，采用C技术语言_course
20190123Problem Description StarCraft 2 (SC2) is a famous game. More and more people fall in love with this game. As a crazy fan of SC2, Ahua (flower fairy) play it day and night. Recently, he found that the most important part of being a top player of SC2 is economic development, which means you should get as much mine as possible by training SCVs (space construction vehicle) to collect mine. Train a SCV at ith second costs Ci units of mine. After training, this SCV can collect Di units of mine each second. Training a SCV needs one second of time. Based on that, he composes a formula to evaluate the development in a time span from xth second to yth second. Assume at xth second, Ahua has no SCV and mine. He trains one SCV at each second during xth second and yth second (the mount of mine can be negative, so that he always can train SCV). Each SCV will collect some amount of mines for Ahua in each second after it was trained. At ith second Ahua has Mi units of mine in total. The development value is defined as sum(Mi) (x ≤ i ≤ y). Now he asks you to help him calculate the development value. To make it more interesting, Ahua can apply following operations: Cost x y z: the cost of training a SCV between xth second to yth second will increase by z units of mine. i.e. Ci for x ≤ i ≤ y will increase by z. Collect x y z: each SCV trained between xth second and yth second can collect z more mines every second after it has been trained. i.e. Di for x ≤ i ≤ y will increase by z. Query x y: output the development value between xth second and yth second. Input First line of the input is a single integer T (T ≤ 10), indicates there are T test cases. For each test case, the first line is an integer N (1 ≤ N ≤ 100000), means the maximum time you should deal with. Following N lines, each contain two integers Ci and Di (0 ≤ Ci, Di ≤ 100000), the cost and collect speed of SCV training in ith second initially as described above. The next line is an integer Q (1 ≤ Q ≤ 10000), the number of operations you should deal with. Then Q lines followed, each line will be “Cost x y z”, "Collect x y z” or “Query x y”. 1 ≤ x ≤ y ≤ N, 0 ≤ z ≤ 100000 Output For each test case, first output “Case k: “ in a single line, k is the number of the test case, from 1 to T. Then for each "Q x y", you should output a single line contains the answer mod 20110911. Sample Input 1 5 1 3 2 3 3 1 2 2 3 3 5 Query 1 3 Cost 2 2 1 Query 1 3 Collect 1 1 5 Query 1 3 Sample Output Case 1: 2 0 15
矩阵和行列式的一种计算的方式的问题，怎么采用 C 程序的语言代码的编写过程去实现呢？_course
20190518Problem Description There is a matrix M that has n rows and m columns (1≤n≤1000,1≤m≤1000).Then we perform q(1≤q≤100,000) operations: 1 x y: Swap row x and row y (1≤x,y≤n); 2 x y: Swap column x and column y (1≤x,y≤m); 3 x y: Add y to all elements in row x (1≤x≤n,1≤y≤10,000); 4 x y: Add y to all elements in column x (1≤x≤m,1≤y≤10,000); Input There are multiple test cases. The first line of input contains an integer T(1≤T≤20) indicating the number of test cases. For each test case: The first line contains three integers n, m and q. The following n lines describe the matrix M.(1≤Mi,j≤10,000) for all (1≤i≤n,1≤j≤m). The following q lines contains three integers a(1≤a≤4), x and y. Output For each test case, output the matrix M after all q operations. Sample Input 2 3 4 2 1 2 3 4 2 3 4 5 3 4 5 6 1 1 2 3 1 10 2 2 2 1 10 10 1 1 1 2 2 1 2 Sample Output 12 13 14 15 1 2 3 4 3 4 5 6 1 10 10 1
数据结构字符串数组的一个运用，怎么采用C程序编写的语言的算法实现的程序？_course
20190502Problem Description Name PK is a funny game on the Internet. The game will calculate character's property based on its name. Now we're simulating a simple Name PK game. Each character has 3 parameters: HP, STR and SPD (HP for health point, STR for strength and SPD for attacking speed). For a name string of length N, Ci is ASCII code (decimal) of the ith char. PK rule: 1. Timer begins to increase from 1. When it is a multiple of (20SPD), the corresponding character A has that SPD attack once, the opposite lose STR(A's) HP. 2. When any side has a HP ≤ 0, the PK is over. Input The input consists of several test cases. The first line of input consists of an integer T, indicating the number of test cases. Each test case is on a separate line, and it consists two strings separated by a whitespace, indicating the name of the characters. Technical Specification 1. Names contain only English letters. 2. The length of each character's name is more than 1 and no more than 20. 3. T ≤ 1000. Output For each test case, output a line consisting of the result of the first character: "win", "lose" or "tie". Sample Input 3 Sylvia xay Ivan Stacy Boyd Greg Sample Output lose win lose
C语言作为一种高级语言，有3个优点，除了代码效率高，还有什么别的优点_course
20191231C语言作为一种高级语言，有3个优点，除了代码效率高，还有什么别的优点
用C语言进行，超大数字的计算问题_course
20190809Description In how many ways can you tile a 2xn rectangle by 2x1 or 2x2 tiles? Here is a sample tiling of a 2x17 rectangle. Input Input is a sequence of lines, each line containing an integer number 0 <= n <= 250. Output For each line of input, output one integer number in a separate line giving the number of possible tilings of a 2xn rectangle. Sample Input 2 8 12 100 200 Sample Output 3 171 2731 845100400152152934331135470251 1071292029505993517027974728227441735014801995855195223534251
图形的旋转问题 Endless Spin_course
20190825Problem Description I spin it again and again,and throw it away finally. So now I have a row of n ball,named from 1 to n,each ball is white initially. At each step I randomly chose a interval [l, r] and paint all ball in this interval to black. It means every interval have a equal chance of being chosen. And I'll stop if all ball are black.What is the expected steps before I stop? Input The first line contains integer T(1<=T<=50). Denoting the number of the test cases. Then T lines follows, each line contains an integer n (1<=n<=50). Output For each test cases,print the answer in a line. Print the answer rounded to 15 decimal places. Sample Input 3 1 2 3 Sample Output 1.000000000000000 2.000000000000000 2.900000000000000
水仙花数怎么计算的，用C语言_course
20190828Problem Description 春天是鲜花的季节，水仙花就是其中最迷人的代表，数学上有个水仙花数，他是这样定义的： “水仙花数”是指一个三位数，它的各位数字的立方和等于其本身，比如：153=1^3+5^3+3^3。 现在要求输出所有在m和n范围内的水仙花数。 Input 输入数据有多组，每组占一行，包括两个整数m和n（100<=m<=n<=999）。 Output 对于每个测试实例，要求输出所有在给定范围内的水仙花数，就是说，输出的水仙花数必须大于等于m,并且小于等于n，如果有多个，则要求从小到大排列在一行内输出，之间用一个空格隔开; 如果给定的范围内不存在水仙花数，则输出no; 每个测试实例的输出占一行。 Sample Input 100 120 300 380 Sample Output no 370 371
水仙花数 C语言_course
20191206Problem Description 春天是鲜花的季节，水仙花就是其中最迷人的代表，数学上有个水仙花数，他是这样定义的： “水仙花数”是指一个三位数，它的各位数字的立方和等于其本身，比如：153=1^3+5^3+3^3。 现在要求输出所有在m和n范围内的水仙花数。 Input 输入数据有多组，每组占一行，包括两个整数m和n（100<=m<=n<=999）。 Output 对于每个测试实例，要求输出所有在给定范围内的水仙花数，就是说，输出的水仙花数必须大于等于m,并且小于等于n，如果有多个，则要求从小到大排列在一行内输出，之间用一个空格隔开; 如果给定的范围内不存在水仙花数，则输出no; 每个测试实例的输出占一行。 Sample Input 100 120 300 380 Sample Output no 370 371
三种颜色的混合的比例问题，RGB混合算法，采用C程序设计的语言的做法_course
20190408Problem Description Chae Yeon is a popular pop female singer who rose to fame with her amazing sexy dance style and the sounds of nature voice she has. She proved to be a great dancer, and she showed off her vocals in her live performances. If you had ever seen her dance, I bet you’d love it. I felt stage lighting interesting when I was enjoying Chae Yeon’s performance. We all know that stage lighting instruments are used for the concerts and other performances taking place in live performance venues. They are also used to light the stages. Actually this is a color mixing process. There are two types of color mixing: Additive and Subtractive. Most stages use the former and in this case there are three primary colors: red, green, and blue. In the absence of color, or when no colors are showing, the stage is black. If all three primary colors are showing, the result is white. When red and green combine together, the result is yellow. When red and blue combine together, the result is magenta. When blue and green combine together, the result is cyan. When two same color combine together, the result is still that color. We have got the coordinate and the primary color of the figure that each Stage Lighting Instrument sent out. For simplicity’s sake, we can just treat the figure as a circle. Of course we’ll know the radius of each colored circle. Some color may be changed based on the Color Mixed Theory we mentioned above. Can you find the area of each color? Input The first line consists of an integer T, indicating the number of test cases. The first line of each case consists of three integers R, G, B, indicating the number of red circles, green circles and blue circles. The next R + G + B lines, each line consists of three integer x, y, r, indicating the coordinate and the radius. The first R lines descript the red circles, the second G lines descript the green circles and the last B lines descript the blue circles. Output Output seven floating numbers, they are the area of red, green, blue, white, yellow, magenta and cyan. Please take each number with two factional digits. Constraints 0 < T <= 20 0 <= R, G, B <= 100 100 <= x, y <= 100; 0 <= r <= 100 Sample Input 1 1 1 1 0 2 3 2 0 3 2 0 3 Sample Output 9.28 15.04 15.04 4.92 7.04 7.04 1.28
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20200501本课程为HoloLens2开发入门教程，讲解部署开发环境，安装VS2019，Unity版本，Windows SDK，创建Unity项目，讲解如何使用MRTK，编辑器模拟手势交互，打包VS工程并编译部署应用到HoloLens上等。
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改进SEIR模型的matlab代码.zip
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