3个回答

select
from fwinfo a
left join a.房屋类型对象 as b
left join a.街道对象 as c
left join a.区县对象 as d
where a.条件=？and b.条件=? and c.条件=?

select
a.fwid,
a.uid,
a.title.........
from fwinfo a left join fwlx b on a.lxid=b.lxid
left join jd c on a.jdid = c.jdid
left join (街道和区县关联表) d on-----
and 街道区县关联表.街道 = ‘’
and a.房屋类型=‘’

Rolling Hongshu 的算法
Problem Description To see his girl friend, sweet potato has to go over thousands of mountains. What make things worse, many bitter potatoes lives in these mountains. They hate sweet potato because they don't have girl friends. In the world of potatoes, friction force does not exist. So the way potatoes travel is very simple: they start with an initial speed, rolling forward and waiting to get to the destination. Bitter potatoes lived in different places. When sweet potato rolls passing their home, they begin to chase him (surely by rolling with an initial speed). If sweet potato is caught by a bitter potato, his date with girl friend has to be canceled. Now sweet potato wants to know the minimum initial speed necessary to see his girl friend. Input First line is an integer T (T ≤ 50), the number of test cases. At the beginning of each case is three integers, N, M and w, indicate the number of peaks in the mountains, the number of bitter potatoes and the weight of sweet potato separately. 2 ≤ N ≤ 1000, 0 ≤ M ≤ 1000, 0 < w < 100000. The next N lines each contain a pair of integers. Each pair of integers xi, hi describe a peak. xi is the horizontal distant between sweet potato's home and the peak. hi is the height of the peak. All xi are different. 0 = x1 < x2 < … < xn ≤ 100000000, -100000000 ≤ hi ≤ 100000000. Between adjacent peaks is a smooth slope. The bitter potatoes are on these slopes. The following M lines each contain 3 integers. Each triple of integers pi, vi, mi describe a bitter potato. pi is the horizontal distant between his home and sweet potato’s home. vi is his initial speed. mi is his weight. 0 < pi < xn, 0 ≤ vi ≤ 100000000, 0 < mi < 100000 The gravitational constant in potatoes' world is 20. Sweet potato's home is at point (x1, h1). His girl friend lives at point (xn, hn). Output For each case, you should output “Case k: ” first. Following a number, the lower bound of sweet potato's initial speed rounded to two decimal places. Sample Input 1 6 2 100 0 0 2 5 3 2 4 1 5 3 8 -2 2 15 100 5 11 100 Sample Output Case 1: 20.62

![图片说明](https://img-ask.csdn.net/upload/202001/29/1580308263_642113.png) User类和Account类 ``` public class User implements Serializable { private Integer id; private String username; private String address; private String sex; private Date birthday; 以及getset方法 } public class Account { private Integer ID; private Integer UID; private Integer MONEY; private User user; 以及getset方法 } ``` mysql数据库中有和上面两个类相同的属性的表（除了数据库中的account没有user属性） ``` @Result(property = "user",column = "UID",one = @One( select = "com.HSY.dao.IUSerDao.findById", fetchType = FetchType.EAGER )) ``` 这个代码中，property是user我能理解，后面的select可以获得User的信息，但是为什么column是UID，求求个位大佬讲解一下其中的流程

hi： 我最近在学习makefile，看到一句话 make会一层又一层地去找文件的依赖关系，直到最终编译出第一个目标文件。在找寻的过程中，如果出现错误，比如最后被依赖的文件找不到，那么make就会直接退出，并报错，**而对于所定义的命令的错误，或是编译不成功，make根本不理**。make只管文件的依赖性，即，如果在我找了依赖关系之后，冒号后面的文件还是不在，那么对不起，我就不工作啦。 我之前在实习时候，用到make，我文件编译错误，也报错了，那是为什么呢？

``` #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; void merge(int* A, int lo, int mi, int hi) { int lb = mi - lo; int* B = new int[lb]; for(int i = 0; i < lb; i++) { B[i] = A[lo + i]; } int* C = A + mi; int lc = hi - mi; for(int i = 0, j = 0, k = 0; (j < lb) || (k < lc);) { if((j < lb) && (k >= lc) || (B[j] <= C[k])) A[i++] = B[j++]; if((k < lc) && (j >= lb) || (C[k] < B[j])) A[i++] = C[k++]; } delete [] B; } void mergeSort (int* A, int lo, int hi) { if(hi - lo < 2) return; int mi = (lo + hi) >> 1; mergeSort(A, lo, mi); mergeSort(A, mi, hi); merge(A, lo, mi, hi); } int main () { int A[10] = {9, 2, 5, 7, 0, 8, 1, 4, 3, 6}; mergeSort(A, 0, 10); for(int i = 0; i < 10; i++) { cout << A[i] << ' '; } return 0; } ```
Super Mario 怎么实现呢
Problem Description Mario is world-famous plumber. His “burly” figure and amazing jumping ability reminded in our memory. Now the poor princess is in trouble again and Mario needs to save his lover. We regard the road to the boss’s castle as a line (the length is n), on every integer point i there is a brick on height hi. Now the question is how many bricks in [L, R] Mario can hit if the maximal height he can jump is H. Input The first line follows an integer T, the number of test data. For each test data: The first line contains two integers n, m (1 <= n <=10^5, 1 <= m <= 10^5), n is the length of the road, m is the number of queries. Next line contains n integers, the height of each brick, the range is [0, 1000000000]. Next m lines, each line contains three integers L, R,H.( 0 <= L <= R < n 0 <= H <= 1000000000.) Output For each case, output "Case X: " (X is the case number starting from 1) followed by m lines, each line contains an integer. The ith integer is the number of bricks Mario can hit for the ith query. Sample Input 1 10 10 0 5 2 7 5 4 3 8 7 7 2 8 6 3 5 0 1 3 1 1 9 4 0 1 0 3 5 5 5 5 1 4 6 3 1 5 7 5 7 3 Sample Output Case 1: 4 0 0 3 1 2 0 1 5 1

select count(*) from BUSINESS_APPLY BA, IND_INFO II, MLHOUSE_INFO HI, EXT_PERFORMANCE EP, ML_VENDOR_M VM, ML_VENDOR_G VG, CUSTOMER_ENSURE CE, APPLY_ADD AA, MLVERBAL_VALUATION MV where BA.Customerid = II.Customerid and BA.Customerid = HI.Serialno and BA.Serialno = EP.Serialno(+) and HI.Bucdeveloper = VM.SERIALNO(+) and BA.Warrantor = VG.SerialNo(+) and BA.SerialNo = CE.ObjectNo(+) and AA.ApplyNo = BA.SerialNo and MV.ObjectNo(+) = BA.SerialNo order by BA.SerialNo2
Snake 的代码实现问题
Problem Description Snake is a popular game , and I believe most of us had played it . The Original game is that you can control a snake to eat the magic bean and after the snake eat one magic bean , the length of the snake’s body will get longer .But today we are talking about a new game. These are the rules of the new Snake Game : 1. The length of the snake’s body won’t change even though it eat a magic bean. 2. Some pairs of the beans have a relation , that is, one of them can not be eaten until another one had been eaten . We call the latter “the key bean” . For example , if A can’t be eaten until B had been eaten ,we say “B is the key bean of A”. (That means when A can’t be eaten , the snake can not move into the grid where A is.) 3. The snake could not move to a wall or its body.Befor it move,it will chooses an adjacent vacant square of its head,which is neither a stone nor occupied by its body. Figure 1 and figure2 shows how the snake move Figure 1 Figure 2 Input The first line contain a integer T (T <= 10).Followed by T cases. Each case contain five parts. The first part: six integers ,H,W,L,K,R,N,(H <= 20 , W <= 20 , L <= 8 , K <= 7 ) means the height of the map , the width of the map , the length of the snake’s body, the number of the magic beans . the number of the relations , the number of the wall respectively. The second part: L lines , each line contain two integer hi ,wi, indicating the original position of each block of snake's body, from B1(h1,w1) to BL(hL,wL) orderly, where 1<=hi<=H, and 1<=wi<=W,1<=i<=L. The third part: K lines ,each line contain two integer hi ,wi , indicating the position of each magic bean , from MB1(h1,w1) to MBK(hK,wK) orderly, where 1<=hi<=H, and 1<=wi<=W,1<=i<=K. The fourth part : R lines , each line contain two integer A ,B means “A is the key bean of B ”. The A and B may appear several times , but “one bean will have only one key bean”. The fifth part: N lines , each line contain two integer hi ,wi , indicating the position of each wall , from W1(h1,w1) to WN(hN,wN) orderly, where 1<=hi<=H, and 1<=wi<=W,1<=i<=N. Output For each case , if the snake could eat all the magic beans , output the minimum step it took. If the snake could not , just output “-1” (without the quotation marks) . Sample Input 1 8 9 5 2 1 8 5 2 6 2 6 3 6 4 6 5 4 2 2 6 2 1 2 5 3 5 4 4 4 5 4 6 5 6 5 7 6 7 Sample Output 21
Daizhenyang's Letter 字符串问题
Problem Description One day, Daizhenyang is wandering in BUPT campus, thinking about having a girlfriend. Out of nowhere, a breath-taking beautiful girl came by, Daizhenyang was shocked by her beauty and was too stunned to say hi. There's an old saying, "Once there was true love for me but I did not cherish it. When it slipped away between my fingers, I began to regret. That is the saddest thing in life. If God could give me a second chance, I would tell the girl 'I love you!'. If there should be a time limit on this love, I wish it would be 10000 years." Regretting is what Daizhengyang is being through. He wants to make everything right, so he figured out her mail address and decides to send a love letter. After thinking again and again for a long time, Daizhenyang picked up some keywords to express his love and emotion. However, Daizhenyang is busy on his eternal lasting enterprise, unifying all kinds of ACM/ICPC problems to network flow, he wants to make this letter as short as possible. You can help him on this holy project, making his dream come true. Input There are multiple cases in the input file. For each test case: Line 1: an single integer N ( 0<N<=12 ) , represents the number of keywords. Line 2 - N+1: a keyword in each line, keywords only consist of uppercase letters ( for sincerity ) , length is less than 56. Output For each test case, output one line, begin with "Case #k: ", k is the case number, followed by the letter required. If there're several solutions output the lexicographically smaller one. Sample Input 3 BUPT TALENT DAIZHENYANG 3 LOVE VERY RYTHM Sample Output Case #1: BUPTALENTDAIZHENYANG Case #2: LOVERYTHM
Man Down 下降问题
Problem Description The Game “Man Down 100 floors” is an famous and interesting game.You can enjoy the game from http://hi.baidu.com/abcdxyzk/blog/item/16398781b4f2a5d1bd3e1eed.html We take a simplified version of this game. We have only two kinds of planks. One kind of the planks contains food and the other one contains nails. And if the man falls on the plank which contains food his energy will increase but if he falls on the plank which contains nails his energy will decrease. The man can only fall down vertically .We assume that the energy he can increase is unlimited and no borders exist on the left and the right. First the man has total energy 100 and stands on the topmost plank of all. Then he can choose to go left or right to fall down. If he falls down from the position (Xi,Yi),he will fall onto the nearest plank which satisfies (xl <= xi <= xr)(xl is the leftmost position of the plank and xr is the rightmost).If no planks satisfies that, the man will fall onto the floor and he finishes his mission. But if the man’s energy is below or equal to 0 , he will die and the game is Over. Now give you the height and position of all planks. And ask you whether the man can falls onto the floor successfully. If he can, try to calculate the maximum energy he can own when he is on the floor.(Assuming that the floor is infinite and its height is 0,and all the planks are located at different height). Input There are multiple test cases. For each test case, The first line contains one integer N (2 <= N <= 100,000) representing the number of planks. Then following N lines representing N planks, each line contain 4 integers (h,xl,xr,value)(h > 0, 0 < xl < xr < 100,000, -1000 <= value <= 1000), h represents the plank’s height, xl is the leftmost position of the plank and xr is the rightmost position. Value represents the energy the man will increase by( if value > 0) or decrease by( if value < 0) when he falls onto this plank. Output If the man can falls onto the floor successfully just output the maximum energy he can own when he is on the floor. But if the man can not fall down onto the floor anyway ,just output “-1”(not including the quote) Sample Input 4 10 5 10 10 5 3 6 -100 4 7 11 20 2 2 1000 10 Sample Output 140

```void sanjitutai::OnTimer(UINT_PTR nIDEvent) { switch (nIDEvent) { case DealData_timer://每秒一次的处理 DealData(); /*KillTimer(DealData_timer);*/ break; case BatteryData_timer: //每分钟一次的处理 ReadBattery(); /* KillTimer(BatteryData_timer);*/ break; default: ; } CDialog::OnTimer(nIDEvent); } BOOL sanjitutai::ReadBattery() { HANDLE hI2C; // I2C设备操作号 I2C_TRANSACTION_INFO i2cInfo; byte InBuffer = 0 ; hI2C = I2C_Open(L"I2C1:"); if(hI2C == INVALID_HANDLE_VALUE) { printf("Open I2C Device fail!\r\n"); I2C_Close(hI2C); return 0; } I2C_Close(hI2C); return 0; } ```
Rolling Hongshu 是怎么写的
Problem Description To see his girl friend, sweet potato has to go over thousands of mountains. What make things worse, many bitter potatoes lives in these mountains. They hate sweet potato because they don't have girl friends. In the world of potatoes, friction force does not exist. So the way potatoes travel is very simple: they start with an initial speed, rolling forward and waiting to get to the destination. Bitter potatoes lived in different places. When sweet potato rolls passing their home, they begin to chase him (surely by rolling with an initial speed). If sweet potato is caught by a bitter potato, his date with girl friend has to be canceled. Now sweet potato wants to know the minimum initial speed necessary to see his girl friend. Input First line is an integer T (T ≤ 50), the number of test cases. At the beginning of each case is three integers, N, M and w, indicate the number of peaks in the mountains, the number of bitter potatoes and the weight of sweet potato separately. 2 ≤ N ≤ 1000, 0 ≤ M ≤ 1000, 0 < w < 100000. The next N lines each contain a pair of integers. Each pair of integers xi, hi describe a peak. xi is the horizontal distant between sweet potato's home and the peak. hi is the height of the peak. All xi are different. 0 = x1 < x2 < … < xn ≤ 100000000, -100000000 ≤ hi ≤ 100000000. Between adjacent peaks is a smooth slope. The bitter potatoes are on these slopes. The following M lines each contain 3 integers. Each triple of integers pi, vi, mi describe a bitter potato. pi is the horizontal distant between his home and sweet potato’s home. vi is his initial speed. mi is his weight. 0 < pi < xn, 0 ≤ vi ≤ 100000000, 0 < mi < 100000 The gravitational constant in potatoes' world is 20. Sweet potato's home is at point (x1, h1). His girl friend lives at point (xn, hn). Output For each case, you should output “Case k: ” first. Following a number, the lower bound of sweet potato's initial speed rounded to two decimal places. Sample Input 1 6 2 100 0 0 2 5 3 2 4 1 5 3 8 -2 2 15 100 5 11 100 Sample Output Case 1: 20.62
Walk 旅行的问题
Problem Description Alice would like to visit Bob. However, they live in a hilly landscape, and Alice doesn’t like to walk in hills. She has a map of the area, showing the height curves. You have to calculate the total altitude climbed, and the total altitude descended, for the route which minimizes these numbers. It does not matter how far she has to walk to achieve this. Since you don’t know what the landscape looks like in between the height curves, you cannot know exactly how much climb and descent she will actually get in practice, but you should calculate the minimum possible under optimal conditions based on what you can deduce from the map. The map is represented as an xy grid. Alice lives in (0, 0), and Bob lives in (100 000, 0). The height curves are represented as polygons, where a polygon cannot intersect itself or another polygon. Furthermore, neither Alice nor Bob lives exactly on a height curve. Second test case from sample input (compressed). Input On the first line one positive number: the number of testcases, at most 100. After that per testcase: One line with 0 ≤ N ≤ 2 500, the number of height curves. One line for each height curve, with 1 ≤ Hi ≤ 1 000 being the height of the curve, 3 ≤ Pi ≤ 2 000 the number of vertices in the polygon, and the vertices x1, y1, …, xPi, yPi having integral values &#8722;300 000 ≤ xi, yi ≤ 300 000. There will be no more than 200 000 polygon vertices in total in all test cases. Output Per testcase: One line with two numbers: the total altitude climbed and the total altitude descended. Sample Input 2 2 20 3 10 10 0 -10 -10 10 25 3 20 20 0 -20 -20 20 3 100 4 -1 1 1 1 1 -1 -1 -1 300 8 -2 2 2 2 2 -2 5 -2 5 1 6 1 6 -3 -2 -3 50 8 3 3 100001 3 100001 -1 7 -1 7 2 4 2 4 -1 3 -1 Sample Output 5 0 200 250 Source
Catch The Heart 的计算
Problem Description Do you know Popeye? He likes spinach(菠菜) very much. Maybe you have played some games about him. Now here comes one game. There are many boards, on one board there is a heart, Popeye wants to catch the heart and send it to his GF, Oliver, so that, she will be happy. At the beginning, Popeye is on one board, and the heart is on another board. Every board is described as h, l, and r, the height, the left end and the right end. You can look it as a line (l, h) (r, h) in XY coordinate system. Popeye can walk on the board, jump from one board to another. Notes: 1.He can walk on a board 1 unit per second. 2.He can only jump vertically(垂直地), and it takes 1s for each jumping. 3.He can jump only 1 unit high, so that he can jump only if the difference of the boards’ height is 1. 4.He can only jump up to the left or right point of the board. 5.He can only jump down from the left or right point of the board. You can look the picture: The problem is, find the shortest time to catch the heart. Input The first line of the input contains an integer T (T <= 20) which means the number of test cases. For each case, first line is an integer m (2 <= m <= 10000) which means the number of the boards. Then comes two lines, the first line contains h0, l0, r0, p0 (l0 <= p0 <= r0), which mean the information of the board and the position that the heart on it. Next line contains h1, l1, r1, p1 (l1 <= p1 <= r1), which mean the information of the board and the position that Popeye on it. Then comes m-2 lines, each contains hi, li, ri mean the ith (2 <= i < m) board’s information. There is no common point or common parts for every two boards. All inputs are integers, for each board, 0 <= h <= 10000, 0 <= l < r <= 100000000. Output For each case, output the shortest time to catch the heart if Popeye can catch it, otherwise output -1. Sample Input 3 3 3 4 5 5 1 0 10 0 4 1 2 2 1 1 99999999 1 0 0 100000000 100000000 5 4 0 5 0 0 0 5 0 1 5 10 2 6 15 3 5 10 Sample Output -1 100000000 24
Hi3518ev300 在HI_MIPI_ENABLE_SENSOR_CLOCK 时失败，报错：__osal_unlocked_ioctl - Input param err,it is null!

Cycling 的寻找问题
Problem Description You want to cycle to a programming contest. The shortest route to the contest might be over the tops of some mountains and through some valleys. From past experience you know that you perform badly in programming contests after experiencing large differences in altitude. Therefore you decide to take the route that minimizes the altitude difference, where the altitude difference of a route is the difference between the maximum and the minimum height on the route. Your job is to write a program that finds this route. You are given: the number of crossings and their altitudes, and the roads by which these crossings are connected. Your program must find the route that minimizes the altitude difference between the highest and the lowest point on the route. If there are multiple possibilities, choose the shortest one. For example: In this case the shortest path from 1 to 7 would be through 2, 3 and 4, but the altitude difference of that path is 8. So, you prefer to go through 5, 6 and 4 for an altitude difference of 2. (Note that going from 6 directly to 7 directly would have the same difference in altitude, but the path would be longer!) Input On the first line an integer t (1 <= t <= 100): the number of test cases. Then for each test case: One line with two integers n (1 <= n <= 100) and m (0 <= m <= 5000): the number of crossings and the number of roads. The crossings are numbered 1..n. n lines with one integer hi (0 <= hi <= 1 000 000 000): the altitude of the i-th crossing. m lines with three integers aj , bj (1 <= aj , bj <= n) and cj (1 <= cj <= 1 000 000): this indicates that there is a two-way road between crossings aj and bj of length cj . You may assume that the altitude on a road between two crossings changes linearly. You start at crossing 1 and the contest is at crossing n. It is guaranteed that it is possible to reach the programming contest from your home. Output For each testcase, output one line with two integers separated by a single space: the minimum altitude difference, and the length of shortest path with this altitude difference. Sample Input 1 7 9 4 9 1 3 3 5 4 1 2 1 2 3 1 3 4 1 4 7 1 1 5 4 5 6 4 6 7 4 5 3 2 6 4 2 Sample Output 2 11
HI3518ev300 HI_MPI_VPSS_SetExtChnAttr 失败 0xA0078003，报错参数设置无
``` /******************************************************************************* *@ Description :配置VPSS的扩展通道 *@ Input :<vpssGrp>:组号 <vpssChn>：扩展通道号 <bindChn>：绑定到目标物理通道号 <width>：扩展通道图片的宽 <height>：扩展通道图片的高 <frmRate>扩展通道图片的帧率 *@ Output : *@ Return :成功：0 失败：错误码 *@ attention :该接口在H3518ev300（hi3516ev200）的SDK中设置不正常 *******************************************************************************/ static int vpss_config_ext_chn(VPSS_GRP vpssGrp, VPSS_CHN vpssChn, VPSS_CHN bindChn, int width, int height, int frmRate) { int ret; VPSS_EXT_CHN_ATTR_S extAttr = {0}; /*---#------------------------------------------------------------*/ printf("----ExtChnAttr:-----------------------------------------\n"); printf("extAttr.s32BindChn = %d\n",extAttr.s32BindChn); printf("extAttr.u32Width = %d\n",extAttr.u32Width); printf("extAttr.u32Height = %d\n",extAttr.u32Height); printf("extAttr.enVideoFormat = %d\n",extAttr.enVideoFormat); printf("extAttr.enPixelFormat = %d\n",extAttr.enPixelFormat); printf("extAttr.enDynamicRange = %d\n",extAttr.enDynamicRange); printf("extAttr.enCompressMode = %d\n",extAttr.enCompressMode); printf("extAttr.u32Depth = %d\n",extAttr.u32Depth); printf("extAttr.stFrameRate.s32SrcFrameRate = %d\n",extAttr.stFrameRate.s32SrcFrameRate); printf("extAttr.stFrameRate.s32DstFrameRate = %d\n",extAttr.stFrameRate.s32DstFrameRate); printf("-------------------------------------------------------\n"); /*---#------------------------------------------------------------*/ extAttr.s32BindChn = bindChn; extAttr.u32Width = width;//960; extAttr.u32Height = height;//540; extAttr.enVideoFormat = VIDEO_FORMAT_LINEAR; extAttr.enPixelFormat = PIXEL_FORMAT_YUV_SEMIPLANAR_420; extAttr.enDynamicRange = DYNAMIC_RANGE_SDR8; extAttr.enCompressMode = COMPRESS_MODE_NONE; extAttr.u32Depth = 0; extAttr.stFrameRate.s32SrcFrameRate = 15; extAttr.stFrameRate.s32DstFrameRate = frmRate; ret = HI_MPI_VPSS_SetExtChnAttr(vpssGrp, vpssChn, &extAttr); if (HI_SUCCESS != ret) { ERROR_LOG("HI_MPI_VPSS_SetExtChnAttr(%d, %d) fail: %#x!\n", vpssGrp, vpssChn, ret); return HLE_RET_ERROR; } ret = HI_MPI_VPSS_EnableChn(vpssGrp, vpssChn); if (HI_SUCCESS != ret) { ERROR_LOG("HI_MPI_VPSS_EnableChn(%d, %d) fail: %#x!\n", vpssGrp, vpssChn, ret); return HLE_RET_ERROR; } return HLE_RET_OK; } ``` ![图片说明](https://img-ask.csdn.net/upload/201909/29/1569723865_767347.jpg) 实现代码如上所述，VPSS 物理通道号是0，已经创建并使能， 扩展通道5绑定到物理通道0，总是报参数设置无效，怎么回事？？？

《奇巧淫技》系列-python！！每天早上八点自动发送天气预报邮件到QQ邮箱

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Python爬虫爬取淘宝，京东商品信息

Java工作4年来应聘要16K最后没要,细节如下。。。

Python爬虫精简步骤1 获取数据

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