ball ball各位大佬，C语言中序表达式树求值问题

1个回答

`````` #include<iostream>
#include<map>
#include<string>
#include<stack>
#include<vector>
using namespace std;
class Logic {
public:
Logic() {}                                      //构造函数
int priority(char);                               //获取运算符优先级
string trans(string);                             //中缀式->后缀式
double calculate();    //逻辑判断
void V_assign();                                  //变量赋值
string M_exp;                            //中缀式
string P_exp;                            //后缀式
map<string, double> variate;               //赋值序列
};
M_exp = str;;
P_exp = trans(M_exp);             //处理数据（表达式转换）
}
int Logic::priority(char ch) {
if (ch == '*'||ch=='/')
return 2;
if (ch == '+'||ch=='-')
return 1;
if (ch == '(')
return -1;
return 0;
}
double Logic::calculate() {
string operators("+-*/");
stack<double> res;            //此栈用作运算
double a, b;
for (int i = 0; i<P_exp.length(); i++) {
if (operators.find(P_exp[i]) == string::npos) {      //遇到操作数，根据“字典”翻译后入栈
res.push(variate[P_exp.substr(i, 1)]);
}
else {
switch (P_exp[i]) {
case '+':
a = res.top();
res.pop();
b = res.top();
res.pop();
res.push(a + b);
break;
case '*':
a = res.top();
res.pop();
b = res.top();
res.pop();
res.push(a * b);
break;
case '-':
a = res.top();
res.pop();
b = res.top();
res.pop();
res.push(b-a);
break;
case '/':
a = res.top();
res.pop();
b = res.top();
res.pop();
res.push(b/a);
break;
}
}
}
return res.top();
}
string Logic::trans(string m_exp) {
string p_exp;
stack<char> stk;
string operators("+-*/(");
for (int i = 0; i < m_exp.length(); i++) {
string one;
if (operators.find(m_exp[i]) != string::npos) {      //出现操作符
if (m_exp[i] == '(')         //栈中添加左括号
stk.push(m_exp[i]);
else {                      //操作符的优先级判断
while ((!stk.empty()) && (priority(m_exp[i]) <= priority(stk.top()))) {    //当栈不为空时，进行优先级判断
p_exp.push_back(stk.top());   //若当前操作符优先级低于栈顶，弹出栈顶，放到后缀式中
stk.pop();
}
stk.push(m_exp[i]);             //将当前操作符入栈
}
}
else if (m_exp[i] == ')') {            //出现右括号时，将栈中元素一直弹出，直至弹出左括号
while (stk.top() != '(') {
p_exp.push_back(stk.top());
stk.pop();
}
stk.pop();                         //弹出左括号
}
else {           //把操作数加入到后缀式中
variate[m_exp.substr(i, 1)] = 0;
p_exp.push_back(m_exp[i]);
}

}
while (!stk.empty()) {    //将栈中剩余操作符放到后缀式中
p_exp.push_back(stk.top());
stk.pop();
}
return p_exp;
}
void Logic::V_assign() {       //公式赋值
int i = 0;
for (auto it = variate.begin(); it != variate.end(); it++) {
cout << "Enter the " << it->first << " : ";
cin >> it->second;
}
}
int main() {
Logic my;
string str;
cin >> str;
cout << "后缀表达式：" << my.P_exp << endl;
cout << "赋值：" << endl;
my.V_assign();
cout<<"结果为："<<my.calculate();
return 0;
}
`````` C语言的实现，Crystal Ball Factory 问题

Problem Description The Astrologically Clairvoyant Manufacturers (ACM),a pioneer in future-predicting technology, just landed a contract to manufacture crystal balls for weather forecasters around the world. Every week, a variable number of crystal balls needs to be delivered; the required amount for each week is specified in the contract. Crystal balls are made from the highest-quality crystal, whose price fluctuates from week to week. Fortunately, the ACM is able to foresee the price of crystal for the coming weeks, thanks to its own future-predicting technology. When the price is low, the ACM would like to buy crystal and manufacture crystal balls, storing any excess in their warehouse. On the other hand, in weeks for which the price is high, ACM would rather use the crystal balls stored in the warehouse to satisfy the demand specified in their contract. However, since there is a also a fixed weekly cost to store each crystal ball in the warehouse, and an initial cost for turning on the manufacturing machines and producing a non-zero quantity of crystal balls, the decision is not always simple. Can you help them fulfill their contract at minimal cost? Input The first line of each test case (representing a contract) will contain the number of weeks for which the contract will last. The next line will contain the non-negative integers b, k and n, where b is the base cost for manufacturing a non-zero quantity of crystal balls on a given week, k is the cost for storing each crystal ball in the warehouse for a week, and n is the maximum capacity of the warehouse. The following lines will describe the weeks specified in the contract in chronological order. Each week is described by a single line which will contain a pair of non-negative integers c and r, where c is the cost for manufacturing a new crystal ball using new crystal bought this week, and r is the number of crystal balls which must be delivered this week. A crystal ball can be manufactured and delivered in the same week if appropriate, in which case it won’t need to be stored in the warehouse at all. The last line of the input will contain the integer 0 and should not be processed. Output For each test case, output the minimum amount which the ACM will have to spend in order to fulfill the entire contract. All the numbers in the input will be at most 1000. Sample Input 4 1 0 1000 1 1 12 4 1 0 1000 1000 2 0 100 1 1 1000 1000 101 0 Sample Output 1007 101101

C语言，Ball Toss

Description Classmates stand in a circle facing inward, each with the direction left or right in mind. One of the students has a ball and begins by tossing it to another student. (It doesn't really matter which one.)When one catches the ball and is thinking left, she throws it back across the circle one place to the left (from her perspective) of the person who threw her the ball. Then she switches from thinking left to thinking right. Similarly, if she is thinking right, she throws the ball to the right of the person who threw it to her and then switches from thinking right to thinking left. There are two exceptions to this rule: If one catches the ball from the classmate to her immediate left and is also thinking left, she passes the ball to the classmate to her immediate right, and then switches to thinking right. Similarly, if she gets the ball from the classmate to her immediate right and is thinking right, she passes the ball to the classmate to her immediate left, and then switches to thinking left.(Note that these rules are given to avoid the problem of tossing the ball to oneself.) No matter what the initial pattern of left and right thinking is and who first gets tossed the ball,everyone will get tossed the ball eventually! In this problem, you will figure out how long it takes. You'll be given the initial directions of n classmates (numbered clockwise), and the classmate to whom classmate 1 initially tosses the ball. (Classmate 1 will always have the ball initially.) Input There will be multiple problem instances. Each problem instance will be of the form n k t1 t2 t3 . . . tn where n (2 <= n <= 30) is the number of classmates, numbered 1 through n clockwise around the circle,k (> 1) is the classmate to whom classmate 1 initially tosses the ball, and ti (i = 1, 2, . . . , n) are eacheither L or R, indicating the initial direction thought by classmate i. (n = 0 indicates end of input.) Output For each problem instance, you should generate one line of output of the form: Classmate m got the ball last after t tosses. where m and t are for you to determine. You may assume that t will be no larger than 100,000. Note that classmate number 1 initially has the ball and tosses it to classmate k. Thus, number 1 has not yet been tossed the ball and so does not switch the direction he is thinking. Sample Input 4 2 L L L L 4 3 R L L R 10 4 R R L R L L R R L R 0 Sample Output Classmate 3 got the ball last after 4 tosses. Classmate 2 got the ball last after 4 tosses. Classmate 9 got the ball last after 69 tosses.

javascipt函数调用过程和值传递

``` <canvas id="canvas" style="border: 1px solid red ;" ></canvas> <script type="text/javascript"> var ball={x:512, y:100, r:20, g:2, vx:-4, vy:4, color:"#00FFFF"}//g重力加速度， window.onload=function(){ var canvas=document.getElementById("canvas"); canvas.width=768; canvas.height=768; var context=canvas.getContext("2d"); setInterval(function(){ render(context); update();},50) } function update(){ ball.x =ball.x+ball.vx; ball.y =ball.y+ball.vy; ball.vy =ball.vy+ball.g; if(ball.y>=768-ball.r)、 //请问这里的条件成立后，ball.y和ball.vy的值是值传递的吗? //两个ball.y和ball.vy传递之后，每调用update()方法是怎样是怎样的执行过程？ { ball.y=768-ball.r; ball.vy=-ball.vy; } } function render(cxt){ cxt.clearRect(0,0,cxt.canvas.height,cxt.canvas.height); cxt.fillStyle=ball.color; cxt.beginPath(); cxt.arc(ball.x,ball.y,ball.r,0,2*Math.PI); cxt.closePath(); cxt.fill(); } </script> ```

sk-learn中KNN算法能动态的构建ball-tree吗

Problem Description Now that the Loonie is hovering about par with the Greenback, you have decided to use your \$1000 entrance scholarship to engage in currency speculation. So you gaze into a crystal ball which predicts the closing exchange rate between Canadian and U.S. dollars for each of the next several days. On any given day, you can switch all of your money from Canadian to U.S. dollars, or vice versa, at the prevailing exchange rate, less a 3% commission, less any fraction of a cent. Assuming your crystal ball is correct, what's the maximum amount of money you can have, in Canadian dollars, when you're done? Input The input contains a number of test cases, followed by a line containing 0. Each test case begins with 0 <d &#8804; 365, the number of days that your crystal ball can predict. d lines follow, giving the price of a U.S. dollar in Canadian dollars, as a real number. Output For each test case, output a line giving the maximum amount of money, in Canadian dollars and cents, that it is possible to have at the end of the last prediction, assuming you may exchange money on any subset of the predicted days, in order. Sample Input 3 1.0500 0.9300 0.9900 2 1.0500 1.1000 0 Sample Output 1001.60 1000.00

![图片说明](https://img-ask.csdn.net/upload/201905/10/1557480619_89273.png) ``` package tabletennis; import java.awt.*; import javax.swing.*; public class BallGame extends JFrame{ //加载图片 Image ball = Toolkit.getDefaultToolkit().getImage("img/ball.png"); Image desk = Toolkit.getDefaultToolkit().getImage("img/desk.jpg"); //画窗口方法 public void paint(Graphics g) { System.out.println("窗口被画了一次!"); g.drawImage(desk,0,0,null); g.drawImage(ball,100,100,null); } //窗口加载 void launchFrame() { setSize(300,300); setLocation(856,500); setVisible(true); } public static void main(String[] args) { BallGame ballgame = new BallGame(); ballgame.launchFrame(); } } ```

Problem Description Matt is playing a naive computer game with his deeply loved pure girl. The playground is a rectangle with walls around. Two balls are put in different positions inside the rectangle. The balls are so tiny that their volume can be ignored. Initially, two balls will move with velocity (1, 1). When a ball collides with any side of the rectangle, it will rebound without loss of energy. The rebound follows the law of refiection (i.e. the angle at which the ball is incident on the wall equals the angle at which it is reflected). After they choose the initial position, Matt wants you to tell him where will the two balls collide for the first time. Input The first line contains only one integer T which indicates the number of test cases. For each test case, the first line contains two integers x and y. The four vertices of the rectangle are (0, 0), (x, 0), (0, y) and (x, y). (1 ≤ x, y ≤ 105) The next line contains four integers x1, y1, x2, y2. The initial position of the two balls is (x1, y1) and (x2, y2). (0 ≤ x1, x2 ≤ x; 0 ≤ y1, y2 ≤ y) Output For each test case, output “Case #x:” in the first line, where x is the case number (starting from 1). In the second line, output “Collision will not happen.” (without quotes) if the collision will never happen. Otherwise, output two real numbers xc and yc, rounded to one decimal place, which indicate the position where the two balls will first collide. Sample Input 3 10 10 1 1 9 9 10 10 0 5 5 10 10 10 1 0 1 10 Sample Output Case #1: 6.0 6.0 Case #2: Collision will not happen. Case #3: 6.0 5.0

Ball Toss

Description Classmates stand in a circle facing inward, each with the direction left or right in mind. One of the students has a ball and begins by tossing it to another student. (It doesn't really matter which one.)When one catches the ball and is thinking left, she throws it back across the circle one place to the left (from her perspective) of the person who threw her the ball. Then she switches from thinking left to thinking right. Similarly, if she is thinking right, she throws the ball to the right of the person who threw it to her and then switches from thinking right to thinking left. There are two exceptions to this rule: If one catches the ball from the classmate to her immediate left and is also thinking left, she passes the ball to the classmate to her immediate right, and then switches to thinking right. Similarly, if she gets the ball from the classmate to her immediate right and is thinking right, she passes the ball to the classmate to her immediate left, and then switches to thinking left.(Note that these rules are given to avoid the problem of tossing the ball to oneself.) No matter what the initial pattern of left and right thinking is and who first gets tossed the ball,everyone will get tossed the ball eventually! In this problem, you will figure out how long it takes. You'll be given the initial directions of n classmates (numbered clockwise), and the classmate to whom classmate 1 initially tosses the ball. (Classmate 1 will always have the ball initially.) Input There will be multiple problem instances. Each problem instance will be of the form n k t1 t2 t3 . . . tn where n (2 <= n <= 30) is the number of classmates, numbered 1 through n clockwise around the circle,k (> 1) is the classmate to whom classmate 1 initially tosses the ball, and ti (i = 1, 2, . . . , n) are eacheither L or R, indicating the initial direction thought by classmate i. (n = 0 indicates end of input.) Output For each problem instance, you should generate one line of output of the form: Classmate m got the ball last after t tosses. where m and t are for you to determine. You may assume that t will be no larger than 100,000. Note that classmate number 1 initially has the ball and tosses it to classmate k. Thus, number 1 has not yet been tossed the ball and so does not switch the direction he is thinking. Sample Input 4 2 L L L L 4 3 R L L R 10 4 R R L R L L R R L R 0 Sample Output Classmate 3 got the ball last after 4 tosses. Classmate 2 got the ball last after 4 tosses. Classmate 9 got the ball last after 69 tosses.

Endless Spin用C语言实现

Problem 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

Problem Description The princess is trapped in a magic place. In this place, there are N magic stones. In order to rescue the princess, you should destroy all the stones. The N stones are in a straight line. We number them as s1, s2, ... sn from left to right. Each stone has a magic strength m1, m2, ... mn. You have a powerful skill that can do some damage to the stones. To release the skill, you should stand to the right of some stone (si). Then you throw a power ball towards left. Initially, this ball has a power of p. When it hits a stone, it will do some damage to the stone and its power will be decreased, and the ball will continue to fly left to the next stone if its power is still positive. Formally, if you stand to the right of si and the power ball's initial power is p, then the ball will do Max(0, p - (i - j) * (i - j)) damage to sj, for each j <= i. So from this formula, we can see that the damage to stone sj is only determined by the initial power of the ball and the number of stones between si and sj. A stone is destroyed if the damage you do is larger than its magic strength. Note that even if a stone is destroyed, it will not disappear; your magic ball will do damage to it and the power will be decreased by that stone. You are not strong enough so that you can release at most k magic balls. It will cost a lot of energy if the power of the magic ball is too high. So what is the minimum value of p with which you can destroy all the magic stones, with no more than k magic balls? You can choose where to release each magic ball as your will, and the power of the ball must be a positive integer. Input The first line is the number of cases T (T ≤ 100). For each case, the first line gives two integers n, k (1 ≤ n ≤ 50000, 1 ≤ k ≤ 100000). The second line are n integers, giving m1, m2, ... mn (1 ≤ m ≤ 109). Output Print minimum possible p in a line. Sample Input 2 1 1 1 3 1 1 4 5 Sample Output 2 6

Ball bearings

Problem Description The Swedish company SKF makes ball bearings. As explained by Britannica Online, a ball bearing is “one of the two types of rolling, or anti friction, bearings (the other is the roller bearing). Its function is to connect two machine members that move relative to one another so that the frictional resistance to motion is minimal. In many applications, one of the members is a rotating shaft and the other a fixed housing. Each ball bearing has three main parts: two grooved, ring like races and a number of balls. The balls fill the space between the two races and roll with negligible friction in the grooves. The balls may be loosely restrained and separated by means of a retainer or cage.” Presumably, the more balls you have inside the outer ring, the smoother the ride will be, but how many can you t within the outer ring? You will be given the inner diameter of the outer ring, the diameter of the balls, and the minimum distance between neighboring balls. Your task is to compute the maximum number of balls that will t on the inside of the outer ring (all balls must touch the outer ring). Input The first line of input contains a positive integer n that indicates the number of test cases. Then follow n lines, each describing a test case. Each test case consists of three positive oating point numbers, D, d, s, where D is the inner diameter of the outer ring, d is the diameter of a ball, and s is the minimum distance between balls. All parameters are in the range [0.0001, 500.0]. Output For each test case output a single integer m on a line by itself, where m is the maximum number of balls that can t in the ball bearing, given the above constraints. There will always be room for at least three balls. Sample Input 2 20 1 0.1 100.0 13.0 0.2 Sample Output 54 20

<div class="post-text" itemprop="text"> <p>I have a function in which I need to replace "byte" with "\$ball". This doesn't seem to work correctly. Here is the program snippet.</p> <pre><code>fun main() { str := []byte("\$apple in a byte ") strReplace := "\$ball" re := regexp.MustCompile("byte") final := re.ReplaceAll(str, []byte(strReplace)) ioutil.WriteFile("testfile.txt", final, 0744) } </code></pre> <p>Expected Output in testfile.txt: \$apple in a \$ball</p> <p>Actual Output in testfile.txt: \$apple in a </p> <p>Any solutions for successfully getting the desired output?</p> </div>

Simulation Wizardry C语言的技术

Problem Description Now that the Loonie is hovering about par with the Greenback, you have decided to use your \$1000 entrance scholarship to engage in currency speculation. So you gaze into a crystal ball which predicts the closing exchange rate between Canadian and U.S. dollars for each of the next several days. On any given day, you can switch all of your money from Canadian to U.S. dollars, or vice versa, at the prevailing exchange rate, less a 3% commission, less any fraction of a cent. Assuming your crystal ball is correct, what's the maximum amount of money you can have, in Canadian dollars, when you're done? Input The input contains a number of test cases, followed by a line containing 0. Each test case begins with 0 <d &#8804; 365, the number of days that your crystal ball can predict. d lines follow, giving the price of a U.S. dollar in Canadian dollars, as a real number. Output For each test case, output a line giving the maximum amount of money, in Canadian dollars and cents, that it is possible to have at the end of the last prediction, assuming you may exchange money on any subset of the predicted days, in order. Sample Input 3 1.0500 0.9300 0.9900 2 1.0500 1.1000 0 Sample Output 1001.60 1000.00

Collision Ball Game

Collision ball game is one of my favorite game when I was young. I was so fond of this game that I nearly decided to be a physical scientist (But I am not now). Firstly I wanted to construct a problem to describe the whole process of this game. But it is too difficult for me and I hate being bsed by others for wrong data. So I decide to make it easier, but maybe I will construct one next time. Assume the game is in a triangle with the lower-right angle k (in degree) and height h. (see the picture) The ball's initial place is in (0, a) and it eventually arrives at (b, 0) with collising the bevel edge only once. The two points (0, a) and (b, 0) will always be on the edge of the triangle. You are to calculate the distance the ball pass. The process of collision obey the physical law. The angle of incidence equals to the angle of reflection. Input The input contains multiple test cases. Each test cases contain four number (all double). The first line is the angle of the triangle k (in degree, 0 < k < 90). The second line is the height h (h > 0). The third line is a. The fourth line is b (a, b >= 0). Process to the end-of-file. Output For each test case print a single line that contains the distance the ball pass (rounded to 2 decimal places). Sample Input 45 2 1 1 30 2 1 2.3094 Sample Output 2.00 2.89

Problem Description Zty has met a big problem.His X Gu Niang was kidnaped by a secret organization.The organization only left Zty a letter: "If you can beat our boss ,we will give her the freedom,or..." The boss of the organization was so secret that no one knows his name.We only know that he was so powerful,and Zty is not powerful enough now,what he need to do is to train and to get more experience(Exp).Then he found a place wonderful for traning,There are N enemies there with different Exp,and Zty has M power ball,the number of power ball he used will effect the probability he beat the enemy.These probabilities are given as percentages pij, where i (with 1 ≤ i ≤ N) is the number of the enemy and j is the quantity of power balls used on it.One power ball can be used only once. Zty has to level up to 99,then he will be able to beat the boss.Of cause he is level 1 at the begining.He want to know weather the maximal expected Exp he can get is enough.The expected Exp is calculated as Sum(P(i)*Exp) where P is the probability. The Exp Zty need to level up one level is K/100 , and K will be given. Notice that: If Zty doesn't used a power ball,the probability he beat the enemy is 0. ^_^ Input The first line contain a T ,then T cases followed.Each test case has the following format: One line with one integer K <= 100000: as the description means. One line with one integer N with 1 ≤ N ≤ 100: the number of enemies. One line with one integer M with 0 ≤ M ≤ 100: the maximal number of available power ball. One line with N integers indicating the Exp of the N enemies. N lines, each line corresponding to a enemy i, containing n integers pi1, pi2, …, pim (the percentages, with 0 ≤ pi1, pi2, …, pim ≤ 100). Output If the maximal expected Exp Zty can get is enough for him to level up tp 99,then ouput "Love you Ten thousand years.",else ouput"Cry,men,not crime." Sample Input 2 1000 2 4 8 975 85 94 93 100 0 0 100 100 1000 1 4 979 0 0 0 100 Sample Output Love you Ten thousand years. Cry,men,not crime.

Problem Description The Swedish company SKF makes ball bearings. As explained by Britannica Online, a ball bearing is “one of the two types of rolling, or anti friction, bearings (the other is the roller bearing). Its function is to connect two machine members that move relative to one another so that the frictional resistance to motion is minimal. In many applications, one of the members is a rotating shaft and the other a fixed housing. Each ball bearing has three main parts: two grooved, ring like races and a number of balls. The balls fill the space between the two races and roll with negligible friction in the grooves. The balls may be loosely restrained and separated by means of a retainer or cage.” Presumably, the more balls you have inside the outer ring, the smoother the ride will be, but how many can you t within the outer ring? You will be given the inner diameter of the outer ring, the diameter of the balls, and the minimum distance between neighboring balls. Your task is to compute the maximum number of balls that will t on the inside of the outer ring (all balls must touch the outer ring). Input The first line of input contains a positive integer n that indicates the number of test cases. Then follow n lines, each describing a test case. Each test case consists of three positive oating point numbers, D, d, s, where D is the inner diameter of the outer ring, d is the diameter of a ball, and s is the minimum distance between balls. All parameters are in the range [0.0001, 500.0]. Output For each test case output a single integer m on a line by itself, where m is the maximum number of balls that can t in the ball bearing, given the above constraints. There will always be room for at least three balls. Sample Input 2 20 1 0.1 100.0 13.0 0.2 Sample Output 54 20

Python数据挖掘简易入门

&nbsp; &nbsp; &nbsp; &nbsp; 本课程为Python数据挖掘方向的入门课程，课程主要以真实数据为基础，详细介绍数据挖掘入门的流程和使用Python实现pandas与numpy在数据挖掘方向的运用，并深入学习如何运用scikit-learn调用常用的数据挖掘算法解决数据挖掘问题，为进一步深入学习数据挖掘打下扎实的基础。

HoloLens2开发入门教程

2019 Python开发者日-培训

Only老K说-爬取妹子图片（简单入门）

2020_五一数学建模_C题_整理后的数据.zip

R语言入门基础

PHP 5.0及以上 + MySQL 5.0及以上 开发的人才招聘系统完全可运行源码，按照操作说明简单配置即可运行。学习PHPWEB应用的完整系统程序源码。

Java基础知识面试题（2020最新版）

python可视化分析（matplotlib、seaborn、ggplot2）

python可视化分析总结（matplotlib、seaborn、ggplot）一、matplotlib库1、基本绘图命令3、图形参数设置4、特殊统计图的绘制4.1 数学函数图4.2 气泡图4.1 三维曲面图二、seaborn库1、常用统计图1.1 箱线图1.2 小提琴图1.3 点图1.4 条图与计数图1.5 分组图1.6 概率分布图2、联合图3、配对图三、ggplot库1、图层画法+常用图形2、快速绘图 一、matplotlib库 1、基本绘图命令 import matplotlib.pyplot as

Vue.js 2.0之全家桶系列视频课程

【大总结2】大学两年，写了这篇几十万字的干货总结

lena全身原图(非256*256版本,而是全身原图)

lena全身原图(非256*256版本,而是全身原图) lena原图很有意思,我们通常所用的256*256图片是在lena原图上截取了头部部分的256*256正方形得到的. 原图是花花公子杂志上的一个

【项目实战】 图书信息管理系统（Maven，mybatis）（第一个自己独立完成的项目）

《程序设计综合训练实践报告》 此项目为图书信息管理系统，是一个采用了mysql+mybatis框架+java编写的maven项目

Python入门视频精讲

Python入门视频培训课程以通俗易懂的方式讲解Python核心技术，Python基础，Python入门。适合初学者的教程，让你少走弯路！ 课程内容包括：1.Python简介和安装 、2.第一个Python程序、PyCharm的使用 、3.Python基础、4.函数、5.高级特性、6.面向对象、7.模块、8.异常处理和IO操作、9.访问数据库MySQL。教学全程采用笔记+代码案例的形式讲解，通俗易懂！！！

20行代码教你用python给证件照换底色

20行代码教你用python给证件照换底色

2018年全国大学生计算机技能应用大赛决赛 大题

2018年全国大学生计算机技能应用大赛决赛大题,程序填空和程序设计（侵删）

MySQL数据库从入门到实战应用

C/C++学习指南全套教程

C/C++学习的全套教程，从基本语法，基本原理，到界面开发、网络开发、Linux开发、安全算法，应用尽用。由毕业于清华大学的业内人士执课，为C/C++编程爱好者的教程。

C/C++跨平台研发从基础到高阶实战系列套餐

<p> 限时福利限时福利，<span>15000+程序员的选择！</span> </p> <p> 购课后添加学习助手（微信号：csdn590），按提示消息领取编程大礼包！并获取讲师答疑服务！ </p> <p> <br> </p> <p> 套餐中一共包含5门程序员必学的数学课程（共47讲） </p> <p> 课程1：《零基础入门微积分》 </p> <p> 课程2：《数理统计与概率论》 </p> <p> 课程3：《代码学习线性代数》 </p> <p> 课程4：《数据处理的最优化》 </p> <p> 课程5：《马尔可夫随机过程》 </p> <p> <br> </p> <p> 哪些人适合学习这门课程？ </p> <p> 1）大学生，平时只学习了数学理论，并未接触如何应用数学解决编程问题； </p> <p> 2）对算法、数据结构掌握程度薄弱的人，数学可以让你更好的理解算法、数据结构原理及应用； </p> <p> 3）看不懂大牛代码设计思想的人，因为所有的程序设计底层逻辑都是数学； </p> <p> 4）想学习新技术，如：人工智能、机器学习、深度学习等，这门课程是你的必修课程； </p> <p> 5）想修炼更好的编程内功，在遇到问题时可以灵活的应用数学思维解决问题。 </p> <p> <br> </p> <p> 在这门「专为程序员设计的数学课」系列课中，我们保证你能收获到这些:<br> <br> <span> </span> </p> <p class="ql-long-24357476"> <span class="ql-author-24357476">①价值300元编程课程大礼包</span> </p> <p class="ql-long-24357476"> <span class="ql-author-24357476">②应用数学优化代码的实操方法</span> </p> <p class="ql-long-24357476"> <span class="ql-author-24357476">③数学理论在编程实战中的应用</span> </p> <p class="ql-long-24357476"> <span class="ql-author-24357476">④程序员必学的5大数学知识</span> </p> <p class="ql-long-24357476"> <span class="ql-author-24357476">⑤人工智能领域必修数学课</span> </p> <p> <br> 备注：此课程只讲程序员所需要的数学，即使你数学基础薄弱，也能听懂，只需要初中的数学知识就足矣。<br> <br> 如何听课？ </p> <p> 1、登录CSDN学院 APP 在我的课程中进行学习； </p> <p> 2、登录CSDN学院官网。 </p> <p> <br> </p> <p> 购课后如何领取免费赠送的编程大礼包和加入答疑群？ </p> <p> 购课后，添加助教微信：<span> csdn590</span>，按提示领取编程大礼包，或观看付费视频的第一节内容扫码进群答疑交流！ </p> <p> <img src="https://img-bss.csdn.net/201912251155398753.jpg" alt=""> </p>

Eclipse archetype-catalog.xml

Eclipse Maven 创建Web 项目报错 Could not resolve archetype org.apache.maven.archetypes:maven-archetype-web

Python代码实现飞机大战