<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[CherryTaylor的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/CherryTaylor</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; CherryTaylor]]></copyright><item><title><![CDATA[【黑马程序员JAVA笔记+完整代码】项目实战2 石头迷阵]]></title><link>https://blog.youkuaiyun.com/CherryTaylor/article/details/147217073</link><guid>https://blog.youkuaiyun.com/CherryTaylor/article/details/147217073</guid><author>CherryTaylor</author><pubDate>Mon, 14 Apr 2025 11:53:08 +0800</pubDate><description><![CDATA[黑马Java网课]]></description><category></category></item><item><title><![CDATA[【黑马程序员JAVA笔记+完整代码】项目实战1 结合GUI编程实现一个黑马人事管理系统]]></title><link>https://blog.youkuaiyun.com/CherryTaylor/article/details/147216387</link><guid>https://blog.youkuaiyun.com/CherryTaylor/article/details/147216387</guid><author>CherryTaylor</author><pubDate>Mon, 14 Apr 2025 11:48:35 +0800</pubDate><description><![CDATA[黑马Java网课]]></description><category></category></item><item><title><![CDATA[【黑马程序员JAVA笔记+完整代码】DAY05-08 面向对象编程]]></title><link>https://blog.youkuaiyun.com/CherryTaylor/article/details/147099062</link><guid>https://blog.youkuaiyun.com/CherryTaylor/article/details/147099062</guid><author>CherryTaylor</author><pubDate>Wed, 09 Apr 2025 20:52:27 +0800</pubDate><description><![CDATA[黑马Java网课]]></description><category></category></item><item><title><![CDATA[【黑马程序员JAVA笔记+完整代码】DAY04 数组]]></title><link>https://blog.youkuaiyun.com/CherryTaylor/article/details/147098960</link><guid>https://blog.youkuaiyun.com/CherryTaylor/article/details/147098960</guid><author>CherryTaylor</author><pubDate>Wed, 09 Apr 2025 17:45:57 +0800</pubDate><description><![CDATA[黑马Java网课笔记]]></description><category></category></item><item><title><![CDATA[【黑马程序员JAVA笔记+完整代码】DAY03 程序流程控制]]></title><link>https://blog.youkuaiyun.com/CherryTaylor/article/details/147049315</link><guid>https://blog.youkuaiyun.com/CherryTaylor/article/details/147049315</guid><author>CherryTaylor</author><pubDate>Mon, 07 Apr 2025 17:56:23 +0800</pubDate><description><![CDATA[黑马Java网课笔记]]></description><category></category></item><item><title><![CDATA[【黑马程序员JAVA笔记+完整代码】DAY02 基础语法]]></title><link>https://blog.youkuaiyun.com/CherryTaylor/article/details/147048566</link><guid>https://blog.youkuaiyun.com/CherryTaylor/article/details/147048566</guid><author>CherryTaylor</author><pubDate>Mon, 07 Apr 2025 17:42:08 +0800</pubDate><description><![CDATA[黑马Java网课笔记]]></description><category></category></item><item><title><![CDATA[【黑马程序员JAVA笔记+完整代码】DAY01 入门知识]]></title><link>https://blog.youkuaiyun.com/CherryTaylor/article/details/147041708</link><guid>https://blog.youkuaiyun.com/CherryTaylor/article/details/147041708</guid><author>CherryTaylor</author><pubDate>Mon, 07 Apr 2025 13:27:04 +0800</pubDate><description><![CDATA[黑马JAVA网课笔记]]></description><category></category></item><item><title><![CDATA[雨课堂同济大学《知识产权法》（2023）期末考试]]></title><link>https://blog.youkuaiyun.com/CherryTaylor/article/details/134781104</link><guid>https://blog.youkuaiyun.com/CherryTaylor/article/details/134781104</guid><author>CherryTaylor</author><pubDate>Mon, 04 Dec 2023 16:39:39 +0800</pubDate><description><![CDATA[韦某开设了“韦老四”煎饼店,在当地颇有名气。经营汽车配件的个体户肖某从外地路过,吃过后赞不绝口。当发现韦某尚未注册商标时,肖某就餐饮服务注册了“韦老四”商标。甲受乙的委托，为乙画了一幅肖像。黄某扫描上传吴某的小说至网站“精品区”,李某下载打印用于出租,学生白某阅读该小说转换为音频,发送至微信群,以下说法对的是?因不正当竞争行为受到损害的经营者的赔偿数额,赔偿数额不包括经营者为制止侵权行为所支付的合理开支。链接：https://pan.baidu.com/s/18vp7WZomN60X1kOp-PCmbA。]]></description><category></category></item><item><title><![CDATA[杭电OJ记录--1014]]></title><link>https://blog.youkuaiyun.com/CherryTaylor/article/details/119941998</link><guid>https://blog.youkuaiyun.com/CherryTaylor/article/details/119941998</guid><author>CherryTaylor</author><pubDate>Thu, 26 Aug 2021 22:46:00 +0800</pubDate><description><![CDATA[1014–伪随机数
要求：

随机输入STEP和MOD（最大至100000），seed(x+1) = [seed(x) + STEP] %MOD，若结果可取到从0至MOD-1，则good，反之则bad。

分析：

seed(x+1) = [seed(x) + STEP] %MOD
当STEP和MOD的最大公约数等于1的时候，seed可以取到从0至MOD-1
问题转换成了判断两个大数的最大公约数是否为1

注意：

输出的对齐格式可以不写  因为要求的数字右对齐就是默认的对齐方式
字母从第25行开始   也]]></description><category></category></item><item><title><![CDATA[杭电OJ记录--1013]]></title><link>https://blog.youkuaiyun.com/CherryTaylor/article/details/119846410</link><guid>https://blog.youkuaiyun.com/CherryTaylor/article/details/119846410</guid><author>CherryTaylor</author><pubDate>Sat, 21 Aug 2021 22:54:49 +0800</pubDate><description><![CDATA[1013–计算数的根

大数，用字符串存储
注意字符串的输入格式  getline(cin, a)

————AC代码————
#include &lt;iostream&gt;
#include &lt;string&gt;
using namespace std;

int main() {
	int len, root = 0;
	string a ;
	while (getline(cin, a) &amp;&amp; a != "0") {						//用双引号
		for (len = a.s]]></description><category></category></item><item><title><![CDATA[杭电OJ记录--1012]]></title><link>https://blog.youkuaiyun.com/CherryTaylor/article/details/119828708</link><guid>https://blog.youkuaiyun.com/CherryTaylor/article/details/119828708</guid><author>CherryTaylor</author><pubDate>Fri, 20 Aug 2021 17:48:21 +0800</pubDate><description><![CDATA[1012–精确计算e

注意输出格式与题目一致
前三行直接输出
之后小数部分保留九位有效数字，不够则补0

————AC代码————
#include &lt;iostream&gt;
#include &lt;iomanip&gt;
using namespace std;

int main() {
	int n = 0, i = 1;
	double fm = 1.0;
	double *a = new double[10];						//存放 1/（i!),i从0到9共10项
	a[0] = 1.]]></description><category></category></item><item><title><![CDATA[杭电OJ记录--1005]]></title><link>https://blog.youkuaiyun.com/CherryTaylor/article/details/119807220</link><guid>https://blog.youkuaiyun.com/CherryTaylor/article/details/119807220</guid><author>CherryTaylor</author><pubDate>Thu, 19 Aug 2021 17:43:17 +0800</pubDate><description><![CDATA[1005–计算f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.
要注意的问题是因为有内存和编译时间的限制
【time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)】
且题目中明确说了n最大到一亿，如果直接用递归必然MLE，所以要找到规律（我MLE那么多次之前怎么就没想过呢……）

规律递推，f(n)可能的取值是6 5 4]]></description><category></category></item><item><title><![CDATA[杭电OJ记录--1004]]></title><link>https://blog.youkuaiyun.com/CherryTaylor/article/details/119806471</link><guid>https://blog.youkuaiyun.com/CherryTaylor/article/details/119806471</guid><author>CherryTaylor</author><pubDate>Thu, 19 Aug 2021 17:16:59 +0800</pubDate><description><![CDATA[1004–最受欢迎的气球

关于new和delete的格式！！！！
数组

 int *a = new int [5];//不能赋初值
 delete [ ] a;//[ ]很重要

单独的一个数据
 int *a = new int; //没有初始值
 int *b = new int(10); //有初始值，注意用小括号
 delete a;
 delete b;


输出数量最多的气球颜色，比较的是气球个数（数组中的数据），修改的是下标（maxindex)
————AC代码————

#include]]></description><category></category></item><item><title><![CDATA[杭电OJ记录--1001]]></title><link>https://blog.youkuaiyun.com/CherryTaylor/article/details/119806364</link><guid>https://blog.youkuaiyun.com/CherryTaylor/article/details/119806364</guid><author>CherryTaylor</author><pubDate>Thu, 19 Aug 2021 17:02:47 +0800</pubDate><description><![CDATA[1001–求和n+(n-1)+……+1
————AC代码————
//In this problem, your task is to calculate SUM(n) = 1 + 2 + 3 + ... + n.
#include &lt;iostream&gt;
using namespace std;

int SUM(int n) {
	if (n &lt;= 1)
		return n;
	else
		return (n + SUM(n - 1));
}

int main() {
	int n]]></description><category></category></item><item><title><![CDATA[杭电OJ记录--1000]]></title><link>https://blog.youkuaiyun.com/CherryTaylor/article/details/119806145</link><guid>https://blog.youkuaiyun.com/CherryTaylor/article/details/119806145</guid><author>CherryTaylor</author><pubDate>Thu, 19 Aug 2021 16:57:29 +0800</pubDate><description><![CDATA[1000–简单求和
————以下AC代码————
#include &lt;iostream&gt;
using namespace std;

int main() {
	int A, B;
	while (cin &gt;&gt; A &gt;&gt; B) {
		cout &lt;&lt; A + B &lt;&lt; endl;
	}
	return 0;
}


]]></description><category></category></item><item><title><![CDATA[杭电OJ记录--1008]]></title><link>https://blog.youkuaiyun.com/CherryTaylor/article/details/119805920</link><guid>https://blog.youkuaiyun.com/CherryTaylor/article/details/119805920</guid><author>CherryTaylor</author><pubDate>Thu, 19 Aug 2021 16:52:35 +0800</pubDate><description><![CDATA[1008--读题要注意：


首先，每一组结束之后要返回底层(0)；
其次，输入的一组数据中，第一个数表示停几次，后面的数字分别表示在几层停
————以下代码————

#include &lt;iostream&gt;
using namespace std;

int main() {
	int to_floor, now_floor;   		//分别表示目的楼层和当前所在楼层
	int n;							//表示一共停几次
	int time;						//表示一共用时；
	while (cin ]]></description><category></category></item></channel></rss>