leijiawu 2周前
### 1-9题 ``` -- 1 select * from student where sage='20' -- 2 select cno,cname,ccredit from course -- 3 select sno,sname from student where ssex='男' -- 4 select student.sno,sname,ssex,sage,sdept,cno,grade from student, sc where student.sno=sc.sno -- 5 select sname,sdept,cno from student,sc where student.sno=sc.sno and student.ssex='女' and sc.grade>80 -- 6 select cno from course where cno not in ( select cno from student,sc where student.sno=sc.sno and student.sname='李勇' ) -- 7 select sname from student where sname not in ( select sname from student,sc,course where student.sno=sc.sno and sc.cno=course.cno and course.cname='操作系统' ) -- 8 select sname from student where sname not in ( select sname from student,sc where student.sno=sc.sno ) -- 9 select student.sno,sname,grade from student,sc,course where student.sno=sc.sno and sc.cno=course.cno and course.cname='数据库' ``` ### 1-7题 ``` -- 1 select * from student where sage<=19 or sdept='cs' -- 2 select sno from sc where cno='1' union select sno from sc where cno='2' -- 3 select * from student where sdept='cs' and sage<='19' -- 4 select sno from sc where cno='1' intersect select sno from sc where cno='2' -- 5 select * from student where sno is null or sname is null or sage is null or ssex is null or sdept is null -- 6 select sno from sc where cno='1' intersect select sno from sc where grade <'60' -- 7 select sno from sc where cno='1' intersect select sno from sc where grade <'60' or grade is null ```
leijiawu 2周前
1.[CTF工具箱](http://www.hiencode.com/)
leijiawu 1个月前
``` http:// net/T php? ``` <p align = "center"> <img src="https://leijiawu.github.io/httpwww.atoolbox.netTool.phpId=699.png" width="400" /> </p> ``` //html转码工具 https://www.lddgo.net/convert/htmlencode ```
leijiawu 1个月前
 <p align = "center"> <img src="https://img.zcool.cn/community/01ab4d58d9bd48a801219c77ae8272.jpg@1280w_1l_2o_100sh.jpg" width="400" /> </p>
leijiawu 1个月前
### 【上一个数据库实验】按住Ctrl单击右边链接👉[22计科2班数据库实验2：数据定义及更新](https://whathecode.fun/blog/29) ``` create database scsjk use scsjk create table student (sno char(9) primary key, sname char(20), ssex char(2), sage smallint, sdept char(20) ); create table course (cno char(4) primary key, cname char(40), cpno char(4), ccredit smallint ); create table sc (sno char(9), cno char(4), grade smallint, primary key(sno,cno) ); insert into student values('95001','李勇','男','20','CS') insert into student values('95002','刘晨','女','19','CS') insert into student values('95003','王敏','女','18','MA') insert into student values('95004','张立','男','19','IS') insert into course values('1','数据库',5,4) insert into course values('2','数学',null,2) insert into course values('3','信息系统',1,4) insert into course values('4','操作系统',6,3) insert into course values('5','数据结构',7,4) insert into course values('6','数据处理',null,2) insert into course values('7','PASCAL语言',6,4) insert into sc values('95001','1',92) insert into sc values('95001','2',85) insert into sc values('95001','3',88) insert into sc values('95002','2',90) insert into sc values('95002','3',80) create view isstudent as select sno,sname,ssex,sage from student where sdept='IS' with check option; create index sc_snograde on sc(sno,grade desc) select*from student select*from course select*from sc --1 select sno"姓名", 1991-sage"出生年份" from student --2 select sno, cno, grade from sc where grade>=70 and grade<=80 --3 select distinct sno from sc --4 select cno, cname from course where ccredit<4 --5-1 select sno,sname,sage from student where sdept='IS' --5-2 select sno,sname,sage from isstudent --6 select sno,sname from student where sname like '刘%' --7 select cno,cname from course where cpno in ('1','2','3') --8 select count(*)计算机人数 from student where sdept='CS' --9 select count(distinct cno)学生选修课程门数 from sc --10 select sdept"系部",count(sno)系部人数 from student group by sdept --11 select sdept"系部",ssex"性别",count(sno)"人数" from student group by sdept, ssex order by sdept asc,ssex desc --12 select max(grade)"课程2最高分", avg(grade)"课程2平均分" from sc where cno='2' --13 select sno"学号", avg(grade)"平均分" from sc group by sno --14 select sno from sc group by sno having avg(grade)>85 --15 select sno"学号",avg(grade)"平均分" from sc group by sno order by avg(grade) desc ```
leijiawu 1个月前
``` <!-- 编写JavaScript程序。实现用户发表评论（出现在最新评论的顶部）。 --> <!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8"> <meta name="viewport" content="width=device-width, initial-scale=1.0"> <title>Document</title> <style> body { margin: 0; padding: 0; } .news { position: relative; width: 800px; background-color: #fff; margin: 20px auto; justify-content: center; align-items: center; /* box-shadow: 0px 0px 10px 5px #888; */ } .image { text-align: center; } hr { width: 800px; } p { font-size: 18px; } button { width: 80px; height: 35px; background-color: teal; color: #fff; border-radius: 5px; cursor: pointer; } button:hover { background-color: rgb(46, 97, 97); box-shadow: 2px 2px 5px #888; } .comment, .comment-list { width: 800px; margin: 0 auto; text-align: center; margin-top: 20px; } .headline { margin-top: 20px; flex-direction: row; justify-content: space-between; } h2, h3 { display: inline; } input { width: 500px; height: 100px; background-color: #F0FFFF; margin-left: 20px; font-size: 18px; } input[text]::placeholder { text-align: left; } .item { width: 600px; font-size: 18px; border: 1px solid #888; margin: 0 auto; margin-top: 10px; text-align: justify; border-radius: 8px; } </style> </head> <body> <div class="news"> <h1>王毅会见美中关系全国委员会负责人</h1> <h4>北京时间2024-03-26 15时00分</h4> <p>  2024年3月26日，中共中央政治局委员、外交部长王毅在北京会见美中 关系全国委员会董事会主席埃文·格林伯格、会长欧伦斯。 </p> <p>   王毅说，去年11月，美中关系全国委员会同其他友好团体一道，克服困难、排除干扰，响应两国人民共同愿望，在旧金山为习近平主席举行隆重欢迎宴会。习主席在演讲中回顾了两国人民友好交往历史，强调中美关系希望在人民、基础在民间、未来在青年、活力在地方。 希望美中关系全国委员会继续为推动两国人民加强交往，推动中美关系健康稳定发展发挥建设性作用。 </p> <div class="image"> <img src="news.jpg"> </div> </div> <!-- 评论区 --> <hr /> <div class="comment"> <h2>网友评论</h2> <input type="text" placeholder="我有话要说..."> <button>发布</button> </div> <div class="comment-list" id="commentlist"> <div class="headline"> <h2>最新评论</h2> <h3 id="comment_count">评论数：0</h3> </div> <!-- .... --> <div class="item"> <p>推动两国人民加强交往</p> <p>今天21:31</p> </div> <div class="item"> <p>合作共赢</p> <p>今天20:22</p> </div> </div> <script> const btn = document.querySelector('button'); const commentlist = document.getElementById("commentlist") var count = 2; btn.addEventListener('click', function () { let newComment = document.createElement('div') let newCommentp1 = document.createElement('p') let newCommentp2 = document.createElement('p') var lastComment = document.querySelector('.item') var commentValue = document.querySelector('input').value newCommentp1.innerHTML = commentValue; newComment.appendChild(newCommentp1) newComment.appendChild(newCommentp2) commentlist.insertBefore(newComment, lastComment) newComment.className = 'item' const comment_count = document.getElementById("comment_count") comment_count.innerHTML = `评论数：${++count}` let dateArr=[ "星期日", "星期一", "星期二", "星期三", "星期四", "星期五", "星期六" ] let nowDate = new Date() let week = dateArr[nowDate.getDay()] let hh = nowDate.getHours()//小时 let mm = nowDate.getMinutes()//计算分钟 let ss = nowDate.getSeconds()//计算秒数 let ampm = hh <= 12 ? '上午' : '下午' hh = ("0" + hh).slice(-2) mm = ("0" + mm).slice(-2) ss = ("0" + ss).slice(-2) newCommentp2.innerHTML = `今天${hh}:${mm}` }) </script> <!-- setInterval(()=>{ let dateArr=[ "星期日", "星期一", "星期二", "星期三", "星期四", "星期五", "星期六" ] let nowDate = new Date() let week = dateArr[nowDate.getDay()] let hh= nowDate.getHours()//小时 let mm = nowDate.getMinutes()//计算分钟 let ss = nowDate.getSeconds()//计算秒数 let ampm = hh <=12? '上午':'下午' hh = ("0" + hh).slice(-2) mm = ("0" + mm).slice(-2) ss = ("0" + ss).slice(-2) document.getElementById("time").innerHTML = `${week} ${hh}时${mm}分${ss}秒 ${ampm}` },1000) --> </body> </html> ```
leijiawu 1个月前
### [递增三元组](https://www.lanqiao.cn/problems/172/learning/?page=1&first_category_id=1&sort=students_count&second_category_id=3&problem_id=172) ``` #include <bits/stdc++.h> using namespace std; int n, a[100005], b[100005], c[100005]; long long ans = 0; // bool cmp(int a, int b) // { // return a > b; // } int main() { cin >> n; for (int i = 1; i <= n; i++) cin >> a[i]; for (int i = 1; i <= n; i++) cin >> b[i]; for (int i = 1; i <= n; i++) cin >> c[i]; sort(a + 1, a + n + 1); sort(b + 1, b + n + 1); sort(c + 1, c + n + 1); // for (int i = 1; i <= n; i++) // { // for (int j = n; j >= 1; j--) // { // for (int k = n; k >= 1; k--) // { // if (a[i] >= b[j]||b[j] >= c[k]) // { // break; // } // if (a[i] < b[j] && b[j] < c[k]) // ans++; // } // if (a[i] >= b[j]) // { // break; // } // } // if(a[i]>=b[n]){ // break; // } // } for(int i=1;i<=n;i++){ //for的是中间的数，比它小的*比它大的等于含有b[i]的组合数 long long x = lower_bound(a+1,a+n+1,b[i])-a-1; long long y = n-(upper_bound(c+1,c+n+1,b[i])-c-1); ans+=x*y; } cout << ans; return 0; } ```
leijiawu 1个月前
优化技巧：先排好序，从大到小遍历，当后面的数都不满足时，就终止当前循环. [1.倍数问题](https://www.lanqiao.cn/problems/168/learning/?page=1&first_category_id=1&sort=students_count&second_category_id=3&problem_id=168) ```cpp #include <iostream> #include <algorithm> using namespace std; int n, k; long arr[100005]; bool cmp(long a, long b) { return a > b; } int main() { cin >> n >> k; for (int i = 1; i <= n; i++) { cin >> arr[i]; } sort(arr + 1, arr + n + 1, cmp); long long maxans = 0; for (int i = 1; i <= n - 2; i++) { for (int j = i + 1; j <= n - 1; j++) { for (int h = j + 1; h <= n; h++) { if ((arr[i] + arr[j] + arr[h]) % k == 0) { if (maxans < arr[i] + arr[j] + arr[h]) { maxans = arr[i] + arr[j] + arr[h]; } if (arr[i] + arr[j] + arr[h] < maxans) { break; } } } if (arr[i] + arr[j] + arr[j + 1] < maxans) { break; } } if (arr[i] + arr[i+1] + arr[i + 2] < maxans) { break; } } cout << maxans; return 0; } ```
leijiawu 1个月前
```javascript <!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8"> <meta name="viewport" content="width=device-width, initial-scale=1.0"> <title>Document</title> <style> body { background-color: #CDCDCD; } .container { position: relative; width: 600px; height: 480px; background-color: #fff; margin: 100px auto; text-align: center; border-radius: 10px; box-shadow: 0px 0px 20px 5px #999; } .container h1 { padding: 20px 10px; } .userinput { margin: 5px 20px; /* padding-top: 20px; */ font-size: 18px; } button { width: 100px; height: 35px; margin-top: 20px; margin-bottom: 20px; margin-right: 20px; background-color: teal; color: #fff; border-radius: 5px; cursor: pointer; } button:hover { background-color: rgb(46, 97, 97); box-shadow: 2px 2px 5px #888; } p { font-size: 12px; color: red; visibility: hidden; } input { width: 300px; height: 35px; border-radius: 5px; border: 1px solid #888; } </style> </head> <body> <div class="container"> <h1>欢迎注册</h1> <div class="userinput"> 手机号： <input type="text" id="phone" placeholder="手机号">* <p id="phone-tip">手机号位数不足或已注册过提示</p> </div> <div class="userinput"> 登录密码： <input type="text" id="pwd1" placeholder="以字母开头，长度在6~18之间，只能包含字母、数字和下划线">* <p id="pwd-tip1">密码位数不足或与要求不匹配提示</p> </div> <div class="userinput"> 确认密码： <input type="text" id="pwd2" placeholder="输入与上一个相同的密码">* <p id="pwd-tip2">密码与上一密码框输入不相同提示</p> </div> <button id="register">注册</button> <button id="cacle">取消</button> </div> <script> var value1 var value2 var value3 var flag1=false var flag2=false var flag3=false var arr= new Array() function checkPhone() { var reg = /^1[3456789]\d{9}$/; value1 = document.getElementById("phone").value; if (reg.test(value1) == false) { document.getElementById("phone-tip").style.visibility = 'visible'; flag1=false } else { document.getElementById("phone-tip").style.visibility = 'hidden'; flag1=true } } function checkPwd1() { var reg = /^[a-zA-Z]\w{5,17}$/; value2 = document.getElementById("pwd1").value; if (reg.test(value2) == false) { document.getElementById("pwd-tip1").style.visibility = 'visible'; flag2=false } else { document.getElementById("pwd-tip1").style.visibility = 'hidden'; flag2=true } } function checkPwd2() { value3 = document.getElementById("pwd2").value; if (value2!=value3) { document.getElementById("pwd-tip2").style.visibility = 'visible'; flag3=false } else { document.getElementById("pwd-tip2").style.visibility = 'hidden'; flag3=true } } document.getElementById("register").addEventListener('click',function(){ checkPhone(); checkPwd1(); checkPwd2(); if(flag1&&flag2&&flag3){ userObj={userphone:value1,password:value2} arr.push(userObj) console.log(arr) } }) document.getElementById("cacle").addEventListener('click',function(){ document.getElementById("phone").value=""; document.getElementById("pwd1").value=""; document.getElementById("pwd2").value=""; document.getElementById("phone-tip").style.visibility = 'hidden'; document.getElementById("pwd-tip1").style.visibility = 'hidden'; document.getElementById("pwd-tip2").style.visibility = 'hidden'; }) document.getElementById("phone").addEventListener('input',function(){ checkPhone(); }) </script> </body> </html> ```
leijiawu 1个月前
题目: 初始5, 一年后变成5+9，二年后变成5+9+17... 输入n年，以及n年后的总数s，让我们求初始值。 所以变成求数列a1的值。由题目得，样例的a1=5，a2=2*a1-1, a3=2*a2-1, a4=2*a3-1。 a(n)=2*a(n-1)-1 两边减1，得 a(n)-1=2*a(n-1)-2, 即a(n)-1=2*(a(n-1)-1) 所以 [a(n)-1]/[a(n-1)-1]=2 设b(n)=a(n)-1 则b(n)为公比q为2的等比数列，即b(n)=b1*q^(n-1), 又b1=a1-1, 所以b(n)=(a1-1)*2^(n-1) 所以a(n)-1=(a1-1)*2^(n-1) 所以a(n)的通项公式a(n)=(a1-1)*2^(n-1)+1。 由于第1年的和s为a(n)的前2项和, 也就是S2=a1+a2, 第2年的和s为a(n)的前3项和, 也就是S3=a1+a2+a3, 第3年的和s为a(n)的前4项和, 也就是S4=a1+a2+a3+a4, 所以输入第n年的和，总数就是S(n+1)。 所以就可以套数列求和公式求解， a(n)的前n项和为S(n)=[(a1-1)*(1-q^n)]/(1-q)+n, 由q=2，得S(n)=(a1-1)*(2^n - 1)+n 所以输入第n年的总数s=S(n+1)=(a1-1)*[2^(n+1) - 1] + (n+1), 即 s=(a1-1)*[2^(n+1) - 1] + (n+1) 所以我们要求的原始数量也就是a1, 为a1 = (s-n-1)/[2^(n+1) - 1]+1, 所以 cout<<(s-n-1)/(pow(2,n+1)-1)+1; ```cpp #include <iostream> #include<math.h> using namespace std; double n; double s; long long ans=0; int main() { cin>>n>>s; cout<<(s-n-1)/(pow(2,n+1)-1)+1; return 0; } ```
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