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vue3--学习技术胖笔记

<template> <div> <a href="https://vitejs.dev" target="_blank"></a> <H2>欢迎光临红浪漫洗浴中心</H2> <div>请选择一位美女</div> <button v-for="(item,index) in girls"

PAT Advanced 1036 Boys vs Girls（25）

题目描述： This time you are asked to tell the difference between the lowest grade of all the male students and the highest grade of all the female students. Input Specification: Each input file contains one test case. Each case contains a positive integer N,

c3p0-config

<c3p0-config> <named-config name="hello">  <property name="driverClass">com.mysql.jdbc.Driver</property>  <property name="jdbcUrl">jdbc:mysql://12

CF253A Boys and Girls 题解

Content 有 \(n\) 个男生、\(m\) 个女生坐在一排，请求出这样一种方案，使得相邻两个座位之间的人的性别不同的次数最多。数据范围：\(1\leqslant n,m\leqslant 100\)。 Solution 这题不难，就是先把能男女组合的尽量组合了，然后剩余有多的直接丢在后面不管（或者也可以先把多的求出来然后丢

好听的女孩英文名大全Girls Names

Hazel Annabelle Di/ Diane Jen/Jena Sage Leighann Hope Martina Katarina Danielle Joy Mabel Elly Vivi Debra Debby / Deb Sylvia/Sylvianne Vi/Violet Viola Charmaine Anya Bea/Beatrice Glee Harmony Cheryl Sofia Jean Sherry Nina Cleo Elsie Georgina Portia Sam/Sa

7.代理模式

package proxyPattern; import sun.net.www.content.image.gif; /** * 普通实现方式 * * 场景： tom 认识 dany dany认识莉莉但是tom不认识莉莉但是但是 tom想追莉莉就想送莉莉礼物 * * @author LiMing E-mail:1151143484@qq.com * @date 2017

18 修改

修改一、修改单表的记录语法： UPDATE 表名 SET 列=新值, 列=新值,... WHERE 筛选条件；案例一：修改beauty表中姓唐的女神的电话为1388 UPDATE girls.beauty SET phone="123213" WHERE NAME LIKE "唐%"；案例二：修改boys表中id为2的名字为张飞，魅力值为100 UPDATE girls.beauty

PAT 甲级 1036 boys and girls 最后两个测试点

PAT 甲级 1036 boys and girls 最后两个测试点一直没过有无大佬解个惑orz #include <stdio.h> struct NODE{ char name[10]; char id[50]; int grade; }temp,m,f; int main() { int n; scanf("%d",&n); m.grade=101; f.grade=-1; for(int i=0;i<n;i++) { char sex; scanf

css3技巧——产品列表之鼠标滑过效果translateY(三)

<div class="main"> <div class="view view-tenth"> <figure> <div class="thumb"><img src="http://tupian.enterdesk.com/2014/lxy/2014/04/11/2/5.jpg" /></div> <div class="mask&q

hdu1069 Girls and Boys

the second year of the university somebody started a study on the romantic relations between the students. The relation “romantically involved” is defined between one girl and one boy. For the study reasons it is necessary to find out the maximum set sati

Girls' research（mancher）

Girls’ research Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 6177 Accepted Submission(s): 2296 Problem Description One day, sailormoon girls are so delighted that they intend to research a

1036 Boys vs Girls (25 分)（PAT甲级）（C语言）

This time you are asked to tell the difference between the lowest grade of all the male students and the highest grade of all the female students. Input Specification: Each input file contains one test case. Each case contains a positive integer N, fol

PAT 甲级 1036 Boys vs Girls (25 分)（简单题）

1036 Boys vs Girls (25 分) This time you are asked to tell the difference between the lowest grade of all the male students and the highest grade of all the female students. Input Specification: Each input file contains one test case. Each case c

PAT甲级——A1036 Boys vs Girls

This time you are asked to tell the difference between the lowest grade of all the male students and the highest grade of all the female students. Input Specification: Each input file contains one test case. Each case contains a positive integer N, follow

poj 1466 Girls and Boys 二分图最大独立集（★★☆☆☆）

http://poj.org/problem?id=1466 题意：从n个男女同学中找出两两没有任何关系的最大人数。使用C++输入输出，WA了好几次，改了用C就过，也不知道怎么回事，好气愤啊！！！ Sample Input 7 0: (3) 4 5 6 1: (2) 4 6 2: (0) 3: (0) 4: (2) 0 1 5: (1) 0 6: (2) 0 1 3 0: (2) 1 2 1: (1) 0 2: (1)

POJ 1466 Girls and Boys 题解《挑战程序设计竞赛》

为什么80%的码农都做不了架构师？>>> POJ 1466 Girls and BoysFFF团：有V个学生，每行数据x: (n) y z ... 表示学生x与另n个人：y和z……有一腿（贵圈真乱）。不知为何，你手中突然多出了火把和汽油。在放火温暖全世界之前，你想豁免单身狗集合，求集合大小。3.5借助水流解决问题的网络

Ckp的约会（xmu oj）贪心算法问题 by C++

Ckp的约会（xmu oj）贪心算法问题 by C++题目描述题目分析代码截图：题目描述 Ckp的约会描述今天Ckp打算去约会。大家都知道Ckp是超级大帅哥，所以和他约会的MM也超级多，她们每个人都和Ckp订了一个约会时间。但是今天Ckp刚打算出门的时候才发现，某几个MM的约会时间有冲突。由于Ckp不

[2018HN省队集训D6T2] girls

[2018HN省队集训D6T2] girls 题意给定一张 \(n\) 个点 \(m\) 条边的无向图, 求选三个不同结点并使它们两两不邻接的所有方案的权值和 \(\bmod 2^{64}\) 的值. 一个方案 \((i,j,k)\) 的权值定义为 \(iA+jB+kC\), 其中 \(A,B,C\) 给定且 \(i<j<k\). 点从 \(0\) 开始标号. \(n,m\le 2

bzoj5407: girls

题目描述：大皮出行，妹子成群，似乎这种事早已经司空见惯了。大皮最终还是厌倦了这样的⽣活，懂得了节制，他希望每次恰好只有三个妹子陪他。大皮有 $n$个妹子，编号为$0$到$n-1$ ，大皮需要次从$n$个中选出三个来愉悦身心。那么问题来了，有许多妹子为了得到大皮的欢心，产生了冲突，这样的冲突

SQLAlchemy的使用---M2M增删改查

from sqlalchemy.orm import sessionmaker from sqlalchemy_M2M import engine, Girls, Boys Session = sessionmaker(engine) db_session = Session() # 1.通过Boy添加Girl relationship 反向 boy = Boys(name='XJ') boy.boy2girl = [Girls(name='G1'), Girl