关键词[c������]相关搜索结果,共搜索到61149条结果

使用curl断点续传下载文件

)C:\>curl--helpUsage:curl[options...]<url>-a,--appendAppendtotargetfilewhenuploading--basicUseHTTPBasicAuthentication-E,--cert<certificate[:password]>Clientcertificatefileandpassword-C,--continue-at<offset>Resumedtransferoffset-b,--cookie<data>Sendcookiesfromstring/file-c,

SwaggerUI看烦了,IGeekFan.AspNetCore.Knife4jUI 帮你换个新皮肤

jUI;🚁ConfigureServices3.服务配置,CustomOperationIds和AddServer是必须的。services.AddSwaggerGen(c=>{c.SwaggerDoc("v1",newOpenApiInfo{Title="APIV1",Version="v1"});c.AddServer(newOpenApiServer(){Url="",Description="vvv"});c.CustomOperationIds(apiDesc=>{varcontrollerAction=apiDesc.ActionDescriptorasControl

03 | python并发网络通信模型(多进程、多线程的网络传输模型以及简单的ftp模型)

handle(c):whileTrue:data=c.recv(1024).decode()ifnotdata:breakprint(data)c.send(b'OK')#创建监听套接字s=socket()s.setsockopt(SOL_SOCKET,SO_REUSEADDR,1)#设置套接字的端口重用,端口可以立即被重用s.bind(ADDR)s.listen(5)#处理僵尸进程signal.signal(signal.SIGCHLD,signal.SIG_IGN)print("Listentheport8888....")whileTrue:#循环等待客户端连接try:c,addr=s

oeasy教您玩转linux010107那啥在哪 whereis

iversemultiverse这条设置的是debian的源的位置摁下esc注意左下角不再是Insert(插入状态),而是退回到了空白依次摁下:wq回车保存并退出vi用apt下载源代码apt是高级包管理的命令运行sudoaptupdate(更新源)sudoaptsourcecoreutils(获得源代码)在当前文件夹下找到coreutilsXXX文件夹进入src源文件文件夹(source)lsls.*找到了ls.c文件(ls的c语言源文件)思考🤔我们真的可以获得ls源代码,这太棒了!👊可是如果有多个版本的命令比如:java8java13那我到底用的是那个版本的java呢?我们下次再说。👋

养猪日记 2022.1.27

Thursday 晴🐖生日快乐!我的乖🐖~晚上和同学吃了顿饭,喝点啤酒,晕乎的,就不学习了。1道算法题,3节MySQL课,3节内存管理课。亲亲🐖~,今天和🐖打了电话~。好困吖,睡觉。🐖妈妈晚安。🐖宝宝晚安。1:12  安达

uoj455 【UER #8】雪灾与外卖

while(p[i].c&&!q1.empty())66{67t=q1.top();v=t.x;68if(p[i].x+p[i].w+v>=0)break;69q1.pop();70t2=min(ll(p[i].c),t.y);71t.y-=t2;p[i].c-=t2;72ans+=t2*(p[i].x+p[i].w+v);73t1+=t2;74q2.push((P){-2*p[i].x-v,t2});75if(t.y)q1.push(t);76}77if(t1)q1.push((P){-p[i].x-p[i].w,t1});78if(p[i].c)q2.push((P

搜索和组合数学P1246 编码

d;chars[10];//ch记录询问的字符串intn,id,c[30][30];//n记录询问字符串长度signedmain(){FOR(i,0,26)c[i][0]=1;FOR(i,1,26)FOR(j,1,26)c[i][j]=c[i-1][j]+c[i-1][j-1];//组合数递推式,scanf("%s",s+1);n=strlen(s+1);//处理长度小于n的字符串编号FOR(i,1,n-1){if(s[i]>=s[i+1]){cout<<0;//不升序排列就直接输出0exit(0);}id+=c[26][i];//从26个字母中选i个,一定可以排成递增的字符

Git 消息提交规范

提交类型Emoji初始提交。🎉:tada:版本标签🔖:bookmark:新功能✨:sparkles:错误修正🐛:bug:文档📚:books:记录源代码💡:bulb:性能提升🐎:racehorse:添加测试✅:white_check_mark:使测试通过✔️:heavy_check_mark:一般更新⚡:zap:改进格式/结构🎨:art:重构代码🔨:hammer:删除代码/文件🔥:fire:升级依赖⬆️:arrow_up:降级依赖⬇️:arrow_down:删除依赖项➖:heavy_minus_sign:添加依赖项➕:heavy_plus_sign:关键修补程序(定期发布的补丁)🚑:ambulance:工作正在进行中🚧:construction:配置文件🔧:wrench:合并分支🔀:twisted_rightwards_arrows:错误的代码/需要改进💩:hankey:还原更改⏪:rewind:重大变化💥:boom:Seealso:https://gist.github.com/parmentf/035de27d6ed1dce0b36a

Educational Codeforces Round 76 (Rated for Div. 2) A. Two Rival Students 水题

A.TwoRivalStudentsThereare𝑛studentsintherow.Andtherearetworivallingstudentsamongthem.Thefirstoneisinposition𝑎,thesecondinposition𝑏.Positionsarenumberedfrom1to𝑛fromlefttoright.Sincetheyarerivals,youwanttomaximizethedistancebetweenthem.Ifstudentsareinpositions𝑝and𝑠respectively,thendistancebetweenthemis|𝑝−𝑠|.Youcandothefollowingoperationatmost𝑥times:choosetwoadjacent(neighbouring)studentsandswapthem.Calculatethemaximumdistancebetweentworivallingstudentsafteratmost𝑥swaps.InputThefirstlinecontainsoneinteger𝑡(1≤𝑡≤100)—thenumberoftestcases.Theonlylineofeachtestcasecontainsfourintegers𝑛,𝑥,𝑎and𝑏(2≤𝑛≤100,0≤𝑥≤100,1≤𝑎,𝑏≤𝑛,𝑎≠𝑏)—thenumberofstudentsintherow,thenumberofswapswhichyoucando,andpositionsoffirstandsecondrivalingstudentsrespectively.OutputForeachtestcaseprintoneinteger—themaximumdistancebetweentworivalingstudentswhichyoucanobtain.Exampleinput351321003310016023output2991NoteInthefirsttestcaseyoucanswapstudentsinpositions3and4.Andthenthedistancebetweentherivalsisequalto|4−2|=2.Inthesecondtestcaseyoudon'thavetoswapstudents.Inthethirdtestcaseyoucan'tswapstudents.题意现在有n个东西排列成一行,a在第a个位置,b在第b个位置,现在每次操作可以使得一个人和周围的人交换位置,问你在操作最多x次的情况下,最多能够使得a和b的距离最远是多少题解每次交换一次,肯定可以使得距离加一,那么答案就是要么就最远,要么就当前的距离+x即可。代码#include<bits/stdc++.h>usingnamespacestd;voidsolve(){intn,x,a,b;cin>>n>>x>>a>>b;if(a>b)swap(a,b);cout<<min(b-a+x,n-1)<<endl;}intmain(){intt;cin>>t;while(t--)solve();}

C博客作业01--分支、顺序结构

这个作业属于哪个班级C语言--网络2011/2012这个作业的地址C博客作业00--顺序分支结构这个作业的目标初步认识C语法,掌握数据表达、printf、scanf语法及分支结构内容0.展示PTA总分顺序结构:分支结构:1.本章学习总结1.1学习内容总结1.1.1顺序结构1.printf与scanf的基本用法:(1)printf语句:printf("输出内容");(2)scanf语句:scanf("输入内容",&输入内容对应的标识符或字符);注:1.输入内容时不要忘记"&";2.输出时可以使用\n进行换行;3.输入字符型数据时要用"char"定义且要注意用"getchar()"

js ES6 Reflect All In One

length,r=c<3?target:desc===null?desc=Object.getOwnPropertyDescriptor(target,key):desc,d;if(typeofReflect==="object"&&typeofReflect.decorate==="function")r=Reflect.decorate(decorators,target,key,desc);elsefor(vari=decorators.length-1;i>=0;i--)if(d=decorators[i])r=(c<3?d(r):c>3?d(t

Collection集合

llection<String>c=newArrayList<String>();/*Ctrl+B可以查看类的信息Alt+7打开一个窗口能看类的所有信息*///add方法永远返回的是true,因为add方法里只有returntruec.add("Hello");c.add("World");System.out.println(c.remove("World"));//存在即移出返回trueSystem.out.println(c.remove("java"));//不存在返回falseSystem.out.println(c.contains("Hello"));//

使用 VSCode 给STM32配置一个串口 printf 工程

实是让单片机连接超级终端用的。既然要连接超级终端那想必除了要实现printf()还要实现scanf()和其他的东东吧。不过我们目前只关心printf(),剩下的以后再收拾。我们需要用到这个工程里的两个文件,main.c和syscalls.c.main.c的重要性没什么可说的,而syscalls翻译过来是系统调用的意思,因此我觉得所有的底层应该都是在这里实现的。首先看他的main.c文件,把所有干扰视线的注释删除掉可以发现其中除了各模块初始化等我们熟悉的代码外就多了以下两段内容。....../*在我们的工程中__GNUC__肯定是定义了的,因此这一段其实就只有#definePUTCHAR_PRO

CF1037H Security (SAM+二维偏序)

符\(c\)肯定比\(T\)字典序要大,我们只需在\(S\)串中找到一个串为\(T+c\),并使\(c\)最小即可。否则,\(T[1,len]\)上与\(S\)匹配,但是\(T[1,len+1]\)不匹配了,我们只需找到一个最小字符\(c\),满足\(c>T[len+1]\)且\(T[1,len]+c\)为\(S\)的子串,那么\(T[1,len]+c\)就是答案了。如果没找到这样的\(c\),将匹配长度减一,重复进行这个操作。\(len=0\)也找不到这样的\(c\),只能输出\(-1\)。可以用反证法证明这种构造方法得到的一定是最小的比\(T\)字典序大的串。而要想知道\(+c\)后

(数据科学学习手札132)Python+Fabric实现远程服务器连接

方法上传至服务器c.put('file_transfer.txt',remote='/home/feffery/')#打印已上传文件内容c.run('cat/home/feffery/file_transfer.txt');2.3.2从服务器下载指定文件到本地相反的,当我们需要从服务器取回指定文件到本地时,就可以使用get()方法:c=Connection(host=f'{user}@{host}',connect_kwargs={'password':password})#向文件末尾追加行c.run('echo"\n2">>file_transfer.txt')c.get('/h

Codeforces Round #599 (Div. 2) B1. Character Swap (Easy Version) 水题

B1.CharacterSwap(EasyVersion)Thisproblemisdifferentfromthehardversion.InthisversionUjanmakesexactlyoneexchange.Youcanhackthisproblemonlyifyousolvebothproblems.Afterstrugglingandfailingmanytimes,Ujandecidedtotrytocleanuphishouseagain.Hedecidedtogethisstringsinorderfirst.Ujanhastwodistinctstrings𝑠and𝑡oflength𝑛consistingofonlyoflowercaseEnglishcharacters.Hewantstomakethemequal.SinceUjanislazy,hewillperformthefollowingoperationexactlyonce:hetakestwopositions𝑖and𝑗(1≤𝑖,𝑗≤𝑛,thevalues𝑖and𝑗canbeequalordifferent),andswapsthecharacters𝑠𝑖and𝑡𝑗.Canhesucceed?Notethathehastoperformthisoperationexactlyonce.Hehastoperformthisoperation.InputThefirstlinecontainsasingleinteger𝑘(1≤𝑘≤10),thenumberoftestcases.Foreachofthetestcases,thefirstlinecontainsasingleinteger𝑛(2≤𝑛≤104),thelengthofthestrings𝑠and𝑡.Eachofthenexttwolinescontainsthestrings𝑠and𝑡,eachhavinglengthexactly𝑛.ThestringsconsistonlyoflowercaseEnglishletters.Itisguaranteedthatstringsaredifferent.OutputForeachtestcase,output"Yes"ifUjancanmakethetwostringsequaland"No"otherwise.Youcanprinteachletterinanycase(upperorlower).Exampleinput45sousehouhe3catdog2aaaz3abcbcaoutputYesNoNoNoNoteInthefirsttestcase,Ujancanswapcharacters𝑠1and𝑡4,obtainingtheword"house".Inthesecondtestcase,itisnotpossibletomakethestringsequalusingexactlyoneswapof𝑠𝑖and𝑡𝑗.题意现在给你两个字符串s和t,然后你可以选择i,j两个位置,交换s[i]和t[j]。问你在只交换一次的情况下,能不能使得两个字符串相同。题解我们遍历两个字符串,我们进行对比,看一共有多少个位置不同。如果所有位置都相同,输出yes如果只有一个位置不同,那么一定是no。如果是两个位置不同,我们就暴力枚举交换s[i],t[j]还是s[j]和t[i]。其他情况都是no代码#include<bits/stdc++.h>usingnamespacestd;intn;strings,t;voidsolve(){cin>>n;cin>>s>>t;vector<int>pos;for(inti=0;i<s.size();i++){if(s[i]!=t[i]){pos.push_back(i);}}if(pos.size()==0){puts("YES");}elseif(pos.size()==1){puts("NO");}elseif(pos.size()==2){swap(s[pos[0]],t[pos[1]]);if(s==t){puts("YES");return;}swap(s[pos[0]],t[pos[1]]);swap(t[pos[0]],s[pos[1]]);if(s==t){puts("YES");return;}puts("NO");}else{puts("NO");}}intmain(){intt;scanf("%d",&t);while(t--)solve();}

SwiftUI 布局汇总

lue:$data.rating)Viewrunningresults🔝StepperStepperisusedtoincreaseordecreasethevalue,example:Stepper(value:$value,step:2,onEditingChanged:{cinprint(c)}){Text("StepperValue:\(self.value)")}.padding(50)Viewrunningresults🔝SegmentedControlisdeprecatedSegmentedControlisusedforsegmentationconditionsele

【Docker Desktop】在 Windows 上安装 Docker Desktop

ws功能"或"为WSL2安装所需的Windows组件"选项。参考链接:WSL2安装使用-简书(jianshu.com)关闭并重启初始化开始设置尝试运行一个容器:复制并粘贴此命令到您的终端,然后返回MicrosoftWindows[版本10.0.19043.1526](c)MicrosoftCorporation。保留所有权利。C:\Users\xybdiy>dockerrun-d-p80:80docker/getting-startedUnabletofindimage'docker/getting-started:latest'locallylatest:Pullingfromdoc

指定Swagger接口文档中参数序列化组件为Newtonsoft.Json

le.AspNetCore.Newtonsoft-Version6.3.02CLI:dotnetaddpackage--version6.3.0Swashbuckle.AspNetCore.Newtonsoft并在服务注册处添加使用 AddSwaggerGenNewtonsoftSupport()方法来指定其使用Newtonsoft.Json1services.AddMvc();23services.AddSwaggerGen(c=>4{5c.SwaggerDoc("v1",newOpenApiInfo{Title="MyAPI",Version="v1"});6});7ser

Codeforces Round #575 (Div. 3) D2. RGB Substring (hard version) 水题

D2.RGBSubstring(hardversion)inputstandardinputoutputstandardoutputTheonlydifferencebetweeneasyandhardversionsisthesizeoftheinput.Youaregivenastring𝑠consistingof𝑛characters,eachcharacteris'R','G'or'B'.Youarealsogivenaninteger𝑘.Yourtaskistochangetheminimumnumberofcharactersintheinitialstring𝑠sothatafterthechangestherewillbeastringoflength𝑘thatisasubstringof𝑠,andisalsoasubstringoftheinfinitestring"RGBRGBRGB...".Astring𝑎isasubstringofstring𝑏ifthereexistsapositiveinteger𝑖suchthat𝑎1=𝑏𝑖,𝑎2=𝑏𝑖+1,𝑎3=𝑏𝑖+2,...,𝑎|𝑎|=𝑏𝑖+|𝑎|−1.Forexample,strings"GBRG","B","BR"aresubstringsoftheinfinitestring"RGBRGBRGB..."while"GR","RGR"and"GGG"arenot.Youhavetoanswer𝑞independentqueries.InputThefirstlineoftheinputcontainsoneinteger𝑞(1≤𝑞≤2⋅105)—thenumberofqueries.Then𝑞queriesfollow.Thefirstlineofthequerycontainstwointegers𝑛and𝑘(1≤𝑘≤𝑛≤2⋅105)—thelengthofthestring𝑠andthelengthofthesubstring.Thesecondlineofthequerycontainsastring𝑠consistingof𝑛characters'R','G'and'B'.Itisguaranteedthatthesumof𝑛overallqueriesdoesnotexceed2⋅105(∑𝑛≤2⋅105).OutputForeachqueryprintoneinteger—theminimumnumberofcharactersyouneedtochangeintheinitialstring𝑠sothatafterchangingtherewillbeasubstringoflength𝑘in𝑠thatisalsoasubstringoftheinfinitestring"RGBRGBRGB...".Exampleinput352BGGGG53RBRGR55BBBRRoutput103NoteInthefirstexample,youcanchangethefirstcharacterto'R'andobtainthesubstring"RG",orchangethesecondcharacterto'R'andobtain"BR",orchangethethird,fourthorfifthcharacterto'B'andobtain"GB".Inthesecondexample,thesubstringis"BRG".题目大意q次询问,每次询问给你一个n个长度的字符串和长度k,要求你在n长度的字符串里面找到一个长度为k的子串,使得修改最少为RGBRGB....的某个字串题解水题,枚举匹配位置,然后暴力匹配即可,然后再求长度为k的前缀和最小值代码#include<bits/stdc++.h>usingnamespacestd;constintmaxn=2e5+7;intn,k;stringori="RGB";strings;inta[maxn],sum[maxn];intsolve(intst){for(inti=0;i<s.size();i++){a[i]=(s[i]==ori[(st+i)%3]?0:1);sum[i]=(i>0?a[i]+sum[i-1]:a[i]);}intres=s.size();for(inti=k-1;i<s.size();i++){res=min(res,sum[i]-(i-k>=0?sum[i-k]:0));}returnres;}voidsolve(){scanf("%d%d",&n,&k);cin>>s;intans=s.size();for(intst=0;st<3;st++){ans=min(ans,solve(st));}printf("%d\n",ans);}intmain(){intt;scanf("%d",&t);while(t--){solve();}return0;}

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