sign

1006 Sign In and Sign Out (25 point(s)) - C语言 PAT 甲级

由 别说谁变了你拦得住时间么 提交于 2019-11-28 03:45:26
1006 Sign In and Sign Out (25 point(s)) At the beginning of every day, the first person who signs in the computer room will unlock the door, and the last one who signs out will lock the door. Given the records of signing in’s and out’s, you are supposed to find the ones who have unlocked and locked the door on that day. Input Specification: Each input file contains one test case. Each case contains the records for one day. The case starts with a positive integer M, which is the total number of records, followed by M lines, each in the format: ID_number Sign_in_time Sign_out_time where times

hdu2087(剪花布条)

由 人盡茶涼 提交于 2019-11-28 02:13:25
http://acm.hdu.edu.cn/showproblem.php?pid=2087 View Code #include < iostream > using namespace std; int main() { char a[ 1001 ],b[ 1001 ]; int i,j; while (cin >> a) { if (a[ 0 ] == ' # ' ) break ; cin >> b; int L1 = strlen(a); int L2 = strlen(b); int sign = 0 , flag = 0 ; i = sign,j = 0 ; while (i < L1) { if (a[i] == b[j]) { i ++ ; j ++ ; if (j == L2) { flag ++ ; j = 0 ; sign = i; } } else { j = 0 ; sign ++ ; i = sign; } } cout << flag << endl; } return 0 ; } 转载于:https://www.cnblogs.com/FCWORLD/archive/2011/05/16/2048084.html 来源： https://blog.csdn.net/weixin_30764883/article/details/99910123

前n项和！

由 空扰寡人 提交于 2019-11-28 02:08:57
就当是练习了（各种高精度）！综合很强！ View Code #include " iostream " #define M 1010 using namespace std; char ch1[M],ch2[M]; int a[M],b[M]; int c[M],d[M]; int num[M]; int sum[M]; int W[M]; int H[M]; int t,k,g,v; int i,j; int La,Lb; int sign = 0 ; int s; int L; void _ADD() { La = 0 ; W[La ++ ] = (sum[L - 1 ] + 1 ) % 10 ; int flag = (sum[L - 1 ] + 1 ) / 10 ; int i; for (i = L - 2 ;i >= 0 ;i -- ) { W[La ++ ] = (sum[i] + flag) % 10 ; flag = (sum[i] + flag) / 10 ; } while (flag) { W[La ++ ] = flag % 10 ; flag /= 10 ; } } void Div( int * xx) { Lb = 0 ; int sign = 0 ; for ( int i = 0 ;i < L;i ++ ) { if ((xx[i] + sign

【Vue常用指令】

由 不问归期 提交于 2019-11-27 20:50:49
目录 v-html v-text v-for v-if v-show v-bind v-on v-model 指令修饰符 计算与侦听属性 自定义属性 获取DOM元素 "@ *** Vue.js官方给自己的定义为==数据模版引擎==，并给出了一套渲染数据的指令。本文将详细介绍Vue.js的常用指令 导入vue.js https://cdnjs.cloudflare.com/ajax/libs/vue/2.5.16/vue.min.js Vue.js使用了基于HTML的模版语法，使用vue最简单的方式就是渲染数据，渲染数据最常见的形式就是使用"Mustache"语法(双大括号)的文本插值。 ==- - - - - - - - - - - - - - - - - - - - - -== 一个简单的示例： <body> <div id="sign">{{ message }}</div> <script> let sign = new Vue({ el: '#sign', data: { message: 'Hello Vue!', }, }); </script> </body> 首先，创建一个vue实例，并在创建实例的过程中传入一个对象。 · 该对象的第一个属性名为el，它的值是我们需要渲染的目标标签，我们通过属性查找定位这个标签。 · 该对象的第二个属性名为data

【jQuery基础】

由 老子叫甜甜 提交于 2019-11-27 20:46:26
" 目录 #. 介绍 1. 优势 2. 版本 3. jQuery对象 #. 查找标签 1. 选择器 /. 基本选择器 /. 层级选择器 /. 基本筛选器 /. 使用jQuery实现弹框 /. 属性选择器 /. 表单常用筛选 2. 筛选器 /. 下一个元素 /. 上一个元素 /. 父亲元素 /. 儿子和兄弟元素 /. 查找与筛选 #. 样式标签 1. 样式操作 /. 样式类 /. CSS 2. 位置 3. 尺寸 4. 文本操作 5. 属性操作 /. 用于ID或自定义类 /. 用于checkbox和radio /. prop与attr的区别 6. 文档处理 /. 添加到指定元素内部 /. 添加到指定元素的外部 /. 移除和清空 /. 替换 /. 克隆 #. 事件 1. 常用事件 2. 绑定/移除 事件 3. 阻止后续事件执行 4. 阻止事件冒泡 5. 事件委托 6. each() 7. data() #. 动画效果 #. 介绍 jQuery是一个轻量级的、兼容多浏览器的JavaScript库。 jQuery使用户能够更方便地处理HTML Document、Events、实现动画效果、方便地进行Ajax交互，能够极大地简化JavaScript编程。它的宗旨就是：" Write less, do more. " jQuery官网 jQuery AP中文文档 BootCDN 1. 优势

Git sign off previous commits?

由 点点圈 提交于 2019-11-27 19:50:02
问题 I was wondering how to sign( -s ) off previous commits that I have made in the past in git? 回答1: To signoff the previous commit, use amend option: git commit --amend --signoff Edit: the amend does signoff only the latest commit. To signoff multiple commits, filter-branch and interpret-trailers as suggested by vonc et. al. should be used. Here is what worked for me. First, configure git to replace the token sign by Signed-off-by . This has to be done only once and is needed in the next step.

How to sign string with private key

由 本秂侑毒 提交于 2019-11-27 18:24:36
How can I get the signature of a string using SHA1withRSA if I already have the Private Key as byte[] or String ? Prabath Siriwardena I guess what you say is you know the key pair before hand and want to sign/verify with that. Please see the following code. import java.security.KeyPair; import java.security.KeyPairGenerator; import java.security.NoSuchAlgorithmException; import java.security.Signature; import sun.misc.BASE64Encoder; public class MainClass { public static void main(String[] args) throws Exception { KeyPair keyPair = getKeyPair(); byte[] data = "test".getBytes("UTF8"); Signature

DELPHI微信支付代码

由 此生再无相见时 提交于 2019-11-27 17:56:28
DELPHI微信支付代码 不管是微信支付还是支付宝支付， 3个最棘手的问题是： 1，如何生成签名 2，支付请求如何提交 3, 如何验证签名 下面就围绕这二个问题来讲。 我使用的是XE3. 先看微信支付： 1，微信支付的签名生成 uses IdHashMessageDigest, NatvieXml; //我使用的是NativeXml4.07 function TMicroPayParamBuilder.GetSign: string; var Digest: TIdHashMessageDigest5; utf8: UTF8String; begin utf8 := ''; Assert(FAppId <> '', '公众账号ID 不能为空！'); utf8 := utf8 + 'appid=' + FAppId; if FAttach <> '' then utf8 := utf8 + '&attach=' + FAttach; Assert(FAuthCode <> '', '授权码 不能为空！'); utf8 := utf8 + '&auth_code=' + FAuthCode; Assert(FBody <> '', '商品描述 不能为空！'); utf8 := utf8 + '&body=' + FBody; if FDetail <> '' then utf8 :=

Simplest way to check if two integers have same sign?

由 百般思念 提交于 2019-11-27 17:47:57
Which is the simplest way to check if two integers have same sign? Is there any short bitwise trick to do this? Here is a version that works in C/C++ that doesn't rely on integer sizes or have the overflow problem (i.e. x*y>=0 doesn't work) bool SameSign(int x, int y) { return (x >= 0) ^ (y < 0); } Of course, you can geek out and template: template <typename valueType> bool SameSign(typename valueType x, typename valueType y) { return (x >= 0) ^ (y < 0); } Note: Since we are using exclusive or, we want the LHS and the RHS to be different when the signs are the same, thus the different check

Javascript: convert a (hex) signed integer to a javascript value

由 怎甘沉沦 提交于 2019-11-27 16:36:42
问题 I have a signed value given as a hex number, by example 0xffeb and want convert it into -21 as a "normal" Javascript integer. I have written some code so far: function toBinary(a) { //: String var r = ''; var binCounter = 0; while (a > 0) { r = a%2 + r; a = Math.floor(a/2); } return r; } function twoscompl(a) { //: int var l = toBinaryFill(a).length; var msb = a >>> (l-1); if (msb == 0) { return a; } a = a-1; var str = toBinary(a); var nstr = ''; for (var i = 0; i < str.length; i++) { nstr +=
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