git pull/push 速度慢

问题:

windows环境下,不知道为什么最近gitpush和pull速度很慢,经常连接超时;

解决方案:

搜了一下解决方案,大概记录一下

~/.ssh/config

Host github.com *.github.com *.codeup.aliyun.com
    User git
    # SSH默认端口22,git://, HTTPS默认端口443,https://
    Port 22
    Hostname %h
    # 这里放你的SSH私钥
    IdentityFile ~\.ssh\id_rsa
    # 设置代理, 127.0.0.1:10808 换成你自己代理软件监听的本地地址
    # HTTPS使用-H,SOCKS使用-S
    ProxyCommand connect -S 127.0.0.1:10808 %h %p

需要用到connect客户端
connect
解压exe文件后放到C:\Windows\System32目录下面

linux下编译安装php8

下载:

https://www.php.net/downloads

configure

从宝塔安装的php看下通用的一些扩展编译参数
# php -i |grep configure

./configure 
--prefix=/www/server/php/74  --with-config-file-path=/www/server/php/74/etc  --enable-fpm  --with-fpm-user=www  --with-fpm-group=www  --enable-mysqlnd  --with-mysqli=mysqlnd  --with-pdo-mysql=mysqlnd  --with-iconv-dir  --with-freetype  --with-jpeg  --with-zlib  --with-libxml-dir=/usr  --enable-xml  --disable-rpath  --enable-bcmath  --enable-shmop  --enable-sysvsem  --enable-inline-optimization  --with-curl  --enable-mbregex  --enable-mbstring  --enable-intl  --enable-ftp  --enable-gd  --with-openssl  --with-mhash  --enable-pcntl  --enable-sockets  --with-xmlrpc  --enable-soap  --with-gettext  --disable-fileinfo  --enable-opcache  --with-sodium  --with-webp

稍微修改以下

--prefix=/usr/local/php/81 --with-config-file-path=/usr/local/php/81/etc  --enable-fpm  --with-fpm-user=www  --with-fpm-group=www  --enable-mysqlnd  --with-mysqli=mysqlnd  --with-pdo-mysql=mysqlnd  --with-iconv-dir  --with-freetype  --with-jpeg  --with-zlib  --with-libxml-dir=/usr  --enable-xml  --disable-rpath  --enable-bcmath  --enable-shmop  --enable-sysvsem  --enable-inline-optimization  --with-curl  --enable-mbregex  --enable-mbstring  --enable-intl  --enable-ftp  --enable-gd  --with-openssl  --with-mhash  --enable-pcntl  --enable-sockets  --with-xmlrpc  --enable-soap  --with-gettext  --disable-fileinfo  --enable-opcache  --with-sodium  --with-webp

configure跑跑看缺什么就装对应软件

缺少 安装
libxml-2.0 libxml2-dev
libcurl libcurl4-openssl-dev
libpng libpng-dev
libwebp libwebp-dev
libjpeg libjpeg-dev
freetype2 libfreetype-dev
oniguruma libonig-dev
libsodium libsodium-dev

....

安装

执行make && make install

RFID卡号格式的常见样式

北京友我科技RFID读写器 转载请注明出处,本篇地址
http://www.youwokeji.com.cn/ywdn/NoteDetails.asp?id=11

由于各个厂家的读卡器译码格式不尽相同,在读卡输出时,读出的二进制或十六进制(Hex)结果应该是唯一的,但是又可以通过以下几种主要换算办法,输出不同结果的十进制卡号(Dec),因此,请您一定在购买卡片或卡片喷号时,注意卡号格式的一致性:

1、格式0:10位十六进制的ASCII字符串,即10 Hex格式。
如:某样卡读出十六进制卡号为:“01026f6c3a”。

2、格式1:将格式0中的后8位,转换为10位十进制卡号,即8H---10D。
即将“ 026f6c3a”转换为:“0040856634”。

此格式喷码喷码较为常见。

3、格式2:将格式0中的后6位,转换为8位十进制卡号,即6H---8D。
即将“ 6f6c3a”转换为:“07302202”。

4、格式3:将格式0中的倒数第5、第6位,转换为3位十进制卡号,再将后4位,转换为5位十进制卡号,中间用“,”分开,即“2H + 4H”。
即将2H“ 6f”转换为:“111”,4H “6c3a”转为“27706”。 最终将2段号连在一起输出为“111,27706”。

此格式为标准的韦根26(V26)格式,只使用最后6位编码,也有许多卡采用此格式喷码。

5、格式4:将格式0中后8位的前4位,转换为5位十进制卡号,再将后4位,转换为5位十进制卡号,中间用“,”分开,即“4Hex + 4Hec”。
照此推算结果为:00623,27706 (4H+4H)

BCC 异或校验 php实现

问题描述

对接一个硬件设备协议要求进行bcc校验,数据包是十六进制表示的,用php处理

实现方式

    public static function dec2hex($data, $wordLength) {
        return str_pad(dechex($data), $wordLength * 2, '0', STR_PAD_LEFT);
    }

    //十六进制数据生成bcc校验码(十六进制)
    public static function bcc($data) {
        $data = str_split($data, 2);
        $length = count($data);
        $result = intval($data[0], 16);
        for($i=0; $i<$length-1; $i++) {
            $result ^= intval($data[$i+1], 16);
        }
        return self::dec2hex($result, 1);
    }

其它

bcc校验工具

dcat admin 多文件上传

问题描述

需要解决dcat admin 表单多附件管理的(附件名和保存文件名分开保存)

解决方案

上传文件管理

  • 首先是新建上传文件表

    Schema::create('uploads', function (Blueprint $table) {
    $table->string('disk', 30)->comment('存储位置');
    $table->string('type', 20)->default('file')->comment('文件类型');
    $table->string('name', 150)->nullable()->comment('文件名称');
    $table->string('path', 150)->comment('文件路径');
    });
  • 新建上传文件用的路由

    $router->any('/upload/handle/{disk}/{dir}', 'UploadController@handle');
  • 上传处理

    public function handle($diskName, $dir) {
        $disk =$this->disk($diskName);
    
        // 判断是否是删除文件请求
        if ($this->isDeleteRequest()) {
            Upload::where('path', request()->key)->delete();
            // 删除文件并响应
            return $this->deleteFileAndResponse($disk);
        }
    
        // 获取上传的文件
        $file = $this->file();
        $newName =  Common::uniqueID().'.'.$file->getClientOriginalExtension();
        $result = $disk->putFileAs($dir, $file, $newName);
        $path = "{$dir}/$newName";
        $upload = new Upload;
        $upload->name = request()->name;
        $upload->path = $path;
        $upload->disk = $diskName;
        $upload->save();
    
        return $result
            ? $this->responseUploaded($path, $disk->url($path))
            : $this->responseErrorMessage('文件上传失败');
    }
  • 扩展字段

    class CustomMultiFile extends MultipleFile
    {
    protected function initialPreviewConfig()
    {
        $previews = [];
        foreach ($this->value() as $path => $name) {
            $previews[] = [
                'id'   => $path,
                'path' => Helper::basename($name),
                'url'  => $this->objectUrl($path),
            ];
        }
    
        return $previews;
    }
    }
  • 注册扩展字段

    Form::extend('customMultiFile', CustomMultiFile::class);
  • 表单字段

    $form->customMultiFile('attachment', '附件')->disk('public')
     ->url('upload/handle/public/attachment')
     ->autoUpload()->autoSave(false)
     ->removable(true)->limit(10);

codesignal practice 4

10

You are given an array of non-negative integers numbers. You are allowed to choose any number from this array and swap any two digits in it. If after the swap operation the number contains leading zeros, they can be omitted and not considered (eg: 010 will be considered just 10).

Your task is to check whether it is possible to apply the swap operation at most once, so that the elements of the resulting array are strictly increasing.

Example

For numbers = [1, 5, 10, 20], the output should be makeIncreasing(numbers) = true.

The initial array is already strictly increasing, so no actions are required.

For numbers = [1, 3, 900, 10], the output should be makeIncreasing(numbers) = true.

By choosing numbers[2] = 900 and swapping its first and third digits, the resulting number 009 is considered to be just 9. So the updated array will look like [1, 3, 9, 10], which is strictly increasing.

For numbers = [13, 31, 30], the output should be makeIncreasing(numbers) = false.

The initial array elements are not increasing.
By swapping the digits of numbers[0] = 13, the array becomes [31, 31, 30] which is not strictly increasing;
By swapping the digits of numbers[1] = 31, the array becomes [13, 13, 30] which is not strictly increasing;
By swapping the digits of numbers[2] = 30, the array becomes [13, 31, 3] which is not strictly increasing;
So, it's not possible to obtain a strictly increasing array, and the answer is false.

Input/Output

[execution time limit] 4 seconds (php)

[input] array.integer numbers

An array of non-negative integers.

Guaranteed constraints:
1 ≤ numbers.length ≤ 103,
0 ≤ numbers[i] ≤ 103.

[output] boolean

Return true if it is possible to obtain a strictly increasing array by applying the digit-swap operation at most once, and false otherwise.

function makeIncreasing($numbers) {
    $hasChanged = false;
    for($i=1; $i< count($numbers); $i++){
        if($numbers[$i] > $numbers[$i-1]){
            continue;
        }
        if($numbers[$i] < 10 && $numbers[$i-1]< 10){
            return false;
        }
        $before = $i >= 2 ? $numbers[$i-2] : 0; 
        if(swapNumberSuccess($before, $numbers[$i-1], $numbers[$i]) && (!$hasChanged)){
            $hasChanged = true;
            continue;
        }else{
            return false;
        }
    }
    return true;
}

function swapNumberSuccess($number0, $number1, $number2){
    echo $number1,',', $number2, PHP_EOL;
    if($number1 == 1000){
        $min_1 = 1;
    }elseif($number1/100 > 1){
        $d1=floor($number1/100);
        $d2=(floor($number1/10)%10);
        $d3=$number1%10;
        $d_min = min($d1, $d2, $d3);
        $d_max = max($d1, $d2, $d3);
        $d_mid = mid($d1, $d2, $d3, $d_min, $d_max);
        $min_1 = 100 * $d_min + 10 * $d_mid + $d_max;
        echo $d_min,',', $d_mid,',',$d_max, PHP_EOL;
    }elseif($number1/10 > 1){
        $d1= floor($number1/10);
        $d2= $number1%10;
        $min_1 = min($d1, $d2)*10 + max($d1, $d2);
    }else{
        $min_1 = $number1;
    }
    echo $min_1, PHP_EOL;
    if($min_1 < $number2 && $min_1 > $number0){
        return true;
    }

    if($number2 == 1000){
        $max2 = 1000;
    }elseif($number2/100 > 1){
        $d1=floor($number2/100);
        $d2=(floor($number2/10)%10);
        $d3=$number2%10;
        $d_min = min($d1, $d2, $d3);
        $d_max = max($d1, $d2, $d3);
        $d_mid = mid($d1, $d2, $d3, $d_min, $d_max);
        $max_2 = 100 * $d_max + 10 * $d_mid + $d_min;
    }elseif($number2/10 > 1){
        $d1= floor($number2/10);
        $d2= $number2%10;
        $max_2 = max($d1, $d2)*10 + min($d1, $d2);
    }else{
        $max_2 = $number2;
    }
    echo $max_2, PHP_EOL;
    if($max_2 > $number1){
        return true;
    }
    return false;

}

function mid($n1, $n2, $n3, $d_min, $d_max){
    if($n1 < $d_max && $n1 > $d_min){
        return $n1;
    }

    if($n2 < $d_max && $n2 > $d_min){
        return $n2;
    }

    return $n3;
}

11

You are given a matrix of integers field of size n × m representing a game field, and also a matrix of integers figure of size 3 × 3 representing a figure. Both matrices contain only 0s and 1s, where 1 means that the cell is occupied, and 0 means that the cell is free.

You choose a position at the top of the game field where you put the figure and then drop it down. The figure falls down until it either reaches the ground (bottom of the field) or lands on an occupied cell, which blocks it from falling further. After the figure has stopped falling, some of the rows in the field may become fully occupied.

demonstration

Your task is to find the dropping position such that at least one full row is formed. As a dropping position you should consider the column index of the cell in game field which matches the top left corner of the figure 3 × 3 matrix. If there are multiple dropping positions satisfying the condition, feel free to return any of them. If there are no such dropping positions, return -1.

Note: When falling, the 3 × 3 matrix of the figure must be entirely inside the game field, even if the figure matrix is not totally occupied.

Example

For

field =  [[0, 0, 0],
          [0, 0, 0],
          [0, 0, 0],
          [1, 0, 0],
          [1, 1, 0]]
and

figure = [[0, 0, 1],
          [0, 1, 1],
          [0, 0, 1]]
the output should be findFullLine(field, figure) = 0.

The figure can be dropped only from position 0. When the figure stops falling, two fully occupied rows are formed, so dropping position 0 satisfies the condition.

example 1

For

field =  [[0, 0, 0, 0, 0],
          [0, 0, 0, 0, 0],
          [0, 0, 0, 0, 0],
          [1, 1, 0, 1, 0],
          [1, 0, 1, 0, 1]]
and

figure = [[1, 1, 1],
          [1, 0, 1],
          [1, 0, 1]]
the output should be findFullLine(field, figure) = 2.

The figure can be dropped from three positions - 0, 1, and 2. As you can see below, a fully occupied row will be formed only when dropping from position 2:

example 2

For

field =  [[0, 0, 0, 0],
          [0, 0, 0, 0],
          [0, 0, 0, 0],
          [1, 0, 0, 1],
          [1, 1, 0, 1]]
and

figure = [[1, 1, 0],
          [1, 0, 0],
          [1, 0, 0]]
the output should be findFullLine(field, figure) = -1.

The figure can be dropped from two positions - 0 and 1, and in both cases, a fully occupied line won't be obtained:

example 3

Note that the figure could technically form a full row if it was dropped one position further to the right, but in that case the figure matrix wouldn't be fully contained with the field.

Input/Output

[execution time limit] 4 seconds (php)

[input] array.array.integer field

A matrix of integers representing the game field. It's guaranteed that at the beginning there are no fully occupied rows on the game field and that the top three rows are fully free.

Guaranteed constraints:
4 ≤ field.length ≤ 100,
3 ≤ field[i].length ≤ 100,
0 ≤ field[i][j] ≤ 1.

[input] array.array.integer figure

A matrix of integers representing the figure. It's guaranteed that the figure's occupied cells are pairwise connected and there is at least one occupied cell at the bottom row of the figure.

Guaranteed constraints:
figure.length = 3,
figure[i].length = 3,
0 ≤ figure[i][j] ≤ 1.

[output] integer

The dropping position such that a full row is formed. If there are multiple possible positions, return any of them. If there is no such position, return -1.

12

You are given three arrays of integers a, b, and c. Your task is to find the longest contiguous subarray of a containing only elements that appear in b but do not appear in c.

Return this longest subarray. If there is more than one longest subarray satisfying these conditions, return any of these possible subarrays.

Example

For a = [2, 1, 7, 1, 1, 5, 3, 5, 2, 1, 1, 1], b = [1, 3, 5], and c = [2, 3], the output can be longestInversionalSubarray(a, b, c) = [1, 1, 5].

There is no contiguous subarray of length 4 satisfying all the requirements:

a[0..3] = [2, 1, 7, 1] contains the number a[2] = 7, which doesn't appear in b;
a[1..4] = [1, 7, 1, 1] contains the number a[2] = 7, which doesn't appear in b;
a[2..5] = [7, 1, 1, 5] contains the number a[2] = 7, which doesn't appear in b;
a[3..6] = [1, 1, 5, 3] contains the number a[6] = 3, which does appear in c (which is prohibited);
a[4..7] = [1, 5, 3, 5] contains the number a[6] = 3, which does appear in c;
a[5..8] = [5, 3, 5, 2] contains the number a[6] = 3, which does appear in c;
a[6..9] = [3, 5, 2, 1] contains the number a[6] = 3, which does appear in c;
a[7..10] = [5, 2, 1, 1] contains the number a[8] = 2, which doesn't appear in b;
a[8..11] = [2, 1, 1, 1] contains the number a[8] = 2, which doesn't appear in b.
There are two possible contiguous subarrays of length 3 satisfying all the requirements:

a[3..5] = [1, 1, 5]: both numbers 1 and 5 appear in b, and both of these numbers don't appear in c.
a[9..11] = [1, 1, 1]: the number 1 appears in b, and doesn't appear in c.
example

As you can see, the longest consecutive subarrays of a that fulfill the conditions are a[3..5] = [1, 1, 5] and a[9..11] = [1, 1, 1]. Both of these answers are acceptable.

For a = [1, 2, 3], b = [], and c = [], the output should be longestInversionalSubarray(a, b, c) = [].

Since b is empty, there are no elements that appear in b and not c. So the answer is [].

Input/Output

[execution time limit] 4 seconds (php)

[input] array.integer a

The first array of integers.

Guaranteed constraints:
0 ≤ a.length ≤ 105,
-109 ≤ a[i] ≤ 109.

[input] array.integer b

The second array of integers.

Guaranteed constraints:
0 ≤ b.length ≤ 105,
-109 ≤ b[i] ≤ 109.

[input] array.integer c

The third array of integers.

Guaranteed constraints:
0 ≤ c.length ≤ 105,
-109 ≤ c[i] ≤ 109.

[output] array.integer

Any of the longest contiguous subarrays of a which consists only of elements from b that don't appear in c.

codesignal practice 3

7

You are given three integers in the form of strings: firstnum, secondnum, and thirdnum. Your task is to check whether it is possible to erase at most one digit from firstnum, so that the resulting string contains at least one digit, has no leading zeros and the value of thirdnum is equal to the sum of the values of firstnum and secondnum.

Return true if it's possible, otherwise return false.

Note: All three strings are provided without leading zeros.

Example

For firstnum = "10534", secondnum = "67", and thirdnum = "1120", the output should be eraseOneDigit(firstnum, secondnum, thirdnum) = true.

By erasing the 5th digit of firstnum, the result is 1053, and 1053 + 67 = 1120. So the answer is true.

For firstnum = "10000", secondnum = "67", and thirdnum = "1120", the output should be eraseOneDigit(firstnum, secondnum, thirdnum) = false.

The only possible modified values of firstnum would be 10000 (nothing was deleted), 0000 (first digit was deleted), and 1000 (any zero was deleted); none of which would produce the required sum, so the answer is false.

For firstnum = "1067", secondnum = "33", and thirdnum = "100", the output should be eraseOneDigit(firstnum, secondnum, thirdnum) = false.

We could delete the first digit of firstnum, resulting in 067 (and 67 + 33 = 100), but since in this case new firstnum value has a leading zero, it's considered invalid. So the answer is false.

For firstnum = "153", secondnum = "153", and thirdnum = "306", the output should be eraseOneDigit(firstnum, secondnum, thirdnum) = true.

Because 153 + 153 = 306, there's no need to delete a digit from firstnum, and the result is true.

Input/Output

[execution time limit] 4 seconds (php)

[input] string firstnum

A string representing an integer.

Guaranteed constraints:
2 ≤ firstnum.length ≤ 9.

[input] string secondnum

A string representing an integer.

Guaranteed constraints:
1 ≤ secondnum.length ≤ 9.

[input] string thirdnum

A string representing an integer.

Guaranteed constraints:
1 ≤ thirdnum.length ≤ 9.

[output] boolean

Return true if it's possible to erase at most one digit from firstnum such that the value of thirdnum is equal to the sum of the values of firstnum and secondnum. Otherwise return false.

8

You are given two arrays of integers a and b, and two integers lower and upper. Your task is to find the number of pairs (i, j) such that lower ≤ a[i] * a[i] + b[j] * b[j] ≤ upper.

Example

For a = [3, -1, 9], b = [100, 5, -2], lower = 7, and upper = 99, the output should be boundedSquareSum(a, b, lower, upper) = 4.

There are only four pairs that satisfy the requirement:

If i = 0 and j = 1, then a[0] = 3, b[1] = 5, and 7 ≤ 3 * 3 + 5 * 5 = 9 + 25 = 36 ≤ 99.
If i = 0 and j = 2, then a[0] = 3, b[2] = -2, and 7 ≤ 3 * 3 + (-2) * (-2) = 9 + 4 = 13 ≤ 99.
If i = 1 and j = 1, then a[1] = -1, b[1] = 5, and 7 ≤ (-1) * (-1) + 5 * 5 = 1 + 25 = 26 ≤ 99.
If i = 2 and j = 2, then a[2] = 9, b[2] = -2, and 7 ≤ 9 * 9 + (-2) * (-2) = 81 + 4 = 85 ≤ 99.
For a = [1, 2, 3, -1, -2, -3], b = [10], lower = 0, and upper = 100, the output should be boundedSquareSum(a, b, lower, upper) = 0.

Since the array b contains only one element 10 and the array a does not contain 0, it is not possible to satisfy 0 ≤ a[i] * a[i] + 10 * 10 ≤ 100.

Input/Output

[execution time limit] 4 seconds (php)

[input] array.integer a

A first array of integers.

Guaranteed constraints:
1 ≤ a.length ≤ 105,
-104 ≤ a[i] ≤ 104.

[input] array.integer b

A second array of integers.

Guaranteed constraints:
1 ≤ b.length ≤ 105,
-104 ≤ b[i] ≤ 104.

[input] integer lower

An integer representing a lower bound of a satisfiable square sum.

Guaranteed constraints:
0 ≤ lower ≤ 109.

[input] integer upper

An integer representing an upper bound of a satisfiable square sum.

Guaranteed constraints:
lower ≤ upper,
0 ≤ upper ≤ 109.

[output] integer

The number of pairs (i, j) such, that lower ≤ a[i] * a[i] + b[j] * b[j] ≤ upper. It is guaranteed that the answer fits in 32-bit value type.

9

Given an integer n and an array a of length n, your task is to apply the following mutation to a:

Array a mutates into a new array b of length n.
For each i from 0 to n - 1, b[i] = a[i - 1] + a[i] + a[i + 1].
If some element in the sum a[i - 1] + a[i] + a[i + 1] does not exist, it should be set to 0. For example, b[0] should be equal to 0 + a[0] + a[1].
Example

For n = 5 and a = [4, 0, 1, -2, 3], the output should be mutateTheArray(n, a) = [4, 5, -1, 2, 1].

b[0] = 0 + a[0] + a[1] = 0 + 4 + 0 = 4
b[1] = a[0] + a[1] + a[2] = 4 + 0 + 1 = 5
b[2] = a[1] + a[2] + a[3] = 0 + 1 + (-2) = -1
b[3] = a[2] + a[3] + a[4] = 1 + (-2) + 3 = 2
b[4] = a[3] + a[4] + 0 = (-2) + 3 + 0 = 1
So, the resulting array after the mutation will be [4, 5, -1, 2, 1].

Input/Output

[execution time limit] 4 seconds (php)

[input] integer n

An integer representing the length of the given array.

Guaranteed constraints:
1 ≤ n ≤ 103.

[input] array.integer a

An array of integers that needs to be mutated.

Guaranteed constraints:
a.length = n,
-103 ≤ a[i] ≤ 103.

[output] array.integer

The resulting array after the mutation.

codesignal practice 2

4

Given an array of integers a, your task is to count the number of pairs i and j (where 0 ≤ i < j < a.length), such that a[i] and a[j] are digit anagrams.

Two integers are considered to be digit anagrams if they contain the same digits. In other words, one can be obtained from the other by rearranging the digits (or trivially, if the numbers are equal). For example, 54275 and 45572 are digit anagrams, but 321 and 782 are not (since they don't contain the same digits). 220 and 22 are also not considered as digit anagrams, since they don't even have the same number of digits.

Example

For a = [25, 35, 872, 228, 53, 278, 872], the output should be digitAnagrams(a) = 4.

There are 4 pairs of digit anagrams:

a[1] = 35 and a[4] = 53 (i = 1 and j = 4),
a[2] = 872 and a[5] = 278 (i = 2 and j = 5),
a[2] = 872 and a[6] = 872 (i = 2 and j = 6),
a[5] = 278 and a[6] = 872 (i = 5 and j = 6).
Input/Output

[execution time limit] 4 seconds (php)

[input] array.integer a

An array of non-negative integers.

Guaranteed constraints:
1 ≤ a.length ≤ 105,
0 ≤ a[i] ≤ 109.

[output] integer64

The number of pairs i and j, such that a[i] and a[j] are digit anagrams.

5

You are given an array of strings arr. Your task is to construct a string from the words in arr, starting with the 0th character from each word (in the order they appear in arr), followed by the 1st character, then the 2nd character, etc. If one of the words doesn't have an ith character, skip that word.

Return the resulting string.

Example

For arr = ["Daisy", "Rose", "Hyacinth", "Poppy"], the output should be readingVertically(arr) = "DRHPaoyoisapsecpyiynth".

First, we append all 0th characters and obtain string "DRHP";
Then we append all 1st characters and obtain string "DRHPaoyo";
Then we append all 2nd characters and obtain string "DRHPaoyoisap";
Then we append all 3rd characters and obtain string "DRHPaoyoisapaecp";
Then we append all 4th characters and obtain string "DRHPaoyoisapaecpyiy";
Finally, only letters in the arr[2] are left, so we append the rest characters and get "DRHPaoyoisapaecpyiynth";
example

For arr = ["E", "M", "I", "L", "Y"], the output should be readingVertically(arr) = "EMILY".

Since each of these strings have only one character, the answer will be concatenation of each string in order, so the answer is EMILY.

Input/Output

[execution time limit] 4 seconds (php)

[input] array.string arr

An array of strings containing alphanumeric characters.

Guaranteed constraints:
1 ≤ arr.length ≤ 100,
1 ≤ arr[i].length ≤ 100.

[output] string

Return the resulting string.

6

Given an integer n and an array a of length n, your task is to apply the following mutation to a:

Array a mutates into a new array b of length n.
For each i from 0 to n - 1, b[i] = a[i - 1] + a[i] + a[i + 1].
If some element in the sum a[i - 1] + a[i] + a[i + 1] does not exist, it should be set to 0. For example, b[0] should be equal to 0 + a[0] + a[1].
Example

For n = 5 and a = [4, 0, 1, -2, 3], the output should be mutateTheArray(n, a) = [4, 5, -1, 2, 1].

b[0] = 0 + a[0] + a[1] = 0 + 4 + 0 = 4
b[1] = a[0] + a[1] + a[2] = 4 + 0 + 1 = 5
b[2] = a[1] + a[2] + a[3] = 0 + 1 + (-2) = -1
b[3] = a[2] + a[3] + a[4] = 1 + (-2) + 3 = 2
b[4] = a[3] + a[4] + 0 = (-2) + 3 + 0 = 1
So, the resulting array after the mutation will be [4, 5, -1, 2, 1].

Input/Output

[execution time limit] 4 seconds (php)

[input] integer n

An integer representing the length of the given array.

Guaranteed constraints:
1 ≤ n ≤ 103.

[input] array.integer a

An array of integers that needs to be mutated.

Guaranteed constraints:
a.length = n,
-103 ≤ a[i] ≤ 103.

[output] array.integer

The resulting array after the mutation.

UUID理解

简单说明

V2、V5可以理解为V1、V3的升级版本

版本 说明
V1/V2 基于timestamp+MAC生成,V2加上域名
V3/V5 基于输入的namespace和值生成,v3(md5)v5(sha1)可重复生成; 如用户根据用户信息生成用户唯一id
V4 根据伪随机数生成

UUID V3/4/5的PHP实现

<?php
class UUID {
  public static function v3($namespace, $name) {
    if(!self::is_valid($namespace)) return false;

    // Get hexadecimal components of namespace
    $nhex = str_replace(array('-','{','}'), '', $namespace);

    // Binary Value
    $nstr = '';

    // Convert Namespace UUID to bits
    for($i = 0; $i < strlen($nhex); $i+=2) {
      $nstr .= chr(hexdec($nhex[$i].$nhex[$i+1]));
    }

    // Calculate hash value
    $hash = md5($nstr . $name);

    return sprintf('%08s-%04s-%04x-%04x-%12s',

      // 32 bits for "time_low"
      substr($hash, 0, 8),

      // 16 bits for "time_mid"
      substr($hash, 8, 4),

      // 16 bits for "time_hi_and_version",
      // four most significant bits holds version number 3
      (hexdec(substr($hash, 12, 4)) & 0x0fff) | 0x3000,

      // 16 bits, 8 bits for "clk_seq_hi_res",
      // 8 bits for "clk_seq_low",
      // two most significant bits holds zero and one for variant DCE1.1
      (hexdec(substr($hash, 16, 4)) & 0x3fff) | 0x8000,

      // 48 bits for "node"
      substr($hash, 20, 12)
    );
  }

  public static function v4() {
    return sprintf('%04x%04x-%04x-%04x-%04x-%04x%04x%04x',

      // 32 bits for "time_low"
      mt_rand(0, 0xffff), mt_rand(0, 0xffff),

      // 16 bits for "time_mid"
      mt_rand(0, 0xffff),

      // 16 bits for "time_hi_and_version",
      // four most significant bits holds version number 4
      mt_rand(0, 0x0fff) | 0x4000,

      // 16 bits, 8 bits for "clk_seq_hi_res",
      // 8 bits for "clk_seq_low",
      // two most significant bits holds zero and one for variant DCE1.1
      mt_rand(0, 0x3fff) | 0x8000,

      // 48 bits for "node"
      mt_rand(0, 0xffff), mt_rand(0, 0xffff), mt_rand(0, 0xffff)
    );
  }

  public static function v5($namespace, $name) {
    if(!self::is_valid($namespace)) return false;

    // Get hexadecimal components of namespace
    $nhex = str_replace(array('-','{','}'), '', $namespace);

    // Binary Value
    $nstr = '';

    // Convert Namespace UUID to bits
    for($i = 0; $i < strlen($nhex); $i+=2) {
      $nstr .= chr(hexdec($nhex[$i].$nhex[$i+1]));
    }

    // Calculate hash value
    $hash = sha1($nstr . $name);

    return sprintf('%08s-%04s-%04x-%04x-%12s',

      // 32 bits for "time_low"
      substr($hash, 0, 8),

      // 16 bits for "time_mid"
      substr($hash, 8, 4),

      // 16 bits for "time_hi_and_version",
      // four most significant bits holds version number 5
      (hexdec(substr($hash, 12, 4)) & 0x0fff) | 0x5000,

      // 16 bits, 8 bits for "clk_seq_hi_res",
      // 8 bits for "clk_seq_low",
      // two most significant bits holds zero and one for variant DCE1.1
      (hexdec(substr($hash, 16, 4)) & 0x3fff) | 0x8000,

      // 48 bits for "node"
      substr($hash, 20, 12)
    );
  }

  public static function is_valid($uuid) {
    return preg_match('/^\{?[0-9a-f]{8}\-?[0-9a-f]{4}\-?[0-9a-f]{4}\-?'.
                      '[0-9a-f]{4}\-?[0-9a-f]{12}\}?$/i', $uuid) === 1;
  }
}

PHP实现随机红包

    //随机红包(总份数、最大金额、最小金额、总数)
    protected function random_redpaper($total, $max, $min, $sum){
        $sum = $sum * 100;
        //按分单位,后面全部用整数金额

        $users = array_fill(0, $total, $min);
        $left = $sum - $min * $total;

        //将剩余金额按rand(1, ceil($left/$total))每次进行随机分配, $left 小于50直接分配完成
        for(;$left > 0;){
            $amount = $left > 50 ? rand(1, ceil($left/$total)) : $left;
            $user_index = rand(0, $total-1);
            if($users[$user_index] + $amount > $max){
                $amount = $max - $users[$user_index];
                $users[$user_index] = $max;
            }else{
                $users[$user_index] += $amount;
            }
            $left -= $amount;
        }

        //恢复按元单位
        foreach($users as &$user){
            $user = $user/100;
        }
        return $users;
    }