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Commit c224922

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修改代码格式和README.md, 新增listSquared
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‎README.md

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# PHPCode
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<h1 align="center">PHP编程题目</h1>
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<p align="center">:rainbow: 每周两道编程题目, 加强自己对PHP理解和编程能力. 欢迎提出好的思路, 相互交流.</p>
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[![Build Status](https://www.codewars.com)](https://www.codewars.com)
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### 编程题目
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- [alphabet_positon:用字母表数字替换字符串的字母](https://github.com/uuk020/PHPCode/blob/master/alphabet_position/answer.php)
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- [cube_add:循环数组中奇数幂运算](https://github.com/uuk020/PHPCode/blob/master/cube_odd/answer.php)
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- [find:在奇数数组中找出偶数或者在偶数数组中找出奇数](https://github.com/uuk020/PHPCode/blob/master/find/answer.php)
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- [format_duration:转换时间格式](https://github.com/uuk020/PHPCode/blob/master/format_duration/answer.php)
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- [longestConsec:返回数组中由k个连续字符串组成的第一个最长字符串。](https://github.com/uuk020/PHPCode/blob/master/longestConsec/answer.php)
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- [nbDig:循环某个范围的数,算出每个数的平方,根据第二个参数中数字,看该数字在得出结果出现几](https://github.com/uuk020/PHPCode/blob/master/nbDig/answer.php)
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- [pascals_triangle:以数组形式输出杨辉三角](https://github.com/uuk020/PHPCode/blob/master/pascals_triangle/answer.php)
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- [series_sum:规律计算题](https://github.com/uuk020/PHPCode/blob/master/series_sum/answer.php)
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- [thirt:13的可分性规则](https://github.com/uuk020/PHPCode/blob/master/thirt/answer.php)
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- [listSquared:找出一个数是它所有整除的数的平方之和的结果,此结果能够开平方.](https://github.com/uuk020/PHPCode/blob/master/listSquared/answer.php)
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## 每天一道CodeWar编程题目,分析题目和思路, 选取觉得好的方法,分析. 提高自己对PHP理解
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‎listSquared/answer.php

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<?php
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/**
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*
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* Divisors of 42 are : 1, 2, 3, 6, 7, 14, 21, 42. These divisors squared are: 1, 4, 9, 36, 49, 196, 441, 1764. The sum of the squared divisors is 2500 which is 50 * 50, a square!
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* Given two integers m, n (1 <= m <= n) we want to find all integers between m and n whose sum of squared divisors is itself a square. 42 is such a number.
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* The result will be an array of arrays or of tuples (in C an array of Pair) or a string, each subarray having two elements, first the number whose squared divisors is a square and then the sum of the squared divisors.
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* #Examples:
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* list_squared(1, 250) --> [[1, 1], [42, 2500], [246, 84100]]
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* list_squared(42, 250) --> [[42, 2500], [246, 84100]]
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*The form of the examples may change according to the language, see Example Tests: for more details.
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*/
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function listSquared($m, $n) {
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$result = [];
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for (;$m <= $n; $m++) {
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$sum = 0;
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// 低效算法, 循环次数过大
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for ($i = 1; $i <= $m; $i++) {
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// 找出能够整除$m的数
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if ($m % $i == 0) {
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$sum += pow($i, 2);
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}
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}
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// 求平方根
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$sqrt = sqrt($sum);
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if (pow($sqrt, 2) == $sum && $sum % $sqrt == 0) {
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$result[] = [$m, $sum];
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}
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}
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return $result;
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}
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function listSquared_clever($m, $n) {
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$results = [];
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for ($candidate = $m; $candidate <= $n; $candidate++) {
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$sum = 0;
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// 算出平方根的数, 进行循环(循环次数少)
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for ($divisor = 1; $divisor <= (int)sqrt($candidate); $divisor++) {
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// 找出能够整除sqrt($candidate)的数 也就是除数的.
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if ($candidate % $divisor === 0) {
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$sum += $divisor ** 2;
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/**
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* 排除两个除数相等, 找出$candidate / $divisor的求商, 补齐商数的平方.
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* 如: $candidate = 4时候, 4能够被1,2,4整除. 此时$sum=1*1+2*2; 缺少了4*4.
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*/
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if ($candidate / $divisor !== $divisor) {
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$sum += ($candidate / $divisor) ** 2;
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}
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}
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}
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$square = sqrt($sum);
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if (filter_var($square, FILTER_VALIDATE_INT)) {
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$results[] = [$candidate, $sum];
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}
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}
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return $results;
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}
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