修改代码格式和README.md, 新增listSquared

Author: uuk020

README.md

@@ -1,3 +1,20 @@
-# PHPCode
+<h1 align="center">PHP编程题目</h1>
+
+<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>
+
+[![Build Status](https://www.codewars.com)](https://www.codewars.com)
+
+
+###  编程题目
+
+- [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)
+- [cube_add:循环数组中奇数幂运算](https://github.com/uuk020/PHPCode/blob/master/cube_odd/answer.php)
+- [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)
+- [format_duration:转换时间格式](https://github.com/uuk020/PHPCode/blob/master/format_duration/answer.php)
+- [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)
+- [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)
+- [pascals_triangle:以数组形式输出杨辉三角](https://github.com/uuk020/PHPCode/blob/master/pascals_triangle/answer.php)
+- [series_sum:规律计算题](https://github.com/uuk020/PHPCode/blob/master/series_sum/answer.php)
+- [thirt:13的可分性规则](https://github.com/uuk020/PHPCode/blob/master/thirt/answer.php) 
+- [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)
 
-## 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

alphabet_position/alphabet_position.php renamed to alphabet_position/answer.php

@@ -0,0 +1,63 @@
+<?php
+/**
+ * 
+ * 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!
+
+ * 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.
+
+ * 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.
+
+ * #Examples:
+ * list_squared(1, 250) --> [[1, 1], [42, 2500], [246, 84100]]
+ * list_squared(42, 250) --> [[42, 2500], [246, 84100]]
+ *The form of the examples may change according to the language, see Example Tests: for more details.
+ */
+
+function listSquared($m, $n) {
+    $result = [];
+    for (;$m <= $n; $m++) {
+        $sum = 0;
+        // 低效算法, 循环次数过大
+        for ($i = 1; $i <= $m; $i++) {
+            // 找出能够整除$m的数
+            if ($m % $i == 0) {
+                $sum += pow($i, 2);
+            }
+        }
+        // 求平方根
+        $sqrt = sqrt($sum);
+        if (pow($sqrt, 2) == $sum && $sum % $sqrt == 0) {
+            $result[] = [$m, $sum];
+        }
+    }
+    return $result;
+}
+
+
+function listSquared_clever($m, $n) {
+    $results = [];
+    for ($candidate = $m; $candidate <= $n; $candidate++) {
+      $sum = 0;
+      // 算出平方根的数, 进行循环(循环次数少)
+      for ($divisor = 1; $divisor <= (int)sqrt($candidate); $divisor++) {
+        //  找出能够整除sqrt($candidate)的数 也就是除数的.
+        if ($candidate % $divisor === 0) {
+          $sum += $divisor ** 2;
+          /**
+           * 排除两个除数相等, 找出$candidate / $divisor的求商, 补齐商数的平方.
+           * 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.
+           */
+          if ($candidate / $divisor !== $divisor) {
+            $sum += ($candidate / $divisor) ** 2;
+          }
+        }
+      }
+      $square = sqrt($sum);
+    
+      if (filter_var($square, FILTER_VALIDATE_INT)) {
+        $results[] = [$candidate, $sum];
+      }
+    }
+    
+    return $results;
+}