


In the previous article "PHP Algorithm Exercise 11: Check whether two given integers are within the specified range", I introduced how to check whether two given integers are within the specified range. Friends in need can learn about it~
This article will continue to bring you a series of PHP algorithm exercises. [Recommended: PHP algorithm exercise series summary (continuously updated~) ]
First of all, let me describe the problem to be solved in this article: "How to write a PHP program to obtain n and 51 The absolute value between. If n is greater than 51, three times the absolute value is returned."
Related introduction: The absolute value refers to the distance from the corresponding point of a number on the number axis to the origin, represented by "| |". |b-a| or |a-b| represents the distance between the point representing a and the point representing b on the number axis.
Let’s go directly to the code:
<?php function test($n) { $x = 51; if ($n > $x) { return ($n - $x)*3; } return $x - $n; } echo test(53)."<br>"; echo test(30)."<br>"; echo test(51)."<br>";
The execution results are as follows:
6 21 0
It’s very simple to implement!
Here we mainly need to master the use of if conditional statements, and then implement the execution process according to the requirements. For example, in this example, if n is greater than 51, it is required to return three times the absolute value, that is, the "($n - $x)*3;" formula must be executed; otherwise, the absolute value of the two numbers is directly returned.
Related introduction:
In PHP, we can use the following conditional statements:
if statement: If the specified condition is true, Then execute the code;
if...else statement: If the condition is true, the code is executed; if the condition is false, the other end of the code is executed;
if...elseif....else statement: execute different code blocks based on two or more conditions;
switch statement: select among multiple code blocks One to execute.
Finally, I would like to recommend the latest and most comprehensive "PHP Video Tutorial"~ Come and learn!
The above is the detailed content of PHP Algorithm Exercise 12: Get the absolute value between n and 51 and return the value as required. For more information, please follow other related articles on the PHP Chinese website!

php把负数转为正整数的方法:1、使用abs()函数将负数转为正数,使用intval()函数对正数取整,转为正整数,语法“intval(abs($number))”;2、利用“~”位运算符将负数取反加一,语法“~$number + 1”。

实现方法:1、使用“sleep(延迟秒数)”语句,可延迟执行函数若干秒;2、使用“time_nanosleep(延迟秒数,延迟纳秒数)”语句,可延迟执行函数若干秒和纳秒;3、使用“time_sleep_until(time()+7)”语句。

php字符串有下标。在PHP中,下标不仅可以应用于数组和对象,还可应用于字符串,利用字符串的下标和中括号“[]”可以访问指定索引位置的字符,并对该字符进行读写,语法“字符串名[下标值]”;字符串的下标值(索引值)只能是整数类型,起始值为0。

php除以100保留两位小数的方法:1、利用“/”运算符进行除法运算,语法“数值 / 100”;2、使用“number_format(除法结果, 2)”或“sprintf("%.2f",除法结果)”语句进行四舍五入的处理值,并保留两位小数。

判断方法:1、使用“strtotime("年-月-日")”语句将给定的年月日转换为时间戳格式;2、用“date("z",时间戳)+1”语句计算指定时间戳是一年的第几天。date()返回的天数是从0开始计算的,因此真实天数需要在此基础上加1。

在php中,可以使用substr()函数来读取字符串后几个字符,只需要将该函数的第二个参数设置为负值,第三个参数省略即可;语法为“substr(字符串,-n)”,表示读取从字符串结尾处向前数第n个字符开始,直到字符串结尾的全部字符。

方法:1、用“str_replace(" ","其他字符",$str)”语句,可将nbsp符替换为其他字符;2、用“preg_replace("/(\s|\ \;||\xc2\xa0)/","其他字符",$str)”语句。

查找方法:1、用strpos(),语法“strpos("字符串值","查找子串")+1”;2、用stripos(),语法“strpos("字符串值","查找子串")+1”。因为字符串是从0开始计数的,因此两个函数获取的位置需要进行加1处理。


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