Home>Article>Backend Development> PHP algorithm exercise 2: Find the absolute difference between n and the specified number
In "PHP Algorithm Exercise 1: Calculate the sum of two numbers and three times the sum", I introduced you to the first question of PHP arithmetic exercise. I believe everyone has mastered it. So the theme of today’s article is somewhat similar to it, I hope everyone can master it quickly!
We used to find the sum, now we find the difference, let’s do it one by one~
The specific question of this article is "Write a PHP program to get the absolute difference between n and the specified value .If n is greater than this specified value, three times the absolute difference is returned."
You can practice it locally, and then look at my method below:
The PHP code is as follows:
$x) { return ($n - $x)*3; } return $x - $n; } echo test(53)."
"; echo test(30)."
"; echo test(50)."
";
The output calculation result is:
In the above code, the specified value we gave is 50, and then the if statement is used to determine whether n is greater than 50. If n is greater than 50, three times the absolute difference is returned, otherwise it is returned its absolute difference. The key arithmetic code parts here are "($n - $x)*3;
" and "$x - $n;
".
In this example, the first n we gave is 53, then 53 is greater than 50, and the result is naturally to return (53-50)*3=9;
The second one n is 30, which is less than 50, and the result is "50-30=20";
The third n is 50, which is equal to 50, and the result is naturally 0.
In the above code, a function is used to customize a test method. Needless to say, it mainly depends on the understanding and mastery of if judgment statements and PHP operators.
A brief introduction to the conditional statements provided in PHP:
if 语句 - 在条件成立时执行代码 if...else 语句 - 在条件成立时执行一块代码,条件不成立时执行另一块代码 if...elseif....else 语句 - 在若干条件之一成立时执行一个代码块 switch 语句 - 在若干条件之一成立时执行一个代码块
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