


<?php /** * 题目2:在斐波那契数列中,找出4百万以下的项中值为偶数的项之和。 * * @autor 花生米 * @date 2015-09-08 * @desc php version 5.4.33 */ $max = 40000000000; /** * 思路1 * 时间复杂度:O(n) * 这是最容易想到的方法 */ if (false) { $a = 1; $b = 2; $s = 0;//总和 while ($b < $max) { //偶数则加 if (($b & 1) == 0) { $s += $b; } $t = $b; $b = $a + $b; $a = $t; } } /** * 思路2:(优) * 时间复杂度:O(n) * 斐波那契数列 * 0 1 1 2 3 5 8 13 21 34 55 89 144 ... * a b c a b c a b c a b c * 仔细观察可知,都是每三项一个偶数,而且 * 8 = 4 * 2 + 0; * 34 = 4 * 8 + 2; * 144 = 4 * 34 + 8; * 故思路2就是这个方式来求解 */ if (false) { $a = 0;//第一项从0开始 $b = 2;//第二项从2开始 $s = 0; while ($b < $max) { if (($b & 1) == 0) { $s += $b; } $t = $b; $b = 4 * $b + $a; $a = $t; } }
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The above introduces question 2: In the Fibonacci sequence, find the sum of the terms with an even number among the terms below 4 million. , including relevant content, I hope it will be helpful to friends who are interested in PHP tutorials.

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”。

iPhone15Pro与iPhone14Pro:规格比较以下是iPhone15ProMax和iPhone14ProMax的规格比较:iPhone15ProMaxiPhone14ProMax显示尺寸6.7英寸6.7英寸显示技术超级视网膜XDROLED超级视网膜XDROLED分辨率2796x1290像素,460ppi2796x1290像素,460ppi刷新率120赫兹120赫兹峰值亮度2,000尼特2,000尼特尺寸6.29x3.02x0.32英寸6.33x3.06x0.31英寸重量221克240克

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。

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。

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。


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