我自己构思了下,实际上程序来解决这个事情,就是一个偏移量的问题。首先看数列::1、1、2、3、5、8、13、21、34数列的下一个数是前2个数字之和,以此类推。
程序处理的话,实际上就是一个FOR语句,传统FOR语句是for($i=1;$i;$count,$i++),这里的偏移量是$i=$i+1.如果处理这个数列的话,这个偏移量就不是1了,是前1个数字。那么当你for的时候,一个变量记录上一个数字,另外一个记录当前数字,偏移量为这上一个数字,然后在循环中重新赋值,将上一个数字记录成当然循环值,以此做下个循环的偏移量。代码其实很简单:
复制代码 代码如下:
$count = 9999999999967543;
$array = array('0′=>1);
for($a=1,$i=2;$i$array[] = $a;
$array[] = $i;
$a = $a +$i;
}
print_r($array);
echo $count.'里有'.count($array).'个斐波那契数列数';
建议哪个无聊人拿这个去phpchina给大白菜职业顶贴去

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