monkey counting problem

n monkeys sit in a circle and take turns reporting 1, 2, and 3, and every monkey that reports 3 will get out of the queue. The last remaining one is the monkey king. Please write a function in PHP. The input is the number of monkeys and the starting position of the count. The return value is the serial number of the monkey king

<?php

function fun($n,$begin)
{
//输入判断
if(!is_int($n) || $n<=0)return false;
if(!is_int($begin) || $begin>$n || $begin<=0)return false;

//初始化数组，使其内部指针指向传进函数的“开始位置”
$arr = array();
for($i=1;$i<=$n;$i++)$arr[] = $i;
for($i=1;$i<$begin;$i++,next($arr));

while(count($arr)>1) //当数组大小不为1时循环报数
{
//报数，往后数两位
for($i=0;$i<2;$i++)
{
if(!next($arr))reset($arr);
}
//获得报数3位置的键、值(此处内部指针会前进一步)
$key = each($arr);

if(!current($arr)) //如果报数到3的位置是数组末端，及通过each后，指针超出了数组的范围
{
reset($arr); //将内部指针重置到数组首部
array_pop($arr); //删除数组末端的键、值
}
else
{
prev($arr); //否则指针回退一格
unset($arr[$key['key']]); //删除报数为3的键、值 
}
}
if(!current($arr))reset($arr); //循环过后，因为each操作，内部指针有可能超越了数组末端，需要重置
return current($arr);
}

echo fun(5,3);
?>

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The above has introduced the monkey counting problem, including aspects of it. I hope it will be helpful to friends who are interested in PHP tutorials.