Question

它的描述如下：

persistence(39) == 3 // because 3*9 = 27, 2*7 = 14, 1*4=4
// and 4 has only one digit

persistence(999) == 4 // because 9*9*9 = 729, 7*2*9 = 126,
// 1*2*6 = 12, and finally 1*2 = 2

persistence(4) == 0 // because 4 is already a one-digit number

我只能这样做：

$array = str_split(39);

    foreach ($array as $key => $value) {
        echo $array[$key]*$array[$key+1];
    }

接下来我很困惑

解决我问题的任何解决方案？

Answer 1

$array = str_split('999'); //Your string
$j=0; //Counter for counting the number of iteration
while (count($array)>1){ //When more than 2 indexes in array
 for($i=0;$i<count($array);$i++){ //Iterate through all permutations
     $array = array_product($array); //Multiplies all numbers in array
     $array = str_split($array); //Split the array up again
     $j++; //Increment counter(as literal as I can sound)
 }
}
echo $j; //Print out the number of times

Answer 2

您也可以使用功能调用，直到它只有一个数字......

$count = 0;
persistence(2,$count);
function persistence($i,&$c){   
     $v = str_split($i);
        if(count($v) > 1){
            $total = array_product($v);
            $c++;
            persistence($total,$c);
        }
}
var_dump($count);

如何乘以数字直到达到一位数字并计算数字？

2 个答案: