如何乘数直到达到一位数的数字并计算数字? [英] How to multiply number until achieve single-digit numbers and counting the number?
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问题描述
$ b $ pre $
persistence(39)== 3 //因为3 * 9 = 27,2 * 7 = 14,1 * 4 = 4
// 4只有一个数字
持久性(999)== 4 //因为9 * 9 * 9 = 729,7 * 2 * 9 = 126,
// 1 * 2 * 6 = 12,最后是1 * 2 = 2
persistence(4)== 0 //因为4已经是单向函数,数字编号
我只能这样做:
$ array = str_split(39);
$ b $ foreach($ array as $ key => $ value){
echo $ array [$ key] * $ array [$ key + 1];
}
接下来我很困惑
任何解决方案来解决我的问题?
解决方案
$ array = str_split ( '999'); //你的字符串
$ j = 0; (count($ array)> 1){//当数组
中有多于2个索引时($ i = 0; $ i $ array = array_product($ array); //将数组中的所有数字相乘
$ array = str_split($ array); //再次拆分数组
$ j ++; //增加计数器(按照我听到的文字)
}
}
echo $ j; //打印次数
its description like this:
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
I can only make like this :
$array = str_split(39);
foreach ($array as $key => $value) {
echo $array[$key]*$array[$key+1];
}
The next I'm confused
Any solution to solve my problem?
解决方案
$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
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