Kotlin algorithm Entry calculation factorization __ algorithm

/* Each non-prime (composite) can be written as a number of primes (also known as prime numbers) in the form of multiplication, and these primes are called the composite's mass factor.
For example, 6 can be decomposed into 2x3, and 24 can be decomposed into 2x2x2x3. Now, your program reads an integer in a [2,100000] range, and then outputs its mass factorization, and when it reads a prime number, it outputs itself */classQualityfactor {/** * because the non-prime number can be calculated by multiplying by the 1 9 in addition to 1 and 2 only if it is possible to continue to be divisible by 2-9 * This statement uses the extraction of the minimum common divisor to calculate * Of course, to avoid an important problem is that when it is a single digit when it is 1, 2, 3, 5, 7 when the direct return * The advantage of this calculation is to avoid the traditional recursion from 1 to n the repeated calculation of more efficient calculation of the number of thousands above the use of data * Also avoids excessive use of this calculation Method (redundant repeatability calculation): * Method of Judging primes: 2 to sqrt (the square root of this number), if divisible, indicates that the number is not a prime, and the inverse is prime this algorithm is faster * avoids redundant redundancy */ FunIsprimenumber (Divisor:int, Number:int): Boolean {if(number% Divisor = = 0)return Falseelse if(number = = 1 | | number = = 2 | | number = = 3 | | number = = 5 | | number = = 7 | | | Number = = 17 | | Number = = 19)return Trueelse if(Number <= 20)return Falseelse if(divisor = = 9) {returnIsprimenumber (one, divisor)}else if(Divisor > 9) {return if(Divisor < Math.sqrt (number.todouble ())) {Isprimenumber (divisor + 1, number)}else if(divisor.todouble () = = Math.sqrt (number.todouble ()))falseElsetrue }returnIsprimenumber (divisor + 1, number)} FunGetqualityfactor (number:int) {varNumber = numbervarout = number.tostring () +"="  if(Isprimenumber (2, number)) out = out + numberElse{ while(number! = 1) { for(jinch2..number) {/* If every number can be divisible by J then let number/=j*/if(number% J = 0) {Number/= J//is divided by the number of primes to avoid the last large prime number and then the calculation of the repeat calculation */if(number! = 1) {out + = j.tostring () +"x" if(Isprimenumber (2, number))
                        {out + = number number = 1} }Else Out + = J Break }}}} println (out)}}