Kotlin algorithm for getting started with a full number of algorithms

/* If a number is exactly equal to the sum of its factors, this number is called the "end number". */classCompletenumber {private VarFirstfactornumber: Int = 0/** * Because the non-prime number can be calculated by multiplying by 1 9 in addition to 1 and 2 only if it is possible to continue divisible by 2-9 * This statement uses the extraction of the minimum common divisor to calculate * Of course to avoid a The important question is that when it's a single digit, it's 1, 2, 3, 5, 7. The advantage of this calculation is to avoid the iterative calculation of the traditional recursion from 1 to n more efficient calculation of the number of tens of thousands of data using * Also avoids excessive use of this algorithm (redundant repeatability Calculation): * Method of judging primes: A number is removed from 2 to sqrt (the square root of this number), if it can be divisible, it indicates that the number is not a prime, the inverse is prime this algorithm faster * avoid duplication of redundancy * * FunIsprimenumber (Divisor:int, Number:int): Boolean {if(number% Divisor = = 0) {Firstfactornumber= Divisorreturn False }if(number = = 1 | | number = = 2 | | number = = 3 | | number = = 5 | | number = = 7 | | | Number = = 17 | | Number = = 19)return Trueelse if(Number <= 20) {Firstfactornumber= 1return False }else if(divisor = = 9) {returnIsprimenumber (one, divisor)}else if(Divisor > 9) {if(Divisor < Math.sqrt (number.todouble ())) {returnIsprimenumber (divisor + 1, number)}else if(divisor.todouble () = = Math.sqrt (number.todouble ())) {Firstfactornumber= Divisorreturn False }Elsereturn True }returnIsprimenumber (divisor + 1, number)}/* Call optimization algorithm */ FunIscompletenumber (number:int) {if(Number < 6) {println ("from 1 to"+ number +"No complete number")
        }Else{ for(Iinch6..number) {/* Prime cannot be a complete number so it can be computed directly skipping * each time the calculation of the number of primes is calculated if not by calculation * is the first factor +1 Calculate the starting point to avoid the start-up repetition calculation */if(Isprimenumber (2, number))Continuevarsum = 0if(Firstfactornumber> 1) Sum + = 1 +Firstfactornumber else if(Firstfactornumber= = 1) sum + = 1 for(jinchFirstfactornumber+ 1 until i) {if(i% j = 0) {sum = sum + j}}if(sum = = i)
        {println (i)}} }}/* Traditional practice */ FunTradition () {varS:int for(Iinch1..1000000) {s = 0 for(jinch1 until I)if(i% j = 0) s = s + jif(s = = i) print (i.tostring () +" ")} println ()}}