Java algorithm for solving the quality factor

Public classQualityfactor {/** * Because no matter how the number of primes can be computed by multiplication in 1 9, except that 1 and 2 only need to continue to be divisible by 2-9. * This argument uses the extraction of the minimum polynomials to compute the result One important thing to avoid, of course, is that when it's a single-digit number, it's 1, direct return at 2, 3, 5, 7 The advantage of this calculation is to avoid the traditional recursive calculation from 1 to n is more efficient to calculate the number of primes facing thousands of data use * Also avoid excessive use of this calculation Method (redundant redundancy calculation): * To determine the number of primes: 2 to sqrt (the square root of this number), if divisible, it indicates that this number is not prime, Vice is prime this algorithm is faster * Avoid redundant redundancy * * *Public BooleanIsprimenumber (intDivisorintNumber) {if(number% divisor = 0)Return false;else if(number = 1 | | | number = 2 | | number = 3 | | number = 5 | | number = 7 | | number = ONE | | | Number = = 17 | | Number = 19)Return true;else if(Number <= 20)Return false;else if(divisor = 9) { returnIsprimenumber (one, divisor); }else if(Divisor > 9) {if(Divisor < MATH.SQRT (number)) { returnIsprimenumber (divisor + 1, number); }else if(Divisor = = math.sqrt (number))Return false;else return True; } returnIsprimenumber (divisor + 1, number); }Public voidGetqualityfactor (intNumber) {String out = number +"=";if(Isprimenumber (2, number)) out = out + number;Else{ while(number!= 1) { for(intj = 2; J <= number; J + +) {/* If every number can be divisible by J let number/=j*/if(number% J = = 0)
                        {Number/= J; /* Divide and finish to determine whether it is prime so as to avoid the last remaining a relatively large prime number and then have to do the calculation of repeated calculation * *if(number!=1) {out = = j +"x";if(Isprimenumber (2,number))
                                {out = = number;
                            Number = 1; }
                        }Elseout + = j; Break; {}}} System. out. println (out); }
}