我正在编写一个程序来计算一个数字因素是主要因素。该计划的工作原理如下:
1)输入您想要考虑的数字(我将引用为“inputNumber”)
2)测试你是否可以将inputNumber除以2,3,5和7(前4个素数)。如果你可以除以这些中的任何一个,那么你可以多次这样做(即12可以除以2两次,之后,剩下的3) note 每当我除以一个数字时,我将该数字存储在一个数组列表中,并保留其余部分以供进一步测试
3)从while循环中的i = 11(下一个素数)开始,我执行以下操作:
while (i < remainder+1) {
divides by i? yes: store i,
repeat until you cant divide by i. i=i+2,
divides by i? yes: store i, repeat until you can't divide by i.
i=i+4, same as before... i=i+2...
and finally stop the loop at i=i+2
}
这样,while循环的每次成功迭代都会将余数除以1,3,7,9之后的数字。我们不需要测试偶数,因为我们已经除以2,也不需要测试以5结尾的数字,因为我们已经除以5.最后,我们
这是一个非常简洁的算法,因为它比通过测试一个数字来反映数字要快得多,事实上,你只需要测试所有数字的40%。
我的问题是:当我尝试分解84738329279时,一个随机选取的数字,它省略了将最后一个素数因子放在列表中,我无法弄明白这一点。出现的唯一因素是41,61和61.有人能帮助我找到我做错的事吗?这是我的代码:
import java.util.Scanner;
import java.math.BigInteger;
public class Test {
public static void main(String[] args) {
// create a scanner object for inputs
Scanner in = new Scanner(System.in);
// prompt user for number to factor
System.out.print("enter a number to factor: ");
String digits = in.next();
BigInteger BigDigits = new BigInteger(digits);
BigInteger[] results = factor.factorThis(BigDigits);
System.out.print("Factors are ");
for (int i=0; i<results.length;i++){
System.out.print(results[i] + " ");
}
System.out.println("");
}
}
import java.util.ArrayList;
import java.math.BigInteger;
public class factor {
// Returns the prime factors of the BigInteger number
public static BigInteger[] factorThis(BigInteger number) {
BigInteger i = new BigInteger("11");
ArrayList<BigInteger> divisors = new ArrayList<BigInteger>(0);
BigInteger[] firstPrimes = new BigInteger[4];
firstPrimes[0] = new BigInteger("2");
firstPrimes[1] = new BigInteger("3");
firstPrimes[2] = new BigInteger("5");
firstPrimes[3] = new BigInteger("7");
// loop that test for first 4 prime numbers
for (int l=0;l<4;l++){
while ((number.mod(firstPrimes[l])).compareTo(BigInteger.ZERO) == 0) {
number = number.divide(firstPrimes[l]);
divisors.add(firstPrimes[l]);
}
}
// loop that factors only numbers finishing by 1,3,7,9
while (i.compareTo(number) == -1){
// check for ending by 1
if ((number.mod(i)).compareTo(BigInteger.ZERO) == 0) {
while (number.mod(i).compareTo(BigInteger.ZERO) == 0){
number = number.divide(i);
divisors.add(i);
}
}
else if ((number.mod(i)).compareTo(BigInteger.ZERO) != 0){
i=i.add(firstPrimes[0]);
}
// check for ending by 3
if ((number.mod(i)).compareTo(BigInteger.ZERO) == 0) {
while (number.mod(i).compareTo(BigInteger.ZERO) == 0){
number = number.divide(i);
divisors.add(i);
}
}
else if ((number.mod(i)).compareTo(BigInteger.ZERO) != 0){
i=i.add(firstPrimes[0].multiply(firstPrimes[0]));
}
//check for ending by 7
if ((number.mod(i)).compareTo(BigInteger.ZERO) == 0) {
while (number.mod(i).compareTo(BigInteger.ZERO) == 0){
number = number.divide(i);
divisors.add(i);
}
}
else if ((number.mod(i)).compareTo(BigInteger.ZERO) != 0){
i=i.add(firstPrimes[0]);
}
// check for ending by 9
if ((number.mod(i)).compareTo(BigInteger.ZERO) == 0) {
while (number.mod(i).compareTo(BigInteger.ZERO) == 0){
number = number.divide(i);
divisors.add(i);
}
}
else if ((number.mod(i)).compareTo(BigInteger.ZERO) != 0){
i=i.add(firstPrimes[0]);
}
}
// store prime factors into a BigInt array
String[] strArrayDivisors = divisors.toString().replaceAll("\\[", "").replaceAll("\\]","").replaceAll("\\s","").split(",");
BigInteger[] BigIntDivisors = new BigInteger[strArrayDivisors.length];
for(int j=0;j<strArrayDivisors.length;j++){
BigIntDivisors[j] = new BigInteger(strArrayDivisors[j]);
}
// returns all factors of "number"
return BigIntDivisors;
}
}
提前致谢。
答案 0 :(得分:1)
首先,84738329279 = 41 * 61 * 61 * 555439
。你的41,61和61是正确的。
但是当你的算法终止时,你的最后一个素数仍然在number
。您需要在最后添加测试number
的代码:如果它是1
,那么您已经完成了,否则需要将其添加到divisors
以便稍后打印