Question

我有一个给定的号码。如何找到该数字的因子（例如，数字15的5和3）？这是我试过的代码：

int factor1 = 2;

while ((a % factor1 != 0) && (a >= factor1)) {
    d++;
}

if (factor1 == a){
    d = 1;
}

但这只给出了最小的因素（即a = 3）。我想得到一组随机因素。

Answer 1

使用模运算符（%）循环遍历从1到N的每个数字。如果是n%currentNumber==0，则它们是一个因素。下面，我使用for循环执行此操作，输出找到的每个因子。

    int number=15;
    for(int i = 1; i <= number; i++){
        if(number%i==0){
            System.out.println("Found factor: " + i);
        }
    }

正如Theo在对此帖子的评论中所说，您也可以使用number/2，并随意包含1和number。

    int number=2229348;
    System.out.println("Found factor: " + 1);
    for(int i = 2; i <= number/2; i++){
        if(number%i==0){
            System.out.println("Found factor: " + i);
        }
    }
    System.out.println("Found factor: " + number);

Answer 2

您可以遍历从2到a/2的数字，并检查给定的数字是否除以a，这是使用%运算符完成的：

int a = 15;
System.out.print("Divisors of " + a + ": ");
for(int i = 2; i <= a/2; ++i) {
    if(a % i == 0) {
        System.out.print(i + " ");
    }
}
System.out.println();

此代码打印a的所有除数。并非你最想忽略1，因为它除了所有整数。此外，您不需要在a之前检查数字，因为大于a / 2的数字实际上除了a之外实际上可以除a之外。

Answer 3

默认值为a = 15且倍数= 2的while循环已经处于无限循环中。你需要纠正这个问题，并在％倍数时检查后续的倍数增量！ = 0

公共阶层因素{

public static void main(String[] args){
    /**
    int multiple1=2,d=0,a=15; //multiple to be rephrased as factor

    while((a%multiple1 != 0)&&(a>=multiple1)){
        multiple1++; //this will increment the factor and check till the number itself
        //System.out.println(multiple1+" is not a factor of a");

    }
    if(multiple1==a){
        d=1;
    }
     *commented the original code
     */ 


    int factor=1,d=0,a=20; //multiple rephrased as factor

    while(a/2>=factor){  //re-arranged your while condition
        factor++;
        if((a%factor==0))
                d++;      //increment factor count whenever a factor is found
    }

    System.out.println("Total number of factors of a : "+(d+2)); // 1 and 'a' are by default factors of number 'a'
}

}

Answer 4

要查找包含1和数字本身的所有因素，您可以执行以下操作：

//Iterate from 2 until n/2 (inclusive) and divide n by each number.
// Return numbers that are factors (i.e. remainder = 0) and are prime
int[] findPrimeFactors(int number) {
  return IntStream.range(2, 1 + number/ 2).filter(factor -> number % factor == 0 && isPrime(factor)).toArray();
}

要找到素数因素，您可以执行以下操作：

//Iterate from 2 until n/2 (inclusive) and divide n by each number. Return false if at least one divisor is found
boolean isPrime(int n) {
 if (n <= 1) throw new RuntimeException("Invalid input");
 return !IntStream.range(2, 1+n/2).filter(x -> ((n % x == 0) && (x != n))).findFirst().isPresent();
}

素性检查的辅助方法：

//Find all factors of a number
public Set<Integer> findFactors(int number) {
    Set<Integer> factors = new TreeSet<>();
    int i = 1;
    factors.add(i);
    while (i++ < 1 + number / 2) {
      if ((number % i) == 0) {
        factors.add(i);
      }
    }
    factors.add(number);
    return factors;
}

如果您不使用Java 8和/或不使用Lambda表达式，则可以使用以下简单的迭代循环：

class Player(object):
    def __init__(self, name, health):
        self._name = name
        self._health = health

    def get_health(self):
        """Return the players health."""
        ## LINE ##

Answer 5

public class Abc{
public static void main(String...args){
    if(args.length<2){
        System.out.println("Usage : java Abc 22 3");
        System.exit(1);
    }
    int no1=Integer.parseInt(args[0]);
    int no=Integer.parseInt(args[1]),temp=0,i;
    for(i=no;i<=no1;i+=no){
        temp++;
    }

    System.out.println("Multiple of "+no+" to get "+no1+" is :--> "+temp);
    //System.out.println(i+"--"+no1+"---"+no);
    System.out.println("Reminder is "+(no1-i+no));
}

}

在Java中如何查找给定数字的随机因子

5 个答案: