blocks|key|479740|text|看起来你不打算在你的while循环中对f或ff做些什么了？如果是这样，表达式f%25ff+!=+0要么为false+(然后它将转到for循环中的下一个)，要么为true，它将在无限循环中结束。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|479741|你确定你需要这样的时间吗？|479742|entityMap^0|J|1|L|2|12|9|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Q|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|R|8|@]|D|@]|E|$]]]|I|$]]

It looks like you are not going to do something with either <code>f</code> or <code>ff</code> in your while loop? If so, the expression <code>f%ff != 0</code> is either false (and then it will go to the next in the for loop), or it is true, and it will end up in an infinite loop.

Are you sure you need the while like this?

blocks|key|479821|text|稍加修改的解决方案:你可以先检查变量x是否可以被变量y整除。如果是，我们将计数为1，并重复这个过程。对于循环计数器，使用x/y，当x变为零但循环尚未结束时，应检查x>0以避免重复。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|479822|+++public+class+Factor+{

++++public+static+void+main(String[]+args)+{

++++++++int+x+=+48;
++++++++int+x1+=+x;
++++++++int+y+=+2;
++++++++int+k+=+x+/+y;
++++++++int+j+=+0;
++++++++for+(int+i+=+1;+i+<+k;+i%2B%2B)+{
++++++++++++if+((x+%25+y)+==+0+&&+x+>+0)
++++++++++++++++j%2B%2B;
++++++++++++x+=+x+/+2;
++++++++}
++++++++System.out.println(%2Bx1+%2B+"+is+a+factor+of+"+%2B+y+%2B+"+for+"+%2B+j
++++++++++++++++%2B+"+times.");

++++}
}|code-block|syntax|javascript|479823|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Slightly modified solution: You can first check if variable x is divisible by variable y. If yes, we will count 1 and will repeat this process. For the loop counter, x/y is used and you should check x>0 to avoid repetition when x becomes zero but loop is not finished yet.

<pre><code> public class Factor {

 public static void main(String[] args) {

 int x = 48;
 int x1 = x;
 int y = 2;
 int k = x / y;
 int j = 0;
 for (int i = 1; i &lt; k; i++) {
 if ((x % y) == 0 &amp;&amp; x &gt; 0)
 j++;
 x = x / 2;
 }
 System.out.println(+x1 + " is a factor of " + y + " for " + j
 + " times.");

 }
}
</code></pre>

blocks|key|695827|text|这段代码将为您提供这些因子。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|695828|ArrayList<Integer>+arr+=+new+ArrayList<>();
++++++++int+x=48;
++++++++int+y=1;
++++++++while(x!=1)
++++++++{
++++++++++++if(x%25y==0)
++++++++++++{
++++++++++++++++x=x/y;
++++++++++++++++arr.add(y);
++++++++++++++++if(y==1)
++++++++++++++++{
++++++++++++++++++++y%2B%2B;
++++++++++++++++}
++++++++++++}
++++++++++++else
++++++++++++{
++++++++++++++++y%2B=1;
++++++++++++}
++++++++}
++++++++System.out.println(arr);|code-block|syntax|javascript|695829|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This code will give you the factors.

<pre><code>ArrayList&lt;Integer&gt; arr = new ArrayList&lt;&gt;();
 int x=48;
 int y=1;
 while(x!=1)
 {
 if(x%y==0)
 {
 x=x/y;
 arr.add(y);
 if(y==1)
 {
 y++;
 }
 }
 else
 {
 y+=1;
 }
 }
 System.out.println(arr);
</code></pre>

blocks|key|479915|text|我用这个得到了所有的因素(我只是修改了问题中的算法)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|479916|int+num1+=+120;

for(int+num2=1;num2<=num1;num2%2B%2B)
{
++if+(num1%25num2+!=+0)
++++System.out.println(num2);
}|code-block|syntax|javascript|479917|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I got all the factors just fine with this (I just modified the algorithm in the question).

<pre><code>int num1 = 120;

for(int num2=1;num2&lt;=num1;num2++)
{
 if (num1%num2 != 0)
 System.out.println(num2);
}
</code></pre>

blocks|key|695902|text|下面是如何获得给定数字的所有因子的方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|695903|public+class+Factors+{

++++public+static+void+main(String[]+args){
++++++++int+n+=+420;

++++++++for(int+i=2;+i<=n;+i%2B%2B){
++++++++++++while(n%25i==0){
++++++++++++++++System.out.println(i+%2B+"%7C+"+%2B+n);
++++++++++++++++System.out.println("+-----");
++++++++++++++++n+=+n/i;
++++++++++++}
++++++++}
++++}
}|code-block|syntax|javascript|695904|输出：|695905|2%7C+420
+-----
2%7C+210
+-----
3%7C+105
+-----
5%7C+35
+-----
7%7C+7
+-----|695906|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Here is how to get all factors of the given number.

<pre><code>public class Factors {

 public static void main(String[] args){
 int n = 420;

 for(int i=2; i&lt;=n; i++){
 while(n%i==0){
 System.out.println(i + "| " + n);
 System.out.println(" -----");
 n = n/i;
 }
 }
 }
}
</code></pre>

Output:

<pre><code>2| 420
 -----
2| 210
 -----
3| 105
 -----
5| 35
 -----
7| 7
 -----
</code></pre>

blocks|key|480002|text|public+class+Solution+{
++++public+ArrayList<Integer>+allFactors(int+a)+{

++++++++int+upperlimit+=+(int)(Math.sqrt(a));
++++++++ArrayList<Integer>+factors+=+new+ArrayList<Integer>();
++++++++for(int+i=1;i+<=+upperlimit;+i%2B=+1){
++++++++++++if(a%25i+==+0){
++++++++++++++++factors.add(i);
++++++++++++++++if(i+!=+a/i){
++++++++++++++++++++factors.add(a/i);
++++++++++++++++}
++++++++++++}
++++++++}
++++++++Collections.sort(factors);
++++++++return+factors;
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|480003|上面的解决方案很简单，就像计算素因数一样。不同的是，对于每个质数因子，我们一直在计算乘积的另一部分，即reqd数。|unstyled|480004|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>public class Solution {
 public ArrayList&lt;Integer&gt; allFactors(int a) {

 int upperlimit = (int)(Math.sqrt(a));
 ArrayList&lt;Integer&gt; factors = new ArrayList&lt;Integer&gt;();
 for(int i=1;i &lt;= upperlimit; i+= 1){
 if(a%i == 0){
 factors.add(i);
 if(i != a/i){
 factors.add(a/i);
 }
 }
 }
 Collections.sort(factors);
 return factors;
 }
}
</code></pre>

The above solution simply works like calculating prime factors. 
The difference being for every prime factor we keep calculating the other part of the product i.e the reqd number.

blocks|key|696009|text|使用递归函数的最简单方法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|696010|public+static+int+factorial(int+n){
++++++++if(n!=1)
++++++++++++return+n*factorial(n-1);
++++++++return+1;+
}+++|code-block|syntax|javascript|696011|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Easiest way using recursive function

<pre><code>public static int factorial(int n){
 if(n!=1)
 return n*factorial(n-1);
 return 1; 
} 
</code></pre>

blocks|key|696020|text|++++import+java.util.Scanner;
++++public+class+Factors+
++++{
++++++++Scanner+scn=new+Scanner(System.in);
++++++++int+num=scn.nextInt();
++++++++public+void+findFactor()
++++++++{
+++++++++++System.out.println("Factors+are");
+++++++++++System.out.println("1");
+++++++++++for(int+i=2;i<=num;i%2B%2B)
+++++++++++{
++++++++++++++++if(num%25i==0)
++++++++++++++++{
++++++++++++++++++++num=num/i;
++++++++++++++++++++System.out.println(i);
++++++++++++++++++++i=2;
++++++++++++++++}
+++++++++++}
++++++++}
public+static+void+main(String[]+args)+
{
++++while(1==1)
++++{
++++++++System.out.println("Enter+a+Number");
++++++++Factors+fct=new+Factors();
++++++++fct.findFactor();
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|696021|}|unstyled|696022|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code> import java.util.Scanner;
 public class Factors 
 {
 Scanner scn=new Scanner(System.in);
 int num=scn.nextInt();
 public void findFactor()
 {
 System.out.println("Factors are");
 System.out.println("1");
 for(int i=2;i&lt;=num;i++)
 {
 if(num%i==0)
 {
 num=num/i;
 System.out.println(i);
 i=2;
 }
 }
 }
public static void main(String[] args) 
{
 while(1==1)
 {
 System.out.println("Enter a Number");
 Factors fct=new Factors();
 fct.findFactor();
 }
}
</code></pre>

}

blocks|key|480120|text|利用Java+8中引入的Streams，下面将打印给定数字的因子。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|480121|int+input+=+1500;
IntStream.rangeClosed(1,+input)
+++++++++.filter(e+->+input+%25+e+==+0)
+++++++++.forEach(System.out::println);|code-block|syntax|javascript|480122|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Utilizing Streams introduced in Java 8, the following will print the factors for a given number.

<pre><code>int input = 1500;
IntStream.rangeClosed(1, input)
 .filter(e -&gt; input % e == 0)
 .forEach(System.out::println);
</code></pre>

blocks|key|480168|text|这就是你如何像老板一样自己写的。需要添加if语句来处理一条和两条语句，但除此之外，此方法非常有用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|480169|++public+static+void+primerize(int+n){
++boolean+reduced+=+false;
++while(n+>+2){
+++++if(n%252+==+0){
++++++++System.out.println(2+%2B+","+%2B+n/2);
++++++++n+/=+2;
+++++}
+++++else{
++++++++int+i+=+isPrime(n);+
++++++++if(i+==+n+&&+reduced+==+false){
+++++++++++System.out.println(1+%2B+","+%2B+n);
+++++++++++n+/=+n;
++++++++}
++++++++else+if(i+==+n){
+++++++++++n/=+n;
++++++++}
++++++++else{
+++++++++++System.out.println(i+%2B+","+%2B+n/i);
+++++++++++n+=+i;
+++++++++++reduced+=+true;
++++++++}
+++++}
++}}



public+static+int+isPrime(int+n){
++for(int+i+=+(n/3);+i+>+0;+i--){
+++++if(i+==+1){
++++++++return+n;
+++++}
+++++else+if(n%25i+==+0){
++++++++return+i;
+++++}
++}
++return+0;}|code-block|syntax|javascript|480170|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This is how you write it yourself like a boss. Needs to add if statements to handle one and two, but besides that; this method is as sexy as it gets 

<pre><code> public static void primerize(int n){
 boolean reduced = false;
 while(n &gt; 2){
 if(n%2 == 0){
 System.out.println(2 + "," + n/2);
 n /= 2;
 }
 else{
 int i = isPrime(n); 
 if(i == n &amp;&amp; reduced == false){
 System.out.println(1 + "," + n);
 n /= n;
 }
 else if(i == n){
 n/= n;
 }
 else{
 System.out.println(i + "," + n/i);
 n = i;
 reduced = true;
 }
 }
 }}



public static int isPrime(int n){
 for(int i = (n/3); i &gt; 0; i--){
 if(i == 1){
 return n;
 }
 else if(n%i == 0){
 return i;
 }
 }
 return 0;}
</code></pre>

blocks|key|480185|text|public+static+void+printFactors(int+number)+{
++++if+(number+<+1+)
++++++++System.out.println("Invalid+Value");

++++for+(int+i+=+1+;+i+<=+number+;+%2B%2Bi)+{
++++++++if+(+number+%25+i+==+0)
++++++++++++++++System.out.println(i);
++++++++}
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|480186|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre class="lang-java prettyprint-override"><code>public static void printFactors(int number) {
 if (number &lt; 1 )
 System.out.println("Invalid Value");

 for (int i = 1 ; i &lt;= number ; ++i) {
 if ( number % i == 0)
 System.out.println(i);
 }
 }
}
</code></pre>

I have something like this down:

<pre><code>int f = 120;
for(int ff = 1; ff &lt;= f; ff++){
 while (f % ff != 0){ 
}
</code></pre>

Is there anything wrong with my loop to find factors? I'm really confused as to the workings of for and while statements, so chances are they are completely wrong.

After this, how would I go about assigning variables to said factors?

Finding factors of a given integer

Java

我有一个类似这样的东西：int f = 120;for(int ff = 1; ff <= f; ff++){ while (f % ff != 0){ }我的查找因子的循环有什么问题吗？我真的对for和while语句的工作原理感到困惑，所以它们很可能是完全错误的。在这之后，我该如何将变量赋...

问查找给定整数的因子
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