Java for循環條件

Question

我想知道是否有可能為此提供最少的代碼：

for (int x = 1; x < 10; x++){    
      /*I want to replace this condition with (x%number == 0) 
        instead of writing out condition for every number, but 
        it did not work with for (int number = 1; number <= 3; number++)
        in (x%number == 0), as it prints out every x and number 
      */
    if ((x%1) == 0 && (x%2) == 0 & (x%3) == 0){
       System.out.println(success!);
    }    
}

Answer 1

我認為
x % a == 0 && x % b == 0 && x % c == 0
等於
x％（a * b * c）== 0

UPDATE
乘法不正確，您需要使用LCM ： x % lcm(a, b, c)

Answer 2

看一看 ：

for (int x = 1; x < 10; x++){
  boolean flag = false;
    for(int num = 1; num <= 3; num++){
       if ((x%num) == 0 ){
          flag = true;
       }else{
          flag = false;
          break;
       }
    }
    if(flag){
            System.out.println(x + " success!");
    }
}

輸出：

6 success!

我知道代碼看起來有些恐懼，但可以用於x和num任何值

Answer 3

這是使一名復合科學教授滿意的條件：

for (int x = 1; x < 10; x++){    
    boolean success = true;
    for (int number = 1; number <= 3; number++) {
        if ((x % number) != 0) {
            success = false;
        }
    }
    if (success) {
       System.out.println("success!");
    }    
}

盡管注意：（x％1）始終為0。

根據我的“避免嵌套循環”規則，這就是讓我高興的條件：

for (int x = 1; x < 10; x++) {
    if (testNumber(x)) 
        System.out.println(x + " success!");
    }
}

private static boolean testNumber(int x) {
    for (int number = 1; number <= 3; number++) {
        if ((x % number) != 0) {
            return false;
        }
    }
    return true;
}