倍数计划

Question

这是我想要解决的编程问题：2520是可以除以1到10之间的每个数字而没有任何余数的最小数字。

从1到20的所有数字均可被整除的最小正数是多少？

到目前为止，这是我的解决方案，但是每次答案都是零，所以我认为我的代码中有错误。任何帮助将不胜感激。

public static boolean isDiv(int num){
    boolean isDiv = false;

    for (int i = 1; i <= 20; i++){
        if (i == 20){

            isDiv = true;

        }
        if ((num % i) == 0){
            continue;

        }
        else  {
            break;
        }

    }


return isDiv;}

public static int smallMulti(int num){
    boolean div = isDiv(num);
    int answer = 0;

    for (int i = num; num < 2520; i--){

        if (div = true){
            answer = i;
        }
    }

return answer;}

Answer 1

你整个问题过于复杂，加上多个逻辑错误。基本上你只需要2个循环。这是一个代码，用于检查每个数字所需的第一个数字，直到Integer.MAX_VALUE。如果你想要更高，你可以采用代码来使用long。

public static int smallMulti(int num) {
    for (int i = 1; num <= Integer.MAX_VALUE; ++i) { // Check every int in the scope of the Integer
        for (int j = 2;j<=num;++j) {
            if(i % j != 0) {
                break; // If i % j is unequal to 0 then this number isn´t valid.
            }
            if(j == num) {
                return i; // If we reached j == num then everything was divisble yet so we can return i as the correct value;
            }
        }
    }
    return -1;
}

这是主要

的示例输出

public static void main(String[] args) {
    for(int i = 2; i <= 20; ++i)
        System.out.println("Smallest Value divisible by 1-"+ i + " = " + smallMulti(i));
}

OutPut

Smallest Value divisible by 1-2 = 2
Smallest Value divisible by 1-3 = 6
Smallest Value divisible by 1-4 = 12
Smallest Value divisible by 1-5 = 60
Smallest Value divisible by 1-6 = 60
Smallest Value divisible by 1-7 = 420
Smallest Value divisible by 1-8 = 840
Smallest Value divisible by 1-9 = 2520
Smallest Value divisible by 1-10 = 2520
Smallest Value divisible by 1-11 = 27720
Smallest Value divisible by 1-12 = 27720
Smallest Value divisible by 1-13 = 360360
Smallest Value divisible by 1-14 = 360360
Smallest Value divisible by 1-15 = 360360
Smallest Value divisible by 1-16 = 720720
Smallest Value divisible by 1-17 = 12252240
Smallest Value divisible by 1-18 = 12252240
Smallest Value divisible by 1-19 = 232792560
Smallest Value divisible by 1-20 = 232792560

Answer 2

我使用lcm（最小公倍数）实现

public static int lcm(int a, int b) {
    return (a*b)/gcd(a, b);
}

public static int gcd(int a, int b) {
    return b == 0 ? a : gcd(b, a % b);
}

public static int smallMulti(int n) {
    int number = 1;

    for (int i = 2; i <= n; i++) {
        number = lcm(number, i);
    }

    return number;
}