Question

问题：给定一个由n个数字组成的数组，找到它的LCM。

由于LCM（a，b）= a * b / GCD（a，b），所以这是我的原始代码：

class GFG {

      // the simple 2 integers recursive to find their GCD.

    static int getGcd(int a, int b){
        while(a != b){
            if(a >b){
                a = a-b;
            }else{
                b= b-a;
            }
        }
        return a;
    }

    // find GCD in an array that has more than two integers.

    static int Gcd(int[] newArray){

     int gcd = newArray[0];
       for(int i=1; i<newArray.length; i++){
        gcd = getGcd(gcd,newArray[i]);
     }

        return gcd;
    }

    static int Lcm(int[] newArray){
        int multi =1;
        for(int i=0; i< newArray.length; i++){
            multi = multi* newArray[i];
        }
        return (multi/gcd);
    }

    public static void main (String[] args) {
        int[] newArray = { 2, 7, 3, 9, 4 };
        System.out.println(Gcd(newArray));
        System.out.println(Lcm(newArray));
    }
}

但是当我运行这段代码时，它显示了一些错误：

prog.java:33: error: cannot find symbol
        return (multi/gcd);
                      ^
  symbol:   variable gcd

我不知道如何解决它。请帮助我更正我的代码...谢谢！！！

Answer 1

Lcm（）函数中未定义gcd。最后定义的gcd在Gcd（）函数中，尽管这不是全局变量Lcm（）无法使用它。

也许你是这个意思？

static int Lcm(int[] newArray){
    int multi =newArray[0];
    int gcd = 0;
    for(int i=1; i< newArray.length; i++){
        gcd = getGcd(multi, newArray[i]);
        if(gcd == 1) {
            multi = multi * newArray[i];
        }else {
            multi = multi * newArray[i] / gcd;
        }
    }
    return multi;
}

使用它可以生成结果。您可以对此有所考虑。我们从2开始，multi为2，下一个值为7。2和7的gcd为1，所以lcd为14。然后我们用14比较3，gcd也为1，所以将它们相乘就是42.然后使用42比较9，我们发现9具有42的gcd是3，所以我们使用42/3是14，然后使用14和9，gcd是1，给出14x9是126。然后使用126和4 ，则gcd为2，因此multi为126/2 = 63，而63和4 gcd为1，所以最终multi为63x4为252。

如果您看到任意两个数字，则无论是[3,7]，[50,5]，[20,10]，[100,10000]，[52,57]，[3,81] ]，则lcm始终为a * b / gcd。在这种情况下，我们可以获得答案。

Answer 2

我希望“本逻辑”对您有所帮助。

// Java Program to find LCM of n elements
public class GFG {

public static long lcm_of_array_elements(int[] element_array)
{
    long lcm_of_array_elements = 1;
    int divisor = 2;

    while (true) {
        int counter = 0;
        boolean divisible = false;

        for (int i = 0; i < element_array.length; i++) {

            // lcm_of_array_elements (n1, n2, ... 0) = 0.
            // For negative number we convert into
            // positive and calculate lcm_of_array_elements.

            if (element_array[i] == 0) {
                return 0;
            }
            else if (element_array[i] < 0) {
                element_array[i] = element_array[i] * (-1);
            }
            if (element_array[i] == 1) {
                counter++;
            }

            // Divide element_array by devisor if complete
            // division i.e. without remainder then replace
            // number with quotient; used for find next factor
            if (element_array[i] % divisor == 0) {
                divisible = true;
                element_array[i] = element_array[i] / divisor;
            }
        }

        // If divisor able to completely divide any number
        // from array multiply with lcm_of_array_elements
        // and store into lcm_of_array_elements and continue
        // to same divisor for next factor finding.
        // else increment divisor
        if (divisible) {
            lcm_of_array_elements = lcm_of_array_elements * divisor;
        }
        else {
            divisor++;
        }

        // Check if all element_array is 1 indicate 
        // we found all factors and terminate while loop.
        if (counter == element_array.length) {
            return lcm_of_array_elements;
        }
    }
}

// Driver Code
public static void main(String[] args)
{
    int[] element_array = { 2, 7, 3, 9, 4 };
    System.out.println(lcm_of_array_elements(element_array));
}
}

Answer 3

对于两个数字的GCD逻辑为，

// Java program to find GCD of two numbers
class Test
{
// Recursive function to return gcd of a and b
static int gcd(int a, int b)
{
    // Everything divides 0 
    if (a == 0 || b == 0)
       return 0;

    // base case
    if (a == b)
        return a;

    // a is greater 
    //u r missing the logic here.....................
    if (a > b)
        return gcd(a-b, b);
    else
        return gcd(a, b-a);
}

// Driver method
public static void main(String[] args) 
{
    int a = 98, b = 56;
    System.out.println("GCD of " + a +" and " + b + " is " + gcd(a, b));
}
}

请查看逻辑，您应该迭代直到a = b，仅重复一次。

请告诉我为什么它错了（LCM＆GCD）

3 个答案: