Question

我编写了一个代码，需要检查一个理论，即除非n=2 a^n + b^n = c^n，否则以下公式永远不会成立。该程序最后告诉我“Meh，Fermat是对的”标志着程序的结束，但之后它抛出了很多异常。

这是代码 -

public class CheckFermats {

    public static void main(String[] args) {
        int a=2;
        int b=2;
        int c=2;
        int n=3;
        int round = 2;
        CheckupA (a,b,c,n,round);

    }

    public static void CheckupA (int a, int b, int c, int n ,int round){
        if (Math.pow(a,n)+Math.pow(b,n)==Math.pow(c, n))
        {
            System.out.println("HOLYSHIT FERMET WAS WRONG!!");
        }
        else
        {
            if (a==9)
            {
                a=round+1;
                CheckupB (a,b,c,n, round);

            }
            else
            {
            a++;
            CheckupA (a,b,c,n, round);
            }
        }
    }

    public static void CheckupB (int a, int b, int c, int n,int round){
        if (Math.pow(a,n)+Math.pow(b,n)==Math.pow(c, n))
        {
            System.out.println("HOLYSHIT FERMET WAS WRONG!!");
        }
        else
        {
            if (b==9)
            {
                b=round+1;
                CheckupC (a,b,c,n, round);

            }
            else
            {
            b++;
            CheckupB (a,b,c,n, round);
            }
        }
    }

    public static void CheckupC (int a, int b, int c, int n,int round){
        if (Math.pow(a,n)+Math.pow(b,n)==Math.pow(c, n))
        {
            System.out.println("HOLYSHIT FERMET WAS WRONG!!");
        }
        else
        {
            if (c==9)
            {
                c=round+1;
                round++;
                CheckIfFinished (n);
                checkIfIncreaseN (a,b,c,round, n);
                CheckupA (a,b,c,n, round);

            }
            else
            {
            c++;
            CheckupC (a,b,c,n, round);
            }
        }
    }

    public static void checkIfIncreaseN(int a, int b, int c, int round, int n){
        if(round==9)
        {
            round = 2;
            a=2;
            b=2;
            c=2;
            n++;
            CheckupA (a,b,c,n,round);

        }
    }

    public static void CheckIfFinished (int n){
        if (n==9)
        {
            System.out.println("Meh, Fermet was right");
            return;
        }
    }
}

Answer 1

即使您已检查Meh, Fermet was right，您的代码也会继续。查看代码CheckIfFinished (n);，即使它已完成，也会继续。

所以你可以在方法中添加一个变量来检查是否满足了检查，或者使用方法的返回值。

E.g。附加参数：

public static void CheckupC (int a, int b, int c, int n,int round, boolean finished)
{
    if (finished) {return;}
    ...
}

或者您使用方法的返回值，以表明它已完成。

if (CheckupC(...)) { return true; }

当你将它用于整个代码时，它看起来像这样（也许它不是最好的风格，但它解决了你的问题）：

public static void main(String[] args) {
    int a=2;
    int b=2;
    int c=2;
    int n=3;
    int round = 2;
    CheckupA (a,b,c,n,round);

}
public static boolean CheckupA (int a, int b, int c, int n ,int round)
{
    if (Math.pow(a,n)+Math.pow(b,n)==Math.pow(c, n))
    {
        System.out.println("HOLYSHIT FERMET WAS WRONG!!");
        return true;
    }
    else
    {
        if (a==9)
        {
            a=round+1;
            return CheckupB (a,b,c,n, round);

        }
        else
        {
            a++;
            return CheckupA (a,b,c,n, round);
        }
    }
}
public static boolean CheckupB (int a, int b, int c, int n,int round)
{
    if (Math.pow(a,n)+Math.pow(b,n)==Math.pow(c, n))
    {
        System.out.println("HOLYSHIT FERMET WAS WRONG!!");
        return true;
    }
    else
    {
        if (b==9)
        {
            b=round+1;
            return CheckupC (a,b,c,n, round);

        }
        else
        {
            b++;
            return CheckupB (a,b,c,n, round);
        }
    }
}
public static boolean CheckupC (int a, int b, int c, int n,int round)
{
    if (Math.pow(a,n)+Math.pow(b,n)==Math.pow(c, n))
    {
        System.out.println("HOLYSHIT FERMET WAS WRONG!!");
        return true;
    }
    else
    {
        if (c==9)
        {
            c=round+1;
            round++;
            if (CheckIfFinished (n)) {
                return true;
            }
            if (checkIfIncreaseN (a,b,c,round, n)) {
                return true;
            }
            return CheckupA (a,b,c,n, round);

        }
        else
        {
            c++;
            return CheckupC (a,b,c,n, round);
        }
    }
}
public static boolean checkIfIncreaseN(int a, int b, int c, int round, int n)
{
    if(round==9)
    {
        round = 2;
        a=2;
        b=2;
        c=2;
        n++;
        return CheckupA (a,b,c,n,round);
    }

    return false;
}
public static boolean CheckIfFinished (int n)
{
    if (n==9)
    {
        System.out.println("Meh, Fermet was right");
        return true;
    }

    return false;
}

使用double进行计算时，还有一些建议不要将双值与==进行比较，而是使用Double.compare。

试图检查费马理论之一的代码正在抛出异常

1 个答案: