如何检查号码是否为Armstrong

Question

我正在尝试编写一个类来检查一个数字是否是阿姆斯壮的数字。我遇到以下代码块时遇到问题。

public boolean checkNum(long num) {
    digits = (int) (Math.log10(num) + 1);
    String number = String.valueOf(num);
    numDigits = number.toCharArray();
    for (int i = 0; i < numDigits.length; i++) {
        digit = numDigits[i] * 1.0;
        power = digits * 1.0;
        sum = sum + (long) (Math.pow(digit, power));
    }
    if (sum == num) {
        return true;
    } else {
        return false;
    }
}

投射似乎不起作用，checkNum每次都返回false。这是一种正确的方法，还有更好的方法吗？

Answer 1

尝试这一点，仅使用算术运算，它适用于具有任意位数的非负整数（只要它们适合long）。

public boolean checkNum(long num) {

    long n = num;
    long sum = 0;

    // find the number of digits
    int power = (int) Math.floor(Math.log10(n == 0 ? 1 : n)) + 1;

    while (n != 0) {
        int digit = (int) n % 10;
        sum += Math.pow(digit, power);
        n /= 10;
    }

    return sum == num;

}

或者（虽然效率较低）您可以将数字转换为字符串并迭代每个字符将其转换为数字。以下是您的预期解决方案的固定版本，并对关键点进行了评论：

public boolean checkNum(long num) {

    String number = String.valueOf(num);
    char[] numDigits = number.toCharArray();

    long sum = 0;
    // a simple way to obtain the number of digits
    int power = numDigits.length;

    for (int i = 0; i < numDigits.length; i++) {
        // this is how we transform a character into a digit
        int digit = Character.digit(numDigits[i], 10);
        // we need to rise digit to the value of power
        sum = sum + (long) Math.pow(digit, power);
    }

    if (sum == num) {
        return true;
    } else {
        return false;
    }

}

例如，使用任一实现来验证以下是Armstrong数字：

checkNum(6)
=> true
checkNum(371)
=> true
checkNum(1634)
=> true

Answer 2

您也可以使用此简单逻辑

public class Armstrong {

    public static void main(String[] args) {

        int number = 371, originalNumber, remainder, result = 0;

        originalNumber = number;

        while (originalNumber != 0)
        {
            remainder = originalNumber % 10;
            result += Math.pow(remainder, 3);
            originalNumber /= 10;
        }

        if(result == number)
            System.out.println(number + " is an Armstrong number.");
        else
            System.out.println(number + " is not an Armstrong number.");
    }
}

Answer 3

猜猜这将起作用：

boolean isArmstrong(int x){
int s=0;
int u=x;
while(x!=0)
   {
       int y=x%10;
       s=s+(y*y*y);
       x=x/10;
    }
    if(u==s)
    return true;
    else 
    return false;
}

Answer 4

如何检查号码是否为Armstrong

null

Answer 5

在Kotlin中，您可以使用：

fun main() {

println("---------------------------------------")

val userInputValues = Scanner(System.`in`)

//* Kotlin Program to Display Armstrong Numbers Between Intervals Using Function
println("* Kotlin Program to Display Armstrong Numbers Between Intervals Using Function\n")

println("Enter your number range")
println("Enter start number of your range \t ")
val startRange = userInputValues.nextInt()

println("Enter end number of your range \t ")
val endRange = userInputValues.nextInt()


println("\n\n------ Armstrong number between $startRange and $endRange ------  ")
for (number in startRange..endRange) {

    var stringNumber : String = number.toString()
    var numberArray = stringNumber.toCharArray()
    var powerOfNumber:Int = numberArray.size;

    var result = 0
    for (digit in numberArray){
        var intDigit:Int = digit.toString().toInt()
        result += intDigit.toDouble().pow(powerOfNumber.toDouble()).toInt()
    }

    if(result == number){
        println( "$number is Armstrong number")
    }

}


println("---------------------------------------")
}

Answer 6

    public static void main(String[] args) {
    Scanner scanner=new Scanner(System.in);
    System.out.print("Enter the number: ");
    int number=scanner.nextInt();
    scanner.close();
    
    int lastDigit=0;
    int reverseNum=0;
    int originalNumber=number;
    while(originalNumber!=0) {
        lastDigit=originalNumber%10;
        reverseNum +=(lastDigit*lastDigit*lastDigit);
        originalNumber /=10;
    }
    if(reverseNum==number) {
        System.out.println("Number is Armstrong");
    }
    else {
        System.out.println("Number is not Armstrong");
    }

}

Answer 7

在这里，我已经完成了一段代码以动态查找阿姆斯壮编号：

import java.util.Scanner;

public class Armstrong {
    public static void main(String[] args) {
        if(isArmstrongNumber(input())) {
            System.out.println("armstrong number");
        } else {
            System.out.println("Not armstrong number");
        }
    }

    private static int input() {
        try(Scanner reader = new Scanner(System.in)) {
            return reader.nextInt();
        }
    }

    private static int digitCount(int num) {
        int count = 0;
        while(num > 0) {
            num = num / 10;
            count++;
        }
        System.out.println("No of digit : " + count);
        return count;
    }

    private static int power(int num, int count) {
        int sum = 0;
        while(num > 0) {
            int result = 1;
            int r2 = num % 10;
            num /= 10;
            for(int digit = count; digit > 0; digit--) {
                result *= r2;
            }
            sum += result;
        }
        System.out.println("Sum : " + sum);
        return sum;
    }

    public static boolean isArmstrongNumber(int num) {
        int count = digitCount(num);
        int sum = power(num, count);
        return sum == num;
    }
}

这是结果：

371
No of digit : 3
Sum : 371
armstrong number

希望此代码对您有所帮助。

用Java检查阿姆斯特朗号码

7 个答案:

如何检查号码是否为Armstrong