Numărul Armstrong în programul JAVA folosind bucla For

prin: James Hartman James Hartman
Hours Actualizat

Ce este numărul Armstrong?

Într-un număr Armstrong, suma puterii cifrelor individuale este egală cu numărul însuși.

Cu alte cuvinte, următoarea ecuație va fi adevărată

xy..z = xn + yn+.....+ zn

n este numărul de cifre în număr

De exemplu, acesta este un număr Armstrong din 3 cifre

370 = 33 + 73 + o3
 = 27 + 343 + 0
 = 370

Exemple de Armstrong Numbers

0, 1, 4, 5, 9, 153, 371, 407, 8208, etc.

Să scriem asta într-un program:

Java Program pentru a verifica dacă un număr este Armstrong Number

//ChecktempNumber is Armstrong or not using while loop
package com.guru99;
 
public class ArmstrongNumber {
	public static void main(String[] args) {
		 int inputArmstrongNumber = 153; //Input number to check armstrong  
		 int tempNumber, digit, digitCubeSum = 0;
	       tempNumber = inputArmstrongNumber;
	        while (tempNumber != 0)
	        {
	        	/* On each iteration, remainder is powered by thetempNumber of digits n
	        	 */
	            System.out.println("Current Number is "+tempNumber);
	            digit =tempNumber % 10;
				System.out.println("Current Digit is "+digit);
	            //sum of cubes of each digits is equal to thetempNumber itself
	            digitCubeSum = digitCubeSum + digit*digit*digit;
				System.out.println("Current digitCubeSum is "+digitCubeSum);
	            tempNumber /= 10;
	        }
	        //check giventempNumber and digitCubeSum is equal to or not 
	        if(digitCubeSum == inputArmstrongNumber)
	            System.out.println(inputArmstrongNumber + " is an Armstrong Number");
	        else
	            System.out.println(inputArmstrongNumber + " is not an Armstrong Number");
	}
}

producție

Current Number is 153
Current Digit is 3
Current digitCubeSum is 27
Current Number is 15
Current Digit is 5
Current digitCubeSum is 152
Current Number is 1
Current Digit is 1
Current digitCubeSum is 153
153 is an Armstrong Number

Java Program pentru a imprima numerele Armstrong de la 0 la 999

//ChecktempNumber is Armstrong or not using while loop
package com.guru99;
public class ArmstrongNumber {
    public static void main(String[] args) {
        int tempNumber, digit, digitCubeSum;
        for (int inputArmstrongNumber = 0; inputArmstrongNumber < 1000; inputArmstrongNumber++) {
            tempNumber = inputArmstrongNumber;
            digitCubeSum = 0;
            while (tempNumber != 0) {
                /* On each iteration, remainder is powered by thetempNumber of digits n
                 */
                digit = tempNumber % 10;
                //sum of cubes of each digits is equal to thetempNumber itself
                digitCubeSum = digitCubeSum + digit * digit * digit;
                tempNumber /= 10;
            }
            //check giventempNumber and digitCubeSum is equal to or not 
            if (digitCubeSum == inputArmstrongNumber)
                System.out.println(inputArmstrongNumber + " is an Armstrong Number");
        }
    }
}

producție

0 is an Armstrong Number
1 is an Armstrong Number
153 is an Armstrong Number
370 is an Armstrong Number
371 is an Armstrong Number
407 is an Armstrong Number