[백준]#14954 Happy Number

SeungBird·2020년 8월 26일
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문제

Consider the following function f defined for any natural number n:

f(n) is the number obtained by summing up the squares of the digits of n in decimal (or base-ten).

If n = 19, for example, then f(19) = 82 because 12 + 92 = 82.

Repeatedly applying this function f, some natural numbers eventually become 1. Such numbers are called happy numbers. For example, 19 is a happy number, because repeatedly applying function f to 19 results in:

  • f(19) = 12 + 92 = 82
  • f(82) = 82 + 22 = 68
  • f(68) = 62 + 82 = 100
  • f(100) = 12 + 02 + 02 = 1

However, not all natural numbers are happy. You could try 5 and you will see that 5 is not a happy number. If n is not a happy number, it has been proved by mathematicians that repeatedly applying function f to n reaches the following cycle:

4 → 16 → 37 → 58 → 89 → 145 → 42 → 20 → 4.

Write a program that decides if a given natural number n is a happy number or not.

입력

Your program is to read from standard input. The input consists of a single line that contains an integer, n (1 ≤ n ≤ 1,000,000,000)

출력

Your program is to write to standard output. Print exactly one line. If the given number n is a happy number, print out HAPPY; otherwise, print out UNHAPPY.

예제 입력 1

19

예제 출력 1

HAPPY

예제 입력 2

5

예제 출력 2

UNHAPPY

풀이

이 문제에서 유의해야 할 점은 값이 4가 나오면 무조건 UNHAPPY Number가 된다는 점이다.

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String str = sc.next();
        String unhap = "4";

        do{
            str = happy(str);

            if(str.equals(unhap)) {
                System.out.println("UNHAPPY");
                return ;
            }

        }while(!str.equals("1"));
        System.out.println("HAPPY");
    }

    public static String happy(String origin) {
        int k = 0;

        for(int i=0; i<origin.length(); i++) {
            k +=  Math.pow(Character.getNumericValue(origin.charAt(i)), 2);
        }

       return Integer.toString(k);
    }
}
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