Problem:
﻿Score: 100 points

Problem Statement

In 2020, AtCoder Inc. with an annual sales of more than one billion yen (the currency of Japan) has started a business in programming education.

One day, there was an exam where a one-year-old child must write a program that prints Hello World, and a two-year-old child must write a program that receives integers A, B and prints A+B.

Takahashi, who is taking this exam, suddenly forgets his age.

He decides to write a program that first receives his age N (1 or 2) as input, then prints Hello World if N=1, and additionally receives integers A, B and prints A+B if N=2.

Write this program for him.

Constraints

N is 1 or 2.

A is an integer between 1 and 9 (inclusive).

B is an integer between 1 and 9 (inclusive).

Input

Input is given from Standard Input in one of the following formats:

1

2
A
B

Output

If N=1, print Hello World; if N=2, print A+B.

Sample Input 1

1

Sample Output 1

Hello World

As N=1, Takahashi is one year old. Thus, we should print Hello World.

Sample Input 2

2
3
5

Sample Output 2

8

As N=2, Takahashi is two years old. Thus, we should print A+B, which is 8 since A=3 and B=5.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)



import java.io.IOException;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) throws IOException {

		Scanner scan = new Scanner(System.in);

		int N_year;
		int A, B;

		N_year = scan.nextInt();

		if (N_year == 1) {
			System.out.println("Hello world");

		} else {


			A = scan.nextInt();

			B = scan.nextInt();

			System.out.println((A + B));

		}

	}

	//***********************以下共通関数************************************************
	/**
	 * 文字列を反転して返す
	 * @param pStr 文字列
	 * @return 反転した文字列
	 */
	public static String myReverse(String pStr) {
		if (pStr.length() < 1) {
			return "";
		}

		StringBuffer sb = new StringBuffer(pStr);
		return sb.reverse().toString();
	}

	/**
	 * 余りを求める。
	 * @param a
	 * @param b
	 * @return 余り。余りがマイナスの時はプラスで返す。
	 */
	public static long myMod(long a, long b) {
		long ret = 0;

		ret = a % b;

		if (ret < 0) {
			ret += b;
		}

		return ret;
	}

	/**
	 * 乗数を求める ※mod対応
	 * @param kisu 基数
	 * @param josu 乗数
	 * @param primeNumMod_0 法 mod値 使わない場合は「0」
	 * @return
	 */
	public static long myPowWithMod(long kisu, long josu, long primeNumMod_0) {

		long ans = 1;

		//乗数が0の場合→1
		if (josu == 0)
			return 1;

		if (primeNumMod_0 == 0) {
			//modを使わないとき

			while (josu > 0) {

				if (josu % 2 == 0) {
					//累乗が偶数の時

					kisu *= kisu;
					josu = josu / 2;
				} else {
					//累乗が奇数の時
					ans *= kisu;
					josu--;
				}
			}

			return ans;

		} else {
			//modを使うとき

			//初めに桁を落とす
			kisu = myMod(kisu, primeNumMod_0);

			while (josu > 0) {

				if (josu % 2 == 0) {
					//累乗が偶数の時

					kisu *= myMod(kisu, primeNumMod_0);
					josu = josu / 2;

				} else {
					//累乗が奇数の時
					ans *= myMod(kisu, primeNumMod_0);
					josu--;

				}
			}

			return myMod(ans, primeNumMod_0);

		}

	}

}