Problem:
Score : 100 points

Problem Statement

There are a total of A + B cats and dogs.
Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs.

Determine if it is possible that there are exactly X cats among these A + B animals.

Constraints

1 \leq A \leq 100

1 \leq B \leq 100

1 \leq X \leq 200

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B X

Output

If it is possible that there are exactly X cats, print YES; if it is impossible, print NO.

Sample Input 1

3 5 4

Sample Output 1

YES

If there are one cat and four dogs among the B = 5 animals, there are X = 4 cats in total.

Sample Input 2

2 2 6

Sample Output 2

NO

Even if all of the B = 2 animals are cats, there are less than X = 6 cats in total.

Sample Input 3

5 3 2

Sample Output 3

NO

Even if all of the B = 3 animals are dogs, there are more than X = 2 cats in total.

Solution in Java (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 A, B, X;
		A = scan.nextInt();
		B = scan.nextInt();
		X = scan.nextInt();

		int wantCat = X - A;

		if (B > wantCat) {
			System.out.println("YES");
		} else {
			System.out.println("NO");
		}
	}

	//***********************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);

		}

	}

}