Problem:
Score : 200 points

Problem Statement

Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second.
The other child is now at coordinate B, and she can travel the distance of W per second.

He can catch her when his coordinate is the same as hers.
Determine whether he can catch her within T seconds (including exactly T seconds later).
We assume that both children move optimally.

Constraints

-10^9 \leq A,B \leq 10^9

1 \leq V,W \leq 10^9

1 \leq T \leq 10^9

A \neq B

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A V
B W
T

Output

If "it" can catch the other child, print YES; otherwise, print NO.

Sample Input 1

1 2
3 1
3

Sample Output 1

YES

Sample Input 2

1 2
3 2
3

Sample Output 2

NO

Sample Input 3

1 2
3 3
3

Sample Output 3

NO

Solution in Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int ogrePoint = getInt(sc);
		int ogreSpeed = getInt(sc);
		int childPoint = getInt(sc);
		int childSpeed = getInt(sc);
		int time = getInt(sc);

		if (childSpeed >= ogreSpeed) {
			System.out.println("NO");
		} else {
			int ogreDistance = ogreSpeed * time;
			int childDistance = childSpeed * time;
			boolean enable = false;
			if (ogrePoint < childPoint && ogrePoint + ogreDistance >= childPoint + childDistance) {
				enable = true;
			} else if (ogrePoint - ogreDistance <= childPoint - childDistance) {
				enable = true;
			}

			if (enable) {
				System.out.println("YES");
			} else {
				System.out.println("NO");
			}
		}

	}

	/**
	 * *汎用メソッド<br>
	 * 入力した配列に対する順列を取得
	 *
	 * @param numArray
	 * @return
	 */
	private static List<Integer[]> getPermutation(int[] numArray) {
		List<Integer[]> permutation = new ArrayList<>();
		Integer[] ans = new Integer[numArray.length];
		boolean[] used = new boolean[numArray.length];

		return buildPermutaion(permutation, numArray, ans, used, 0);
	}

	/**
	 * *汎用メソッド(内部)<br>
	 * 入力した配列に対する順列を作成
	 *
	 * @param list
	 * @param numArray
	 * @param ans
	 * @param used
	 * @param depth
	 * @return
	 */
	private static List<Integer[]> buildPermutaion(List<Integer[]> list, int[] numArray, Integer[] ans, boolean[] used,
			int depth) {

		// 全部使い切った時=>できた順列を返す
		if (depth == numArray.length) {
			list.add(ans.clone());
			return list;
		}
		// 使い切っていない場合

		for (int i = 0; i < numArray.length; i++) {
			if (used[i]) {
				continue;
			}
			ans[depth] = numArray[i];
			used[i] = true;
			list = buildPermutaion(list, numArray, ans, used, depth + 1);

			used[i] = false;
		}

		return list;
	}

	/**
	 * *汎用メソッド <br>
	 * スキャナーからの入力値をlongで返却
	 *
	 * @return
	 */
	public static long getLong(Scanner sc) {

		return Long.parseLong(sc.next());
	}

	/**
	 * *汎用メソッド <br>
	 * スキャナーからの入力値をintで返却
	 *
	 * @return
	 */
	public static int getInt(Scanner sc) {

		return Integer.parseInt(sc.next());
	}

	/**
	 * *汎用メソッド <br>
	 * スキャナーからの入力値をStringで返却
	 *
	 * @return
	 */
	public static String getString(Scanner sc) {

		return sc.next();
	}

	/**
	 * *汎用メソッド <br>
	 * スキャナーからの入力値をint配列で返却
	 *
	 * @return
	 */
	public static int[] getIntArray(Scanner sc) {
		String line = "";
		// 直前にnext()を実行していると改行コードだけ取ってしまうのでその対策
		while (line.equals("")) {
			line = sc.nextLine();
		}

		String[] strArray = line.split(" ");
		int[] intArray = new int[strArray.length];
		for (int i = 0; i < strArray.length; i++) {
			intArray[i] = Integer.parseInt(strArray[i]);
		}
		return intArray;
	}

}