Problem:
Score : 600 points

Problem Statement

Silver Fox is fighting with N monsters.

The monsters are standing in a row, and we can assume them to be standing on a number line. The i-th monster, standing at the coordinate X_i, has the health of H_i.

Silver Fox can use bombs to attack the monsters.
Using a bomb at the coordinate x decreases the healths of all monsters between the coordinates x-D and x+D (inclusive) by A.
There is no way other than bombs to decrease the monster's health.

Silver Fox wins when all the monsters' healths become 0 or below.

Find the minimum number of bombs needed to win.

Constraints

1 \leq N \leq 2 \times 10^5

0 \leq D \leq 10^9

1 \leq A \leq 10^9

0 \leq X_i \leq 10^9

1 \leq H_i \leq 10^9

X_i are distinct.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N D A
X_1 H_1
:
X_N H_N

Output

Print the minimum number of bombs needed to win.

Sample Input 1

3 3 2
1 2
5 4
9 2

Sample Output 1

2

First, let us use a bomb at the coordinate 4 to decrease the first and second monsters' health by 2.

Then, use a bomb at the coordinate 6 to decrease the second and third monsters' health by 2.

Now, all the monsters' healths are 0.
We cannot make all the monsters' health drop to 0 or below with just one bomb.

Sample Input 2

9 4 1
1 5
2 4
3 3
4 2
5 1
6 2
7 3
8 4
9 5

Sample Output 2

5

We should use five bombs at the coordinate 5.

Sample Input 3

3 0 1
300000000 1000000000
100000000 1000000000
200000000 1000000000

Sample Output 3

3000000000

Watch out for overflow.

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

import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		long d = sc.nextLong();
		long a = sc.nextLong();
		pair[] p = new pair[n];
		for (int i = 0; i < n; i++) {
			p[i] = new pair(sc.nextLong(), sc.nextLong());
		}
		Arrays.sort(p, new Comparator<pair>() {
			public int compare(pair p1, pair p2) {
				return p1.x>p2.x?1:-1;
			}
		});
		long now = p[0].x;
		long max = (p[0].h%a==0) ? (p[0].h/a) : (p[0].h/a+1);
		long ans = 0;
		if(n > 1) {
			for (int i = 1; i < n; i++) {
				long times = (p[i].h%a==0) ? (p[i].h/a) : (p[i].h/a+1);
				if(p[i].x > now+2*d) {
					ans += max;
					now = p[i].x;
					max = times;
				}else {
					if(times > max) {
						ans += max;
						max = times-max;
						now = p[i].x;
					}
				}
			}
		}
		System.out.println(ans+max);
		sc.close();
	}
	public static class pair{
		long x;
		long h;
		pair(long x, long h){
			this.x = x;
			this.h = h;
		}
	}
}