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: Accepted)

import java.util.*;

public class Main {

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        long d = sc.nextInt();
        long a = sc.nextInt();
        E[] e = new E[n];
        for(int i=0 ;i<n; i++) {
            int x = sc.nextInt();
            int y = sc.nextInt();
            e[i] = new E(x, (y+a-1)/a);
        }
        Arrays.sort(e);
        long res = 0;
        long pow = 0;
        Queue<E> q = new LinkedList<>();
        for(int i=0; i<n; i++) {
            while (!q.isEmpty()) {
                E entry = q.peek();
                if (e[i].x - entry.x > d * 2) {
                    pow -= entry.h;
                    q.poll();
                } else {
                    break;
                }
            } if (e[i].h > pow) {
                res += e[i].h - pow;
                q.add(new E(e[i].x, e[i].h - pow));
                pow += e[i].h - pow;
            }
        }


        System.out.println(res);


    }

    static class E implements Comparable {
        long x;
        long h;
        public E (long x, long h) {
            this.x = x;
            this.h = h;
        }

        @Override public int compareTo(Object o) {
            return Long.compare(x, ((E)o).x);
        }
    }

}