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.nextLong();
		long a=sc.nextLong();
		long[][] xh=new long[n][2];
		for(int i=0; i<n; i++){
			xh[i][0]=sc.nextLong();
			xh[i][1]=sc.nextLong();
		}
		Arrays.sort(xh,(x,y)->Long.compare(x[0],y[0]));
		long[] im=new long[n+1];
		int state=1;
		long ans=0;
		for(int i=0; i<n; i++){
			long dis=xh[i][0]+2*d;
			while(state<n && dis>=xh[state][0]){
				state++;
			}
			ans+=Math.max(0,(xh[i][1]-im[i]+a-1)/a);
			im[i+1]+=im[i]+Math.max(0,(xh[i][1]-im[i]+a-1)/a*a);
			im[state]-=Math.max(0,(xh[i][1]-im[i]+a-1)/a*a);
		}
		System.out.println(ans);
	}
}