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 Go (1.6) (judge verdict: Time Limit Exceeded)

package main

import (
	"fmt"
	"sort"
)

type Pair struct {
	x, h int
}

type Pairs []Pair

func (p Pairs) Len() int {
	return len(p)
}

func (p Pairs) Less(i, j int) bool {
	return p[i].x < p[j].x
}

func (p Pairs) Swap(i, j int) {
	p[i], p[j] = p[j], p[i]
}

func main() {
	var N, D, A int
	fmt.Scan(&N, &D, &A)

	P := make(Pairs, N)
	for i := 0; i < N; i++ {
		var x, h int
		fmt.Scan(&x, &h)
		P[i] = Pair{x, h}
	}

	sort.Sort(Pairs(P))

	count := 0
	for i := 0; i < N; i++ {
		if P[i].h <= 0 {
			continue
		}

		bomb := P[i].x + D
		throw := P[i].h / A
		if P[i].h%A != 0 {
			throw++
		}

		dmg := throw * A
		for j := i + 1; j < N; j++ {
			if bomb+D < P[j].x {
				break
			}
			P[j].h -= dmg
		}

		count += throw
	}

	fmt.Println(count)
}