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 Kotlin (1.0.0) (judge verdict: Accepted)

import java.util.Comparator

fun main(args: Array<String>) {
    val (n, d, a) = readLine()!!.split(' ').map { it.toLong() }
    val ms = (0 until n).map {
        val (x, h) = readLine()!!.split(' ').map { it.toLong() }
        x to h
    }.sortedBy { it.first }

    var ans = 0L
    val hs = LongArray(n.toInt() + 2)

    for (i in 0 until n.toInt()) {
        val j = ms.binarySearch(
                ms[i].first + 2 * d to 0,
                Comparator { o1, o2 -> if (o1.first > o2.first) 1 else -1 }
        ).let { -(it + 1) }

        hs[i + 1] += hs[i]
        val h = Math.max(ms[i].second + hs[i + 1], 0)
        val c = (h + a - 1) / a

        ans += c
        hs[i + 1] -= a * c
        hs[j + 1] += a * c
    }

    println(ans)
}