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: Runtime Error)

import java.util.*
fun main(args:Array<String>) {
    data class Monster(val x:Long=0L, var hp:Long=0L)
    var (n,d,a) = readLine()!!.split(" ").map { it.toInt() }
    var monster = Array(n){Monster(0,0)}



    val map = HashMap<Long,Int>()
    val treeSet = TreeSet<Long>()

    for(i in 0 until n) {
        val (x,h) = readLine()!!.split(" ").map { it.toInt() }
        monster[i] = Monster(x.toLong(), h.toLong())
    }
//    createTestSet()

    monster.sortBy { it.x }

    for(i in monster.indices) {
        map[monster[i].x] = i
        treeSet.add(monster[i].x)
    }

    var answer = 0L
    for(i in 0 until n) {
        val (x, h) = monster[i]
        if(h <= 0) continue
        val dst = x + 2 * d
        var cnt = h / a //+ min(h%a, 1)
        if((h%a) >= 1) {
            cnt += 1
        }
        val next = treeSet.lower(dst)
        val nextIdx = map[next]!!
        for(j in i .. nextIdx) {
            monster[j].hp -= cnt * a
        }
        answer += cnt
    }
    println(answer)
    // x 의 범위가 무지막지하게 크다.
    //


}