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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

import math
from collections import deque
n,d,a = map(int,input().split())
lst1 = [list(map(int,input().split())) for i in range(n)]
lst1.sort()

#解法1 - 貪欲法
"""
キューを使った貪欲法で間に合わせる。
que = [[ダメージ量,そのダメージがどこまで使えるか],...]
とすることにより、高速にダメージ管理をすることができる。
"""

que = deque()
ans = 0
dmg = 0

for i in range(n):
    while que:
        lim = que.popleft()
        if lst1[i][0] > lim[1]:
            dmg -= lim[0]
        else:
            break
    
    if lst1[i][1] > dmg:
        need = math.ceil((lst1[i][1]-dmg)/a)
        ans += need
        dmg += need*a
        que.append([need*a,lst1[i][0]+2*d])

print(ans)