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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

using LL = long long;

int main() {

	ios_base :: sync_with_stdio(0); cin.tie(NULL);
	
	int n, d, a;
	cin >> n >> d >> a;	//distance, atack

	vector < pair<int,int> > v;

	for(int i = 0, x, h; i < n; ++i) {
		cin >> x >> h;
		v.push_back({x, h});
	}

	sort(v.begin(), v.end());

	LL delta = 0, ans = 0;

	queue< pair<LL,LL> > q;

	for(int i = 0; i < n; ++i) {

		int x = v[i].first;
		int h = v[i].second;

		while(!q.empty() && (LL)x > q.front().first) {
			delta -= q.front().second;
			q.pop(); 
		}

		LL c = h / a + (h % a != 0) - delta; // number of bombs [x, x + 2*d - 1]
		
		if(c > 0) {
			q.push({x + 2LL * d, (LL)c});
			ans += c;
			delta += c;
		}

		
	}

	cout << ans << '\n';


	return 0;
}