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>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;
using lint=long long;

int main(){
	lint n,d,a; scanf("%lld%lld%lld",&n,&d,&a);
	vector<pair<lint,lint>> mon(n);
	rep(i,n){
		lint x,hp; scanf("%lld%lld",&x,&hp);
		mon[i]={x,hp};
	}

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

	lint ans=0,power=0;
	queue<pair<lint,lint>> Q; // rightmost point, damage
	rep(i,n){
		lint x,hp; tie(x,hp)=mon[i];

		while(!Q.empty() && Q.front().first<=x){
			power-=Q.front().second;
			Q.pop();
		}

		if(power>=hp) continue;

		lint cnt=(hp-power+a-1)/a;
		ans+=cnt;
		power+=cnt*a;
		Q.emplace(x+2*d+1,cnt*a);
	}
	printf("%lld\n",ans);

	return 0;
}