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: Wrong Answer)

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
pair<ll,ll> a[200010];
int main(){
	ll n,d,m;
	cin>>n>>d>>m;
	for(int i=0;i<n;i++)cin>>a[i].first>>a[i].second;
	sort(a,a+n);
	for(int i=0;i<n;i++){
		if(a[i].second%m==0){
			a[i].second=a[i].second/m;
		}
		else a[i].second=a[i].second/m+1;
	}
	ll ans=0LL;
	ll r=0LL;
	for(ll i=0;i<n;i++){
		if(a[i].second==0LL)continue;
		ll mx=0LL;
//		cout<<i<<endl;
		while(r<n){ 
//			cout<<r<<endl; 
			if(d*2<a[r].first-a[i].first){
				r--;
				break;
			}
			mx=max(mx,a[r].second);
			if(r==n-1){
				ans+=mx;
				cout<<ans<<endl;
				return 0;
			}
			r++;
		}
		if(i==r){
			ans+=a[i].second;
			a[i].second=0;
			r++;
			continue;
		}
		ans+=a[i].second;
		for(ll j=i+1;j<=r;j++){
			a[j].second-=min(a[j].second,a[i].second);
		}
		a[i].second=0LL;
		r++;
	}
	cout<<ans<<endl;
	return 0;
}