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 <iostream>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
long long n,d,a;
long long x,h;
vector<pair<long long,long long>> v;
queue<pair<long long,long long>> l;
int main()
{
    ios_base::sync_with_stdio(false);
    cin>>n>>d>>a;
    for(int i=1;i<=n;i++)
    {
        cin>>x>>h;
        v.push_back({x,h});
    }
    sort(v.begin(),v.end());
    long long ans=0;
    long long dif=0;
    for(int i=0;i<n;i++)
    {
        while(!l.empty() && v[i].first>l.front().first)
        {
            dif-=l.front().second;
            l.pop();
        }
        if(v[i].second-dif<=0) continue;
		int aux=v[i].second-dif;
        ans+=(aux+a-1)/a;
        dif+=((aux+a-1)/a)*a;
        l.push({v[i].first+d+d,((aux+a-1)/a)*a});
    }
    cout<<ans;
    return 0;
}