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 <map>
#include <set>
#include <math.h>
#include <queue>
using namespace std;

int main() {
    long long N, D, A, X, H;
    cin >> N >> D >> A;
    vector<pair<long long, long long> > vec;
    vec.reserve(N);
    for (long long i = 0;i < N;i ++) {
        cin >> X >> H;
        vec.push_back(pair<long long, long long>(X, H));
    }
    sort(vec.begin(), vec.end());
    long long dist = D * 2 + 1;
    long long ret = 0;
    long long total_hit(0), damege(0), end(0);
    queue<pair<long long, long long> > q;
    for (long long i = 0;i < N;i ++) {
        while(!q.empty() && vec[i].first >= q.front().first) {
            total_hit -= q.front().second;
            q.pop();
        }
        long long hp = vec[i].second - total_hit * A;
        if (hp < 1) {
            continue;
        }
        long long need = (hp + A - 1) / A;
        ret += need;
        q.push(pair<long long, long long>(vec[i].first + dist, need));
        total_hit += need;
    }
    cout << ret << endl;

    return 0;
}