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;

typedef long long ll;

const int inf = 1 << 30;

int main() {
    ios_base::sync_with_stdio(false);

    int n, d, a;
    cin >> n >> d >> a;

    vector<pair<int, int> > P;

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

        P.push_back({x, h});
    }

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

    vector<int> prefix(n + 1);
    prefix[0] = (P[0].second + a - 1) / a;

    ll ans = prefix[0];

    for(int i = 1; i < n; i++) {
        auto it = lower_bound(P.begin(), P.end(), make_pair(P[i].first - 2 * d, -inf));
        int l = it - P.begin();

//        cout << "l " << l << endl;
//        cout << "x " << (*it).first << endl;

        ll sum = prefix[i - 1] - (l > 0 ? prefix[l - 1] : 0);

//        cout << "sum " << sum << endl;

        P[i].second = max(0LL, P[i].second - sum * a);

        ll t = (P[i].second + a - 1) / a;

//        cout << "t " << t << endl;

        prefix[i] = prefix[i - 1] + t;

        ans += t;

//        cout << t << " ";
//        cout << endl;
    }

    cout << ans << endl;
}

/**
* [x - d, x + d]
*/