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;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
#define rep(i, n) for(int i = 0; i < n; i++)
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define fi first
#define se second

void solve() {
    int n, d, a;
    cin >> n >> d >> a;
    vector<pll> X(n);
    rep(i, n) {
        ll x, h;
        cin >> x >> h;
        X[i] = {x, h};
    }
    sort(all(X));
    queue<pll> que;
    ll acc = 0;
    ll ans = 0;
    rep(i, n) {
        while(que.size() && X[i].fi > que.front().fi) {
            acc -= que.front().se;
            que.pop();
        }
        X[i].se -= acc * a;
        ll damage = max(0LL, (X[i].se + a - 1) / a);
        acc += damage;
        ans += damage;
        que.push({X[i].fi + 2 * d, damage});
    }
    cout << ans << endl;
    return ;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    solve();
    return 0;
}