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>
using namespace std;
using ll = long long;

ll N, D, A;

ll calc(vector<pair<ll, ll>> P) {

  vector<ll> status(N);
  for (int i = 0; i < N; i++) status[i] = 1;

  ll ans = 0;
  ll l = 0;  
  while (true) {
    if (status[N - 1] == 0) break;
    if (status[l] == 0) {
      l++; continue;
    }
    pair<ll, ll> p = P[l];
    ll x = p.first, h = p.second;
    ans += h;
    for (int i = l; i < N; i++) {
      pair<ll, ll> n = P[i];
      ll nx = n.first, nh = n.second;
      if (nx - x > 2 * D) break;

      ll newH = max(nh - h, 0ll);
      P[i] = make_pair(nx, newH);
      if (newH == 0) status[i] = 0;
    }
  }
  return ans;
}

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

  cin >> N >> D >> A;
  vector<pair<ll, ll>> P(N);
  for (int i = 0; i < N; i++) {
    ll X, H;
    cin >> X >> H;
    P[i] = make_pair(X, H);
  }
  sort(P.begin(), P.end());

  ll ans = calc(P);

  for (int i = 0; i < N; i++) {
    pair<ll, ll> p = P[i];
    P[i] = make_pair(-p.first, p.second);
  }
  sort(P.begin(), P.end());

  ans = min(ans, calc(P));

  cout << ans << '\n';
  return 0;
}