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: Time Limit Exceeded)

#include <bits/stdc++.h>
#include <math.h>
#define rep(i, n) for(int64_t i = 0; i < n; ++i)
using namespace std;

int main() {
  int64_t n, d, a, s = 0;
  cin >> n >> d >> a;
  vector<pair<int64_t, int64_t>> v(n);
  rep(i, n) cin >> v[i].first >> v[i].second;
  rep(i, n) v[i].second = (v[i].second - 1) / a + 1;
  sort(v.begin(), v.end());
  if(v[n-1].first - v[0].first <= 2 * d){
    rep(i, n) s = max(s, v[i].second);
  }
  else if(d == 0){
    rep(i, n) s += v[i].second;
  }
  else{
    int64_t tmp = v[0].first + 2*d, c = v[0].second, k = 0, ma = 0;
    while(tmp < v[n-1].first){
      bool b = 1;
      int64_t j;
      rep(i, n-k){
        if(v[i+k].first > tmp) break;
        else{
          v[i+k].second -= c;
          if(v[i+k].second > 0 && b){
            j = i+k;
            b = 0;
          }
        }
      }
      k = j;
      s += c;
      tmp = v[k].first + 2*d;
      c = v[k].second;
    }
    rep(i, n-k){
      ma = max(ma, v[i+k].second);
    }
    s += ma;
  }
  cout << s << endl;
}