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>
#define int long long
using namespace std;
using ll = long long;

    
signed main() {
    int n, d, a;
    cin >> n >> d >> a;
    vector<pair<int, int>> v(n);
    for(auto& i: v) {
        cin >> i.first >> i.second;
    }
    v.push_back({LLONG_MAX >> 2, LLONG_MAX >> 2});
    sort(begin(v), end(v));
    int i = 0, j = 0;
    ll ans = 0;
    while(j <= n) {
        if(v[j].first - v[i].first <= 2 * d) {
            ++j;
        } else {
            int req = (v[i].second + a - 1) / a;
            ans += req;
            // cout << req << " ";
            for(int k = i; k < j; ++k) {
                // cout << k << " ";
                v[k].second -= req * a;
                v[k].second = max(0LL, v[k].second);
            }
            // cout << "\n";
            while(v[i].second == 0) ++i;
        }
    }
    cout << ans << "\n";
    return 0;
}