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 <iostream>
#include <string>
#include <map>
#include <vector>
#include <algorithm>
#include <set>
#include <stdio.h>
#include <cmath>
#include <queue>
typedef long long ll;
using namespace std;

int main(void){
    int N;
    ll D, A;
    cin >> N >> D >> A;
    vector<pair<ll, ll> > XH(N);
    for(int i = 0; i < N; i++) {
        cin >> XH[i].first >> XH[i].second;
        XH[i].second = (XH[i].second + A - 1 ) / A;
    }
    
    sort(XH.begin(), XH.end());
  
    
    ll res = 0;
    ll num = 0;
    queue<pair<ll, ll> > que;
    for(int i = 0; i < N; i++){
        while(!que.empty() && que.front().first < XH[i].first) {
            num -= que.front().second;
            que.pop();
        }
        
        XH[i].second -= num;
        if(XH[i].second <= 0) continue;
        res += XH[i].second;
        que.push(make_pair(XH[i].first + 2*D, XH[i].second));
        num+=XH[i].second;
    }

    cout << res << endl;
}