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 <iostream>
#include <vector>
#include <algorithm>
#include <functional>
#include <cmath>
#include <map>
#include <queue>

#define reps(i,s,n) for(int (i) = (s); (i) < (n); (i)++)
#define rep(i,n) reps(i,0,n)
using namespace std;
using ll = long long;
using pll = pair<ll,ll>;

bool operator<(pll &a, pll &b){
    if(a.second > b.second){
        return true;
    }
    return false;
}

int main(){
    int n,d,a;
    vector<pll> vec;
    cin >> n >> d >> a;
    int x,h;
    rep(i,n){
        cin >> x >> h;
        vec.push_back(make_pair(x,h));
    }
    queue<pll> qu;

    sort(vec.begin(),vec.end());
    
    ll ans = 0;
    ll pos = 0;
    ll dmg = 0;

    rep(i,n){
        while(pos < vec[i].first && qu.size() > 0){
            dmg -= qu.front().second;
            pos = qu.front().first;
            qu.pop();
        }

        if(dmg < vec[i].second){
            pos = vec[i].first;
            int tmp2 = (vec[i].second-dmg);
            tmp2 = (tmp2 + a - 1)/a;
            ans += tmp2;
            dmg = tmp2*a;
            qu.push(make_pair(pos+2*d,tmp2*a) );
            pos = pos + 2*d; 
        }
    }
    
    cout << ans << endl;
    return 0;
}