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 <bits/stdc++.h>
#define rep(i,n) for ( int (i) = 0; (i) < (n); (i)++ )
using namespace std;
using ll = long long int;
const long long int INF = 1001001001001001LL;
const int inf = 1000000007;
long long int MOD = 1000000007LL;
double PI = 3.1415926535897932;

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

    ll n, d, a;
    cin >> n >> d >> a;
    vector<pair<ll, ll>> p(n);
    rep(i, n) cin >> p[i].first >> p[i].second;
    sort( p.begin(), p.end() );

    vector<ll> imos(n+1, 0);
    ll cnt = 0;
    rep(i, n) {
        if ( i > 0 ) imos[i] += imos[i-1];
        ll h = p[i].second - imos[i];

        if ( h > 0 )
        {
            ll c = (h+a-1) / a;
            ll damage = c*a;

            ll out = upper_bound( p.begin()+i, p.end(), make_pair( p[i].first+2*d, INF ) ) - p.begin();
            
            imos[i] += damage;
            imos[out] -= damage;
            cnt += c;
        }
    }

    cout << cnt << endl;
    
    return 0;
}