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 <bits/stdc++.h>
using namespace std;
#define int long long int
mt19937 rng(std::chrono::duration_cast<std::chrono::nanoseconds>(chrono::high_resolution_clock::now().time_since_epoch()).count());
#define mp make_pair
#define pb push_back
#define F first
#define S second
const int N=400015;
#define M 1000000007
#define double long double
#define BINF 100000000000000
#define init(arr,val) memset(arr,val,sizeof(arr))
#define MAXN 15000001
#define deb(x) cout << #x << " " << x << "\n";
const int LG = 22;

struct Fenwick {
    int n; vector<int> t;
    Fenwick(int n_): n(n_), t(n_ + 1) {}

    //range update point query
    int query(int i) { int v = 0; for (; i > 0; i -= (i & -i)) v += t[i]; return v; }
    void update(int i, int v) { for (; i <= n; i += (i & -i)) t[i] += v; }
    void range_update(int l, int r, int v) { update(l,v); update(r + 1, -v); }
};

#undef int
int main() {
#define int long long int
ios_base::sync_with_stdio(false); 
cin.tie(0); 
cout.tie(0);

	
    int n;

    cin >> n;
    int d, a;
    cin >> d >> a;

    vector<pair<int, int>>v(n + 1);
    v[0] = {-1, -1};

    for(int i = 1; i <= n; i++)
    	cin >> v[i].F >> v[i].S;
    sort(v.begin(), v.end());

    Fenwick bit(n);
    for(int i = 1; i <= n; i++){
    	bit.range_update(i, i, v[i].S);
    }

    vector<int>pos;
    for(auto i : v)
    	pos.push_back(i.F);
    sort(pos.begin(), pos.end());

    int ans = 0;

    for(int i = 1; i <= n; i++){
    	int cur = bit.query(i);
    	if(cur <= 0) continue;

    	int need = (cur / a) + (cur % a != 0);
    	ans += need;

    	int x = v[i].F;
    	//cout << x + 2 * d << endl;

    	int l = i;
    	int r = lower_bound(pos.begin(), pos.end(), x + 2 * d) - pos.begin();

    	//cout << l << ' ' << r << endl;
    	

    	bit.range_update(l, r, - (need * a));

    }

    cout << ans << endl;
	
	
 
return 0;
}