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>

using namespace std;

typedef long long int ll;

struct LazySegmentTree {
private:
    int n;
    vector<ll> node, lazy;

public:
    LazySegmentTree(vector<ll> v) {
        int sz = (int)v.size();
        n = 1; while(n < sz) n *= 2;
        node.resize(2*n-1);
        lazy.resize(2*n-1, 0);

        for(int i=0; i<sz; i++) node[i+n-1] = v[i];
        for(int i=n-2; i>=0; i--) node[i] = node[i*2+1] + node[i*2+2];
    }

    void eval(int k, int l, int r) {
        if(lazy[k] != 0) {
            node[k] += lazy[k];
            if(r - l > 1) {
                lazy[2*k+1] += lazy[k] / 2;
                lazy[2*k+2] += lazy[k] / 2;
            }
            lazy[k] = 0;
        }
    }

    void add(int a, int b, ll x, int k=0, int l=0, int r=-1) {
        if(r < 0) r = n;
        eval(k, l, r);
        if(b <= l || r <= a) return;
        if(a <= l && r <= b) {
            lazy[k] += (r - l) * x;
            eval(k, l, r);
        }
        else {
            add(a, b, x, 2*k+1, l, (l+r)/2);
            add(a, b, x, 2*k+2, (l+r)/2, r);
            node[k] = node[2*k+1] + node[2*k+2];
        }
    }

    ll getsum(int a, int b, int k=0, int l=0, int r=-1) {
        if(r < 0) r = n;
        eval(k, l, r);
        if(b <= l || r <= a) return 0;
        if(a <= l && r <= b) return node[k];
        ll vl = getsum(a, b, 2*k+1, l, (l+r)/2);
        ll vr = getsum(a, b, 2*k+2, (l+r)/2, r);
        return vl + vr;
    }
};

int main(){
    int64_t n, d, a;
    cin >> n >> d >> a;
    vector<array<int64_t, 2>> X(n);
    for (int i = 0; i < n; i++){
        cin >> X[i][0] >> X[i][1];
    }
    sort(X.begin(), X.end());
    LazySegmentTree seg(vector<ll>(n, 0));

    for (int i = 0; i < n; i++){
        seg.add(i, i + 1, X[i][1]);
    }

    int64_t ans = 0;
    int64_t r = 0;
    for (int i = 0; i < n; i++){
        if (seg.getsum(i, i + 1) <= 0)
            continue;
        while (r < n and X[r][0] <= X[i][0] + 2 * d)
            r++;
        int64_t m = (seg.getsum(i, i + 1) + a - 1) / a;
        seg.add(i, r, -a * m);
        ans += m;
    }

    cout << ans << endl;
}