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 <fstream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <bitset>
#include <algorithm>
#include <complex>
#include <array>
#include <iomanip>
using namespace std;

#define REP(i,n) for(int i=0; i<n; ++i)
#define FOR(i,a,b) for(int i=a; i<=b; ++i)
#define FORR(i,a,b) for (int i=a; i>=b; --i)
#define ALL(c) (c).begin(), (c).end()

typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<long double> VD;
typedef vector<VI> VVI;
typedef vector<VL> VVL;
typedef vector<VD> VVD;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;

template<typename T> void chmin(T &a, T b) { if (a > b) a = b; }
template<typename T> void chmax(T &a, T b) { if (a < b) a = b; }

int in() { int x; scanf("%d", &x); return x; }
ll lin() { ll x; scanf("%lld", &x); return x; }
#define INF 1LL<<60

struct enemy {
    ll x;
    ll h;

    bool operator<(const enemy& other) const {
        if (x < other.x) return true;
        else return false;
    }
};

bool cmp(const enemy & a, const enemy & b) {
    return a.x > b.x; 
}

int main() {
    int N;
    ll D, A;
    cin >> N >> D >> A;
    vector<enemy> E(N);
    map<ll, int> cord_idx;
    REP(i, N) {
        cin >> E[i].x >> E[i].h;
        cord_idx[E[i].x] = i;
    }

    sort(ALL(E));

    vector<int> correspond_right_side(N);
    REP(i, N) {
        enemy tmp = {E[i].x+2*D, 0};
        auto itr = lower_bound(E.rbegin(), E.rend(), tmp, cmp);
        correspond_right_side[i] = (*itr).x;
        // cerr << "i=" << i << " right_size=" << (*itr).x << endl;
    }

    ll ans = 0;
    int head = 0;
    vector<ll> damages(N);
    while(head < N) {
        if(E[head].h <= damages[head]) {
            if(head+1<N && E[head+1].x <= E[head].x+2*D)
                damages[head+1] = damages[head];
            head++;
            continue;
        }
        ll num = ceil(double(E[head].h - damages[head]) / A);
        damages[head] += A * num;
        if(head+1<N && E[head+1].x <= E[head].x+2*D)
            damages[head+1] += A * num;
        ans += num;
        head++;
    }

    // REP(i, N) {
    //     cerr << "damages[" << i << "] = " << damages[i] << endl;
    // }

    cout << ans << endl;

    return 0;
}