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;

#define fr(i,n) for(int i=0; i<n; i++)
#define rep(i, st, en) for(int i=st; i<=en; i++)
#define repn(i, st, en) for(int i=st; i>=en; i--)
#define sq(a) (a*a)
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
ll mod = 1e9+7;

ll gcdll(ll a, ll b){
    if(!b) return a;
    return gcdll(b, a%b);
}

int gcd(int a, int b){
    if(!b) return a;
    return  gcd(b, a%b);
}

void seive(int n){
    vector<bool> is_prime(n+1, true);
    is_prime[0] = is_prime[1] = false;

    for(int i=2; i*i<=n; i++){
        if(is_prime[i]){
            for(int j=i*i; j<=n; j+=i){
                is_prime[j] = false;
            }
        }
    }
}

vector<int> constructZFunc(string s){
    int n = s.size();
    vector<int> z(n);
    int l, r;
    l = r = 0;
    z[0] = 0;
    rep(i, 1, n-1){
        z[i] = 0;
        if(i<=r){
            z[i] = min(r-i+1, z[i-l]);  /// initialization, req. for linear time approach.
        }
        while(i+z[i] < n && s[z[i]] == s[i+z[i]])
            ++z[i];
        if(i+z[i]-1 > r){
            r = i+z[i]-1;
            l = i;
        }
    }
    return z;
}

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

    ll n, d, a, h, x;
    cin>>n>>d>>a;
    vector<pll> v;
    fr(i, n){
        cin>>x>>h;
        v.push_back({x, h});
    }
    sort(v.begin(), v.end());

    ll pref[n+1] = {0ll};
    ll cnt = 0ll;
    vector<ll> s;
    ll bombs = 0ll;
    ll i = 1;
    ll tot = 0ll;
    ll temp = 0ll;
    for(auto it: v){
        auto lb = lower_bound(s.begin(), s.end(), it.first) - s.begin();
        if(lb == v.size()){
            temp = 0ll;
        } else {
            temp = tot - pref[lb - 1 + 1];
        }
        it.second = max(0ll, it.second - temp);
        bombs = (it.second+a-1)/a;
        s.push_back(it.first + 2ll*d);
        pref[i] = pref[i-1ll] + (bombs*a);
        tot = pref[i];
        cnt += bombs;
        i++;
    }

    cout<<cnt;

    return 0;
}