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;
using ll = long long;
template<class T,class U> using P = pair<T,U>;
template<class T> using vec = vector<T>;
template<class T> using vvec = vector<vec<T>>;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll N,D,A;
    cin >> N >> D >> A;
    vec<ll> v;
    vec<ll> X(N),H(N);
    map<ll,ll> mp;
    vec<int> idx(N);
    iota(idx.begin(),idx.end(),0);
    for(int i=0;i<N;i++){
        cin >> X[i] >> H[i];
        mp[X[i]] += H[i];
    }

    set<P<ll,ll>> s;
    ll ans = 0,now = 0;
    for(auto p:mp){
        ll x = p.first,h = p.second;
        if(now<h){
            ll c = (h-now+A-1)/A;
            ans += c;
            s.insert({x+2*D,c*A});
            now += c*A;
        }
        set<P<ll,ll>>::iterator it = s.begin();
        while(it!=s.end()){
            if((*it).first>x) break;
            now -= (*it).second;
            it = s.erase(it);
        }
    }
    cout << ans << endl;
}