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>
#define INF 1e9
using namespace std;

#define REPR(i,n) for(int i=(n); i >= 0; --i)
#define FOR(i, m, n) for(int i = (m); i < (n); ++i)
#define REP(i, n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define ALL(a)  (a).begin(),(a).end()

template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return true; } return false; }
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return true; } return false; }
int gcd(int a,int b){return b?gcd(b,a%b):a;}
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}


int main() {
    ll N,D,A;
    cin >> N >> D >> A;
    // pos,life
    vector<pair<ll,ll>> monster(N);
    // モンスターの座標表(二分探索用)
    vector<ll> poss(N);
    REP(i,N) {
        int x,h;
        cin >> x >> h;
        monster[i] = make_pair(x,h);
        poss[i] = x;
    }

    // 座標ソート
    sort(ALL(monster));
    sort(ALL(poss));

    if (N==1) {
        cout << ceil((double)monster[0].second/A) << endl;
        return 0;
    }

    // N体目で重なりが終わる線のかず それを含む
    vector<ll> end(N,0);

    ll ans = 0;
    ll kasanari = 0;
    REP(i,N) {

        // endに入っていた場合重なりを解除
        if (end[i] != 0) {
            kasanari -= end[i];
            end[i] = 0;
        }

        // ダメージ足りないなら爆弾の追加
        if ( kasanari * A < monster[i].second ) {
            ll need = ceil((double)(monster[i].second - (kasanari * A) )/ A );
            kasanari += need;
            ans += need;
            // 二分探索でどこまで影響があるかを調べる。
            ll key = poss[i] + (2 * D);
            auto itr = upper_bound(ALL(poss),key);
            ll index;
            // 最後まで影響ある場合は無視で良い
            if (itr == poss.end()) {
                continue;
            }
            end[ itr - poss.begin()] += need;

            // cout << index << endl;
        }


        if (kasanari < 0) {
            cout << "Error " << endl;
            exit(1);
        }
    }

    cout << ans << endl;

    return 0;
}