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 ll;

int main(void){
    ll N,D,A;
    cin >> N >> D >> A;
    vector<pair<ll,ll>> XH(N);
    for(ll i = 0; i < N; i++){
        cin >> XH.at(i).first >> XH.at(i).second;
        XH.at(i).second = 1 + (XH.at(i).second -1) / A;
    }
    sort(XH.begin(), XH.end());
    
    ll zero = 0;
    vector<ll> range(N); //iを左端にしたとき、range[i]までダメージを受ける。
    for(ll i = 0; i < N; i++){
        range.at(i) = lower_bound(XH.begin(), XH.end(), make_pair(XH.at(i).first + 2*D + 1, zero)) - XH.begin();
        range.at(i)--;
    }
    vector<ll> leftrange(N); //iを右端にしたとき、leftrange[i]+1までダメージを受ける。
    for(ll i = 0; i < N; i++){
        leftrange.at(i) = lower_bound(XH.begin(), XH.end(), make_pair(XH.at(i).first - 2*D, zero)) - XH.begin();
        leftrange.at(i) = leftrange.at(i) - 1;
    }

    ll tmp = 0;
    vector<ll> bomb(N+1,0); //i+1を左端にして爆発させる爆弾の数の累積和
    for(ll i = 1; i < N+1; i++){
        bomb.at(i) = bomb.at(i-1);
        if(XH.at(i-1).second <= bomb.at(i) - bomb.at(leftrange.at(i-1) + 1)){
            continue;
        }
        else{
            tmp = XH.at(i-1).second - bomb.at(i) + bomb.at(leftrange.at(i-1) + 1);
            bomb.at(i) += tmp;
        }
    }

    
    printf("%lld\n", bomb.at(N));
}