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 <algorithm>
#include <cmath>
#include <iomanip>
#include <iostream>
#include <limits>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <string>
#include <vector>

#define rep(i, x, n) for(int i = x; i < n; i++)
using namespace std;
const long long INF = 1LL << 60;
using ipair = pair<long long, long long>;
bool operator<(const ipair a, const ipair b) {
    if(a.first == b.first) {
        return a.second < b.second;
    } else {
        return a.first > b.first;
    }
}; //演算子
// bool operator>(const ipair a, const ipair b) {
//         return a.second < b.second;
// };
// bool operator>(const ipair a, const ipair b) {
//     if(a.first == b.first) {
//         return a.second > b.second;
//     } else {
//         return a.first > b.first;
//     }
// }; //演算子オーバーロード
int main() {
    long long N, D, A;
    cin >> N >> D >> A;
    vector<pair<long long, long long>> m(N);
    rep(i, 0, N) {
        cin >> m[i].first;
        cin >> m[i].second;
    }
    queue<pair<long long, long long>> que;
    rep(i, 0, N) { que.push(m[i]); }
    // sort(m.begin(),m.end());
    long long count = 0;
    long long tmp = 0;
    long long res = 0;
    while(count < N) {
        pair<long long, long long> p = que.front();
        que.pop();
        //cout << "f:" << p.first << "se:" << p.second << endl;
        if(p.second > 0) {
            if(p.second - tmp > 0) {
                long long kai = (p.second - tmp + A - 1) / A;
                res += kai;
                //cout << "x;" << p.first << "tmp:" << tmp << ",kai:" << kai
                  //   << endl;
                long long d = kai * A;
                tmp += d;
                que.push(make_pair(p.first + D + D, -1 * d));
            }
            count++;
        } else {
            tmp += p.second;
        }
    }
    cout << res << endl;
}