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<iostream>
#include<vector>
#include<algorithm>
#include<cmath>
using lli = long long int;
using pii = std::pair<lli, lli>;

int main(){
  lli n, d, a;
  std::cin >> n >> d >> a;
  std::vector<pii> xh(n);
  for (int i = 0; i < n; i++) {
    std::cin >> xh[i].first >> xh[i].second;
  }
  std::sort(xh.begin(), xh.end());
  // for (int i = 0; i < n; i++) {
  //   std::cout << xh[i].first << " " << xh[i].second << std::endl;
  // }
  lli ans = 0, pos = 0;
  lli sq[n];
  lli sum[n];
  for (int i = 0; i < n; i++) sq[i] = sum[i] = 0;
  int div = std::sqrt(n) + 1;
  // std::cout << "div:" << div << std::endl;
  for (int i = 0; i < n; i++) {
    while(i < n and xh[i].second - (sum[i] + sq[i/div])<= 0)i++;
    if(i == n)break;
    lli hp = (xh[i].second - (sum[i] + sq[i/div]));
    lli times = hp/a;
    // std::cout << times << std::endl;
    // std::cout << "pos:" << pos << " i:" << i << std::endl;
    if(hp % a != 0)times++;
    lli damege = a*times;
    while(pos < n and xh[pos].first <= xh[i].first + 2*d)pos++;
    for (int j = i; j % div != div - 1 and j < pos; j++) {
      sum[j] += damege;
    }
    for (int j = i/div; j*div < pos; j++) sq[j] += damege;
    for (int j = pos/div * div; pos/div * div > i/div * div and j < pos; j++) {
      // std::cout << "hoge" << std::endl;
      sum[j] += damege;
    }
    ans += times;
  }
  std::cout << ans << std::endl;
  return 0;
}