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;
#define rep(i,n) for (int i = 0; i < (int)(n); ++i)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); i++)


int main() {
  int N,D,A;
  cin >> N>>D>>A;
 
 
  vector<vector<int>> x(N,vector<int>(2));
  rep(i,N) {
    rep(j,2){
      cin >> x.at(i).at(j);
    }
  }
  sort(x.begin(),x.end());
  
  long ans=0;
  rep(i,N){
    if(x.at(i).at(1)>0){
      ans=ans+(x.at(i).at(1)-1)/A+1;
      int lb=i-1,ub=N-1;
      while(ub-lb>1){
        int mid=(lb+ub)/2;
        if(x.at(mid).at(0)>x.at(i).at(0)+2*D){
          ub=mid;
        }
        else{
          lb=mid;
        }
      }
      rep2(j,i+1,ub+1){         
        x.at(j).at(1)=max(0,x.at(j).at(1)-((x.at(i).at(1)-1)/A+1)*A);
      }    
    }
  }    
  cout<<ans<<endl;
}