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;
typedef pair<int,int> pii;
int n,d,a,x,h,idx;long long diff,toadd,out=0;
vector<pii> vii;
vector<int>bombs;
vector<long long>sums;
int bsearch(int val)
{
    int b,e,mid,ans=-1;
    b=0,e=bombs.size()-1;

    while(b<=e)
    {
        mid=b+(e-b)/2;

        if(bombs[mid]==val)
        {
            ans=mid;
            return ans;
        }
        else if(bombs[mid]<val)
        {
            b=mid+1;
        }
        else
        {
            ans=mid;
            e=mid-1;
        }
    }
    return ans;
}
int main(){
   cin>>n>>d>>a;
   for(int i=0;i<n;i++)
   {
       cin>>x>>h;
       vii.push_back(make_pair(x,h));
   }
   sort(vii.begin(),vii.end());

   for(int i=0;i<n;i++)
   {
       x= vii[i].first;
       h= vii[i].second;
       idx = bsearch(x-2*d);
       if(idx>=0&&bombs[idx]>=x-2*d)
       {
           diff= sums[sums.size()-1]-(idx==0?0:sums[idx-1]);
           h-=(diff*a);
           if(h>0)
           {
               bombs.push_back(x);
               out += (long long)ceil(h/a);
               sums.push_back(out);
           }
       }
       else
       {
           bombs.push_back(x);
           out += (long long)ceil(h/a);
           sums.push_back(out);
       }
   }
   cout<<out<<endl;
	return 0;
}