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 int long long
#define PI pair<int,int>
#define ff first
#define ss second
#define endl '\n'
#define boost ios_base::sync_with_stdio(false);cin.tie(NULL)
//#include "debug.cpp"
const int inf=1e9;
int32_t main()
{
	boost;
	
	int n,d,s;
	cin>>n>>d>>s;
	vector<PI>a(n);
	for(int i=0;i<n;i++){
		cin>>a[i].ff>>a[i].ss;
	}

	sort(a.begin(),a.end());
	int ans=0;
	int dp[n+1]={0};
	for(int i=0;i<n;i++){
		if(i>0)
			dp[i]+=dp[i-1];
		a[i].ss-=dp[i]*s;
		//cout<<a[i].ss<<" "<<dp[i]<<endl;
		if(a[i].ss<=0)
			continue;
		int c=a[i].ss/s;
		if(a[i].ss%s!=0)
			c++;
		dp[i]+=c;
		int l=i,r=n-1,aa=i;
		while(l<=r){
			int m=(l+r)/2;
			if(a[m].ss<=a[i].ff+2*d){
				l=m+1;
				aa=m;
			}
			else
				r=m-1;
		}
		//cout<<aa<<endl;
		dp[aa+1]-=c;
		ans+=c;
	}
	cout<<ans<<endl;
}