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 long long ll;
struct node{
	int l;
	int r;
	int h;
	int num;
}a[200000];
bool cmp(struct node a1,struct node a2)
{
	return a1.l<a2.l;
}
int main(void)
{
	int n,d,health;
	scanf("%d%d%d",&n,&d,&health);
	int i;
	for(i=0;i<n;i++)
	{
		int x,h;
		scanf("%d%d",&x,&h);
		a[i].l=x-d;
		a[i].r=x+d;
		int c=h/health;
		if(h%health!=0)
		{
			c++;
		}
		a[i].h=c;
		a[i].num=i+1;
	}
	sort(a,a+n,cmp);
	ll sum=0;
//	for(i=0;i<n;i++)
//	{
//		printf("%d %d %d\n",a[i].l,a[i].r,a[i].h);
//	}
	int j;
	for(i=0;i<n;i++)
	{
		if(a[i].h<=0)
		{
			continue;
		}
		else
		{
			for(j=i+1;j<n&&a[j].l<=a[i].r;j=a[j].num)
			{
				a[j].h=a[j].h-a[i].h;
				if(a[j].h<=0)
				{
					int i1;
					for(i1=j;i1<n&&a[i1].h<=0;i1++);
					a[j].num=i1;
				}
			}
			sum+=a[i].h;
		}
	}
	printf("%lld",sum);
}