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>

#define x first
#define h second

using namespace std;

const int N = (int)2e5 + 5;

int n,d,a;

long long ans,curDam,b[N];

pair<int,int> m[N];

int main(){
	ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    cin >> n >> d >> a;
    for(int i = 1; i <= n; i++){
    	cin >> m[i].x >> m[i].h;
    }
    sort(m + 1,m + 1 + n);
    for(int i = 1; i <= n; i++){
    	curDam -= b[i];
    	m[i].h -= curDam;
    	int need = (m[i].h + a - 1) / a;
    	ans += need;
    	int R = m[i].x + 2 * d,l = i,r = n,curR = i;
    	while(l <= r){
    		int mid = r + l >> 1;
    		if(m[mid].x <= R){
    			curR = mid;
    			l = mid + 1;
    		}	
    		else r = mid - 1;
    	}	
    	curDam += need * a;
    	b[curR + 1] += need * a;
	}
    cout << ans;
 	return 0;  
}