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 <iostream>
#include <algorithm>
#include <vector>

using namespace std;


int shori(int N,int D,int A,vector<vector<int>> M) {
	int length = 2 * D,count,num=0,kaisuu;
	sort(M.begin(), M.end());
	while (num<N) {
		while (M.at(num).at(1) > 0) {
			kaisuu = M.at(num).at(1) / A + !!(M.at(num).at(1) % A);
			//cout << kaisuu << endl;
			count += kaisuu;
			for (int i = num; i < N; i++) {
				if (M.at(i).at(0) > M.at(num).at(0) + length)
					break;
				for (int j = 0; j < kaisuu && M.at(i).at(1)>0; j++)
					M.at(i).at(1) -= A;//?
			}
		}
		num++;
	}
	//cout << length<<num << endl;
	return(count);
}

int main() {
	int N, D, A,ans;
	cin >> N >> D >> A;
	vector<vector<int>> M(N,vector<int>(2));//0:X,1:HP
	for (int i = 0; i < N; i++) {
		cin >> M.at(i).at(0) >> M.at(i).at(1);
	}
	ans=shori(N,D,A,M);
	cout << ans << endl;
	return 0;
}