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 <string>
#include <vector>
#include <algorithm>
using namespace std;
int main(){
    int N, D, A; cin >> N >> D >> A;
    vector<pair<int, int>> P(N);
    for(int i = 0; i < N; i++){
        int x, h; cin >> x >> h;
        P[i] = make_pair(x, h);
    }
    sort(P.begin(), P.end());
    int idx = 0;
    long long ans = 0;
    while(1){
        int sx = P[idx].first;
        int sh = P[idx].first;
        int Dx = sx + 2*D;
        //cout << "Dx : " << Dx << endl;
        //cout << "A : " << A << endl;
        bool flag = true;
        int id;
        ans++;
        for(int i = idx; i < N; i++){
            if(i == N-1){
                if(flag){
                    cout << ans << endl;
                    return 0;
                }
            }
            if(P[i].first > Dx) { if(flag)id = i; break; }
            P[i].second = P[i].second - A;
            if(flag)if(P[i].second > 0) { idx = i; flag = false;} //倒せないのがあったらそこをidxにする

        }
        if(flag) idx = id;
        //cout << "idx : " << idx << endl;
        for(int i = 0; i < N; i++){
            //cout << P[i].first << " " << P[i].second << endl;
        }
    }
}