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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define int long long
typedef pair<int,int> P;
int INF = 1e9+7;
int mod = 1e9+7;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
signed main() {
    int N,D,A;
    cin >> N >> D >> A;
    vector<P>cnt(N);
    for(int i = 0; i < N; i++) {
        cin >> cnt[i].first >> cnt[i].second;
    }
    sort(cnt.begin(),cnt.end());
    vector<int>sum(N+2);
    int ans = 0;
    for(int i = 0; i < N; i++) {
        sum[i+1]+=sum[i];
        cnt[i].second-=sum[i+1];
        if(cnt[i].second <= 0) {
            continue;
        }
        ans += (cnt[i].second+A-1)/A;
        sum[i+1] += (cnt[i].second+A-1)/A*A;
        sum[upper_bound(cnt.begin(),cnt.end(),make_pair(cnt[i].first+D*2,INF))-cnt.begin()+1]-=(cnt[i].second+A-1)/A*A;
    }
    cout << ans << endl;
}