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 REP(i, n) for(int i=0; i<(int)(n); i++)
#define REP1(i,a,b) for(int i=a; i<=(int)(b); i++)
#define ALL(x) begin(x),end(x)
#define PB push_back
using namespace std;
typedef int64_t LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
 
LL n, d, a;
vector<pair<LL, LL>> A;
queue<pair<LL, LL>> q;
 
int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> n >> d >> a;
    A.resize(n);
    REP(i, n) {
        cin >> A[i].first >> A[i].second;
    }
    sort(A.begin(), A.end());
    LL ans = 0, cnt=0;
    d = d*2;
    for(auto p:A) {
        LL i=p.first, j=p.second;
        while(!q.empty() && q.front().first<i) {
            cnt-=q.front().second; q.pop();
        }
        j -= cnt;
        if(j<=0) continue;
        LL k = (j+a-1)/a;
        ans += k;
        cnt += k*a;
        q.push({i+d, cnt});
    }
    cout << ans << endl;
    return 0;
}