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 ll long long
#define all(c) (c).begin(),(c).end()
#define rall(c) (c).rbegin(),(c).rend()
using namespace std;
const int tINF = (int)INT_MAX;
const long long lINF = (long long)LLONG_MAX;

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	#ifdef DEBUG
			assert(freopen("inp", "r", stdin));
			assert(freopen("out", "w",stdout));
	#else
			#define cerr if(false)cerr
	#endif
    int n;
    ll d,a;
    cin >> n >> d >> a;
    vector <pair<ll,ll>> ar(n);
    for(auto &i : ar)
        cin >> i.first >> i.second;
    sort(all(ar));
    ll pos;
    ll ans = 0;
    int ni = 0;
    for(int i = 0; i < n;){
        pos = ar[i].first + 2*d; 
        bool f = 1;
        ll cntBombs = ceil((long double)ar[i].second / a);
        ans += cntBombs;
        for(int j = i + 1; j < n; ++j){
            if(ar[j].first > pos){
                ni = j;
                f = 0;
                break;
            }
            ll curBombs = ceil((long double)ar[j].second / a);
            if(curBombs <= cntBombs)
                continue;
            else{
                ar[j].second -= cntBombs*a;
                f = 0;
                ni = j;
                break;
            }
        }
        if(f)
            break;
        i = ni;
    }
    ll ans2 = ans;
    ans = 0;
    ni = 0;
    reverse(all(ar));
    for(int i = 0; i < n;){
        pos = ar[i].first - 2*d; 
        bool f = 1;
        ll cntBombs = ceil((long double)ar[i].second / a);
        ans += cntBombs;
        for(int j = i + 1; j < n; ++j){
            if(ar[j].first < pos){
                ni = j;
                f = 0;
                break;
            }
            ll curBombs = ceil((long double)ar[j].second / a);
            if(curBombs <= cntBombs)
                continue;
            else{
                ar[j].second -= cntBombs*a;
                f = 0;
                ni = j;
                break;
            }
        }
        if(f)
            break;
        i = ni;
    }
    ans2 = min(ans,ans2);
    cout << ans2 << '\n';
   cerr << "Time execute: " << (double)clock() / (double)CLOCKS_PER_SEC 
    << " sec" << endl;
}