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 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<limits.h>

#define rep(i,begin,end) for(int i=begin; i<end; i++)
#define revrep(i,begin,end) for(int i=begin; i>end; i--)
#define lld long long int

typedef struct node{
    lld x;
    lld h;
    lld imo;
}node;

int cmpAsc(const void* a, const void* b){
    lld x = ((node*)a)->x, y = ((node*)b)->x;
    if(x > y){
        return 1;
    }
    if(x < y){
        return -1;
    }
    return 0;
}

int main(){
    int n;
    lld d;
    lld a;
    scanf("%d %lld %lld", &n, &d, &a);
    d = 2*d;
    node m[n];
    rep(i, 0, n){
        lld x;
        lld h;
        scanf("%lld %lld", &x, &h);
        m[i].x = x;
        m[i].h = h;
        m[i].imo = 0;
    }
    qsort(m, n, sizeof(node), cmpAsc);
    int l, r, p;
    lld ans = 0;
    ans += (lld)(m[0].h + a - 1) / a;
    m[0].imo += (lld)(m[0].h + a - 1) / a * a;
    l = 0; r = n-1; p = (l+r+1)/2;
    while(l < r){
        p = (l + r +1)/2;
        if(m[p].x <= m[0].x + d){
            l = p;
        }else{
            r = p-1;
        }
    }
    p = (l + r +1)/2;
    if(p < n-1){
        m[p+1].imo -= (lld)(m[0].h + a - 1) / a * a;
    }

    /*printf("\n");
    rep(i, 0, n){
        printf("%lld ", m[i].imo);
    }
    printf("\n");*/

    rep(i, 1, n){
        m[i].imo += m[i-1].imo;
        if(m[i].h <= m[i].imo){
            continue;
        }else{
            //printf("--%d--\n", i);
            ans += (lld)(m[i].h - m[i].imo + a - 1) / a;
            m[i].imo += (lld)(m[i].h - m[i].imo + a - 1) / a * a;
            l = 0; r = n-1; p = (l+r+1)/2;
            while(l < r){
                p = (l + r +1)/2;
                if(m[p].x <= m[i].x + d){
                    l = p;
                }else{
                    r = p-1;
                }
            }
            p = (l + r +1)/2;
            if(p < n-1){
                m[p+1].imo -= (lld)(m[0].h + a - 1) / a * a;
            }
        }
    }
    printf("%lld", ans);

    /*printf("\n");
    rep(i, 0, n){
        printf("%lld ", m[i].imo);
    }*/

    return 0;
}