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 <stdlib.h>
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define abs(a) ((a) > 0 ? (a) : (-1 * (a)))
#define rep(i, n) for ((i) = 0; (i) < (int)(n); (i)++)
#define rep2(i, s, e) for ((i) = (s); (i) < (int)(e); (i)++)

void arg_marge(int *, int *, int);

int main(void)
{

    int n, d, a, i, j, k, l, *x, *h;
    long long int at = 0;

    scanf("%d %d %d", &n, &d, &a);
    x = (int *)malloc(n * sizeof(int));
    h = (int *)malloc(n * sizeof(int));
    rep(i, n)
    {
        scanf("%d %d", x + i, h + i);
        h[i] = (h[i] + a - 1) / a;
    }
    //arg_marge(x, h, n);

    rep(i, n)
    {
        j = h[i];
        if (j <= 0)
            continue;
        at += j;
        k = 1;
        l = x[i] + d + d;
        while (i + k < n && l >= x[i + k])
            h[i + k++] -= j;
    }
    printf("%lld", at);
}

void arg_marge(int *a, int *b, int n)
{

    if (n == 1)
        return;

    int *l2, *r2, *tmp, *arg_l2, *arg_r2, *arg_tmp;
    l2 = (int *)malloc((n / 2) * sizeof(int));
    r2 = (int *)malloc((n - n / 2) * sizeof(int));
    tmp = (int *)malloc((n) * sizeof(int));
    arg_l2 = (int *)malloc((n / 2) * sizeof(int));
    arg_r2 = (int *)malloc((n - n / 2) * sizeof(int));
    arg_tmp = (int *)malloc((n) * sizeof(int));

    for (int i = 0; i < n / 2; i++)
    {
        l2[i] = a[i];
        arg_l2[i] = b[i];
    }
    for (int i = 0; i < n - n / 2; i++)
    {
        r2[i] = a[i + n / 2];
        arg_r2[i] = b[i + n / 2];
    }

    arg_marge(l2, arg_l2, n / 2);
    arg_marge(r2, arg_r2, n - n / 2);

    for (int i = 0; i < n / 2; i++)
    {
        tmp[i] = l2[i];
        arg_tmp[i] = arg_l2[i];
    }
    for (int i = n / 2; i < n; i++)
    {
        tmp[i] = r2[n - i - 1];
        arg_tmp[i] = arg_r2[n - i - 1];
    }

    int lp = 0, rp = n - 1, p = 0;
    int l3, r3;
    while (p < n)
    {
        l3 = tmp[lp];
        r3 = tmp[rp];
        if (l3 < r3)
        {
            a[p] = l3;
            b[p] = arg_tmp[lp];
            lp++;
            p++;
        }
        else
        {
            a[p] = r3;
            b[p] = arg_tmp[rp];
            rp--;
            p++;
        }
    }

    return;
}