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>
#include<string.h>

#define ll long long
#define INF 1<<31-1
#define LNF 9223372036854775807
#define PI 3.14159265358979
#define MIN(x,y) ((x)<(y)?(x):(y))
#define MAX(x,y) ((x)<(y)?(y):(x))
#define FOR(i,a,n) for(i=a;i<n;i++)
#define MOD 1000000007
//#define MOD 998244353
#define NUM 200010
 
void swap(int* a,int* b){int tmp=*a;*a=*b;*b=tmp;}
int abs(int a){if(a<0)return -a;return a;}
int dup(int a,int d){if(a%d==0)return a/d;else return a/d+1;}
int mod(int a,int m){if(a<0)return a%m+m;else return a%m;}

void swapl(ll* a,ll* b){ll tmp=*a;*a=*b;*b=tmp;}
ll absl(ll a){if(a<0)return -a;return a;}
ll dupl(ll a,ll d){if(a%d==0)return a/d;else return a/d+1;}
ll modl(ll a,ll m){if(a<0)return a%m+m;else return a%m;}

ll GCD(ll a,ll b){ll c=1,tmp=MAX(a,b);b=MIN(a,b);a=tmp;while(c!=0){c=a%b;a=b;b=c;}return a;}
ll LCM(ll a,ll b){return a*b/GCD(a,b);}

ll factorial(ll n, ll m) {return n < m ? 1 : n*factorial(n-1, m);}

int upll(const void*a, const void*b){return*(ll*)a<*(ll*)b?-1:*(ll*)a>*(ll*)b?1:0;}
int downll(const void*a, const void*b){return*(ll*)a<*(ll*)b?1:*(ll*)a>*(ll*)b?-1:0;}
void sortup(ll*a,int n){qsort(a,n,sizeof(ll),upll);}
void sortdown(ll*a,int n){qsort(a,n,sizeof(ll),downll);}

typedef struct
{
    ll X;
    ll H;
} XF;

int xf_upll(const void*a, const void*b){return((XF*)a)->X<((XF*)b)->X?-1:((XF*)a)->X>((XF*)b)->X?1:0;}

ll bearchm(ll x,XF a[],ll n){
    ll st=0,fi=n-1,te;
    while(st<=fi){
        te=(st+fi)/2;
        if(a[te].X<x){st=te+1;}else{fi=te-1;}
    }
    return st;
}

int main(){
  ll N, D, A;
  ll sum=0, w, m;
  XF list[200000];
  ll k[200001]={0};
  

  scanf("%lld%lld%lld", &N, &D, &A);

  for (int i = 0; i < N; i++)
  {
      scanf("%lld%lld", &(list+i)->X,&(list+i)->H);
  }

  list[N].X = list[N-1].X + 1;

  qsort(list, N, sizeof(XF), xf_upll);

    for (int i = 0; i < N; i++)
    {
        w -= k[i];
        m = (list[i].H - w + A -1)/A;
        m = m < 0 ? 0 : m;
        w += m*A;
        sum += m;
        
        k[bearchm(list[i].X + 2*D + 1 , list, N+1)] = m*A;
    }
    printf("%lld", sum);

  return 0;
}