Problem:
Score : 200 points

Problem Statement

You are given nonnegative integers a and b (a ≤ b), and a positive integer x.
Among the integers between a and b, inclusive, how many are divisible by x?

Constraints

0 ≤ a ≤ b ≤ 10^{18}

1 ≤ x ≤ 10^{18}

Input

The input is given from Standard Input in the following format:

a b x

Output

Print the number of the integers between a and b, inclusive, that are divisible by x.

Sample Input 1

4 8 2

Sample Output 1

3

There are three integers between 4 and 8, inclusive, that are divisible by 2: 4, 6 and 8.

Sample Input 2

0 5 1

Sample Output 2

6

There are six integers between 0 and 5, inclusive, that are divisible by 1: 0, 1, 2, 3, 4 and 5.

Sample Input 3

9 9 2

Sample Output 3

0

There are no integer between 9 and 9, inclusive, that is divisible by 2.

Sample Input 4

1 1000000000000000000 3

Sample Output 4

333333333333333333

Watch out for integer overflows.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
         
typedef long long ll;
using namespace std;
        
#define F first
#define S second
#define sqr(x) x*x
#define pb push_back
#define mp make_pair
#define sz(x) ((int) (x).size())
#define all(x) (x).begin(), (x).end()
#define allr(x) (x).rbegin(), (x).rend()
#define repit(i, n) for(auto it = n.begin(); it != n.end(); ++it)
       
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0);for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b){if(b==0)return a;else return gcd(b,a%b);}
int lcm(int a,int b){return abs(a*b)/gcd(a,b);}

int main(){ios_base::sync_with_stdio(0);cin.tie(0);
    ll a , b, x;
    cin >> a >> b >> x;
    ll ans = b / x - ((a - 1) / x);
    if (a == 0){
        cout << ans + 1;
    }else{
        cout << ans;
    }
}