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>
# define rep(i, n) for(int i=0, i##_len=(n); i<i##_len; ++i)
# define reps(i, n) for(int i=1, i##_len=(n); i<=i##_len; ++i)
# define rrep(i, n) for(int i=((int)(n)-1); i>=0; --i)
# define rreps(i, n) for(int i=((int)(n)); i>0; --i)
# define ALL(x) (x).begin(), (x).end()
# define SZ(x) ((int)(x).size())
# define pb push_back
# define optimize_cin() cin.tie(0); ios::sync_with_stdio(false)
using namespace std;
using lint = long long;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

int main()
{
    lint a, b, x;
    cin >> a >> b >> x;

    lint ans = (b - a) / x; 
    if (!(a % x == 0 && b % x == 0))
    {
    }
    else
    {
        ans++;
    }

    cout << ans << "\n";

    return 0;
}