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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < int(n); ++i)
#define rep1(i, n) for (int i = 1; i < int(n); ++i)
#define repx(i, x, n) for (int i = int(x); i < int(n); ++i)
#define rrep(i, n) for (int i = int(n) - 1; i >= 0; --i)
#define ALL(n) begin(n), end(n)
#define IN(a, x, b) ((a) <= (x) && (x) < (b))
#define OUT(a, x, b) ((x) < (a) || (x) < (b))
typedef long long ll;
typedef long double ld;
const ll INF = 1e18;
const ll MOD = 1000000007;

ll a, b, x;

ll f(ll n)
{
    if (n == -1)
    {
        return 0;
    }
    else
    {
        return ll(n / x) + 1;
    }
}

int main()
{
    int a_flg = 0, b_flg = 0;
    cin >> a >> b >> x;
    cout << f(b) - f(a - 1) << endl;
    return 0;
}