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 <cstdio>
#include <vector>
#include <algorithm>
#include <iostream>
#include <cstring>
using namespace std ;

typedef long long ll ;
#define rep (i , n ) for ( int i =0; i < n ; i++)
#define _sort(arg) sort(begin(arg), end(arg))
#define MOD 1000000007
#define pb push_back
#define DEBUG(x) cout << #x << ": " << x << endl;

/*
__attribute__((constructor))
void initial()
{
	cin.tie(NULL);
	ios::sync_with_stdio(false);
}
*/

int main()
{
	cin.tie(NULL);
	ios::sync_with_stdio(false);

	ll A, B, C;
	
	cin >> A >> B >> C;

	/*
	ll count = 0;
	for (ll i = A; i <= B; i++)
	{
		if (i % C == 0)
		{
			count++;
		}
	}
	*/

	ll start = 0, end = 0;

	if (A % C == 0)
	{
		start = A / C;
	}
	else 
	{
		start = A / C + 1;
	}

	if (B % C == 0)
	{
		end = B / C;
	}
	else 
	{
		end = B / C ;
	}

	cout << end - start + 1 << endl;
	
	return 0;
}