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>
#include<unordered_map>	
#include<unordered_set>
#define pb emplace_back
#define rt(v) return cout<<v,0
#define sz(v) (int)(v.size())
#define all(a) a.begin(),a.end()
#define endl '\n'
#define space " "
typedef long long ll;
typedef unsigned long long ull;
const int N = 10 + 1e5;
const int mod = 1e9 + 7;
/*

███╗   ███╗███████╗███╗   ███╗███████╗    ██╗      ██████╗ ██████╗ ██████╗
████╗ ████║██╔════╝████╗ ████║██╔════╝    ██║     ██╔═══██╗██╔══██╗██╔══██╗
██╔████╔██║█████╗  ██╔████╔██║█████╗      ██║     ██║   ██║██████╔╝██║  ██║
██║╚██╔╝██║██╔══╝  ██║╚██╔╝██║██╔══╝      ██║     ██║   ██║██╔══██╗██║  ██║
██║ ╚═╝ ██║███████╗██║ ╚═╝ ██║███████╗    ███████╗╚██████╔╝██║  ██║██████╔╝
╚═╝     ╚═╝╚══════╝╚═╝     ╚═╝╚══════╝    ╚══════╝ ╚═════╝ ╚═╝  ╚═╝╚═════╝

*/

using namespace std;
void Ferrari() {
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
}
char img[55][55];
char temp[55][55];
char ans[55][55];
int n, m;
bool check() {
	bool ok = 1;
	for(int i = 0 ; i < m; ++i)
		for(int j = 0 ; j < m ; ++j)
			if (ans[i][j] != temp[i][j])
			{
				ok = 0; break;
			}
	return ok;
}
int main() {
	Ferrari();
	long long a, b, x; cin >> a >> b >> x;
	long long len = b - a;
	long long ans = len / x;
	long long rem = len % x;
	long long firstm;
	if (x >= a) firstm = x;
	else firstm = ((a + (x - 1)) / x) * x;
	if (firstm - a <= rem) ++ans;
	cout << ans;
	return 0;
}





/* stuff you should look for
	* keepin sharp mind.
	* int overflow, array bounds.
	* special cases (n=1?), set tle , make sure you understand the proble correctly.
	* don`t stuck in a problem.
	* test the idea on samples.
	*/