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>
using namespace std;
#define rep(i, n) for(int i=0;i<(int)(n);i++)
#define rrep(i, n) for(int i=((int)(n));i>0;i--)
#define FOR(i, m, n) for(int i = m; i < n; i++)	
#define SORT(c) sort((c).begin(),(c).end())
#define REVERSE(c) reverse((c).begin(),(c).end())
#define REVSORT(c) SORT(c);REVERSE(c)
#define ALL(x) (x).begin(),(x).end()
#define INF 1e9
const long long MOD = 10e9 + 7;
typedef long long ll;

int main(){
	ll a, b, x;
	cin >> a >> b >> x;
	ll R = b - a + 1;
	ll ans = R / x;
	if(a % x == 0 && b % x == 0)ans++;
	cout << ans << endl;
}