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>
typedef long long lint;
using namespace std;

lint gcd(lint x, lint y) {
  if(x == 0) { return y; }
  else { return gcd(y%x,x); }
}

lint lcm(lint x, lint y) { return x/gcd(x,y)*y; }

lint C(lint n, lint k) {
  if(n == k) { return 1; }
  else if(n < k) { return 0; }
  else if(k == 0) { return 1; }
  else if(k == 1) { return n; }
  else return C(n-1,k-1) + C(n-1,k);
}

lint P(lint n, lint k) {
  if(k == 1) { return n; }
  return (n*(P(n-1,k-1)%1000000007)%1000000007);
}


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

  lint ans=0;
  if(a%x == 0) { ans++; }
  ans += (b-a)/x-1;
  if((b-a)%x == 0) { ans++; }

  cout << ans << endl;
  return 0;
}