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 <algorithm>
#include <cmath>
#include <cstdio>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <sstream>
#include <stack>
#include <string>
#include <unordered_set>
#include <vector>

using namespace std;

#define ll long long
#define P pair<int, int>
#define FOR(i, N) for (int i = 0; i < (int)N; i++)
#define FORIN(i, a, b) for (int i = a; i < (int)b; i++)
#define ALL(x) (x).begin(), (x).end()
#define GI(name) \
  int(name);     \
  scanf("%d", &(name))
#define GI2(name1, name2) \
  int(name1), (name2);    \
  scanf("%d %d", &(name1), &(name2))
#define GI3(name1, name2, name3) \
  int(name1), (name2), (name3);  \
  scanf("%d %d %d", &(name1), &(name2), &(name3))
#define GVI(name, size)    \
  vector<int>(name)(size); \
  FOR(i, (size)) scanf("%d", &(name)[i])
#define GS(name) \
  string(name);  \
  cin >> (name);
#define MOD 1000000007
#define DEBUG(...) debug(__LINE__, ":" __VA_ARGS__)

string to_string(string s) {
  return s;
}

template <class T>
string to_string(vector<T> v) {
  string ret = "{";
  FOR(i, v.size() - 1) { ret += to_string(v[i]) + ","; }
  if (v.size() > 0) {
    ret += to_string(v.back());
  }
  ret += "}";
  return ret;
}

void debug() {
  cerr << endl;
}

template <class Head, class... Tail>
void debug(Head head, Tail... tail) {
  cerr << to_string(head) << " ";
  debug(tail...);
}

void print() {
  cout << endl;
}

template <class Head, class... Tail>
void print(Head head, Tail... tail) {
  cout << to_string(head);
  print(tail...);
}

int main() {
  ll a, b, x;
  cin >> a >> b >> x;
  ll n1 = (a - 1) / x, n2 = b / x;
  print(n2 - n1);
  return 0;
}