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;

typedef long long ll;
typedef unsigned long long ull;
const long double PI = (acos(-1));
const long long MOD = 1000000007;
struct Edge
{
    long long to;
    long long cost;
};
using Graph = vector<vector<Edge>>;
using P = pair<ll, ll>;
const long long INF = 1LL << 60;

long long modpow(long long a, long long n, long long m)
{
    long long ans = 1;
    while (n)
    {
        if (n & 1)
        {
            ans = (ans * a) % m;
        }
        a = (a * a) % m;
        n >>= 1;
    }
    return ans;
}
long long combi(long long n, long long a)
{
    long long ans = 1, ans1 = 1;
    for (long long i = n - a + 1; i <= n; i++)
    {
        ans *= i % MOD;
        ans %= MOD;
    }

    for (long long i = 2; i <= a; i++)
        ans1 = (ans1 * i) % MOD;
    ans1 = modpow(ans1, MOD - 2, MOD);
    return ((ans % MOD) * ans1) % MOD;
}

template <typename T>
bool next_combination(const T first, const T last, int k)
{
    const T subset = first + k;
    // empty container | k = 0 | k == n
    if (first == last || first == subset || last == subset)
    {
        return false;
    }
    T src = subset;
    while (first != src)
    {
        src--;
        if (*src < *(last - 1))
        {
            T dest = subset;
            while (*src >= *dest)
            {
                dest++;
            }
            iter_swap(src, dest);
            rotate(src + 1, dest + 1, last);
            rotate(subset, subset + (last - dest) - 1, last);
            return true;
        }
    }
    // restore
    rotate(first, subset, last);
    return false;
}

void dfs(vector<ll> s, ll mx, ll N, vector<vector<ll>> &arr)
{
    if (s.size() == (size_t)N)
    {
        //cout << s.c_str() << endl;
        arr.push_back(s);
    }
    else
    {
        for (ll c = s.size() > 0 ? 0 : 0; c <= mx; c++)
        {
            s.push_back(c);
            dfs(s, mx, N, arr);
            s.pop_back();
        }
    }

    return;
}

int bfs(int sx, int sy, int gx, int gy, int h, int w, vector<vector<char>> tmp)
{
    int ans[101][101] = {0};
    char c[101][101] = {0};
    queue<pair<int, int>> s;

    for (int i = 0; i < h; i++)
    {
        for (int j = 0; j < w; j++)
        {
            c[i][j] = tmp[i][j];
        }
    }

    s.push(make_pair(sx, sy));

    while (s.size() > 0)
    {

        pair<int, int> tmp = s.front();
        s.pop();
        c[tmp.first][tmp.second] = '#';
        for (int i = -1; i <= 1; i++)
        {
            for (int j = -1; j <= 1; j++)
            {
                if (tmp.first + i < 1 || tmp.first + i > h)
                {
                    continue;
                }
                if (tmp.second + j < 1 || tmp.second + j > w)
                {
                    continue;
                }
                if (i != 0 && j != 0)
                {
                    continue;
                }
                if (i == 0 && j == 0)
                {
                    continue;
                }
                if (c[tmp.first + i][tmp.second + j] == '#')
                {
                    continue;
                }
                c[tmp.first + i][tmp.second + j] = '#';
                if (ans[tmp.first + i][tmp.second + j] == 0)
                {
                    ans[tmp.first + i][tmp.second + j] = ans[tmp.first][tmp.second] + 1;
                }
                else
                {
                    ans[tmp.first + i][tmp.second + j] = min(ans[tmp.first + i][tmp.second + j], ans[tmp.first][tmp.second] + 1);
                }
                s.push(make_pair(tmp.first + i, tmp.second + j));
            }
        }
    }

    return ans[gy][gx];
}

ll modfactorial(ll a)
{
    if (a == 1)
        return 1;

    return (a % MOD) * (modfactorial(a - 1) % MOD);
}

ll gcd(ll a, ll b)
{
    if (a % b == 0)
    {
        return (b);
    }
    else
    {
        return (gcd(b, a % b));
    }
}

ll lcm(ll a, ll b)
{
    return (a / gcd(a, b)) * b;
}

map<ll, ll> prime_factor(ll n)
{
    map<ll, ll> ret;
    for (ll i = 2; i * i <= n; i++)
    {
        while (n % i == 0)
        {
            ret[i]++;
            n /= i;
        }
    }
    if (n != 1)
        ret[n] = 1;
    return ret;
}

/* dijkstra(G,s,dis)
    入力：グラフ G, 開始点 s, 距離を格納する dis
    計算量：O(|E|log|V|)
    副作用：dis が書き換えられる
*/
void dijkstra(const Graph &G, int s, vector<ll> &dis, vector<ll> &prev)
{
    int N = G.size();
    dis.resize(N, INF);
    prev.resize(N, -1); // 初期化
    priority_queue<P, vector<P>, greater<P>> pq;
    dis[s] = 0;
    pq.emplace(dis[s], s);
    while (!pq.empty())
    {
        P p = pq.top();
        pq.pop();
        int v = p.second;
        if (dis[v] < p.first)
        {
            continue;
        }
        for (auto &e : G[v])
        {
            if (dis[e.to] > dis[v] + e.cost)
            {
                dis[e.to] = dis[v] + e.cost;
                prev[e.to] = v; // 頂点 v を通って e.to にたどり着いた
                pq.emplace(dis[e.to], e.to);
            }
        }
    }
}

vector<ll> get_path(const vector<ll> &prev, ll t)
{
    vector<ll> path;
    for (ll cur = t; cur != -1; cur = prev[cur])
    {
        path.push_back(cur);
    }
    reverse(path.begin(), path.end()); // 逆順なのでひっくり返す
    return path;
}

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

    if(a==b){
        cout << 0 << endl;
        return 0;
    }

    if(a%x != 0){
        a = ((a + 1) / x) * x;
    }
    if(b%x != 0){
        b = ((b + 1) / x) * x;
    }

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

    cout << ans << endl;

    return 0;
}
//cout << std::fixed << std::setprecision(15)