Problem:
Score : 100 points

Problem Statement

Serval is fighting with a monster.

The health of the monster is H.

In one attack, Serval can decrease the monster's health by A.
There is no other way to decrease the monster's health.

Serval wins when the monster's health becomes 0 or below.

Find the number of attacks Serval needs to make before winning.

Constraints

1 \leq H \leq 10^4

1 \leq A \leq 10^4

All values in input are integers.

Input

Input is given from Standard Input in the following format:

H A

Output

Print the number of attacks Serval needs to make before winning.

Sample Input 1

10 4

Sample Output 1

3

After one attack, the monster's health will be 6.

After two attacks, the monster's health will be 2.

After three attacks, the monster's health will be -2.

Thus, Serval needs to make three attacks to win.

Sample Input 2

1 10000

Sample Output 2

1

Sample Input 3

10000 1

Sample Output 3

10000

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

// Problem Statement
// Given is a positive integer N
// Find the number of pairs
// (A,B) of positive integers not greater than N hat satisfy the following condition:
// When A and B are written in base ten without leading zeros, the last digit of A is equal to the first digit of B, and the first digit of
// A is equal to the last digit of B.

#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
#define ll long long
#define ld long double
#define MOD 1000000007
#define nl cout << bl
#define rt ll Q[100005] = {0}
#define mp make_pair
#define test() \
    ull t;     \
    cin >> t;  \
    while (t--)
#define pb push_back
#define ff first
#define ss second
#define all(v) v.begin(), v.end()
#define s(ar, n) sort(ar, ar + n)
#define rs(ar, n) sort(ar, ar + n, greater<ll>())
#define oa(a, n)               \
    for (ll i = 0; i < n; i++) \
        cout << a[i] << " ";
#define cn(a, n)               \
    for (ll i = 0; i < n; i++) \
        cin >> a[i];
#define ov(a, n)                    \
    for (ll i = n - 1; i >= 0; i--) \
        cout << a[i] << " ";        \
    nl
#define ovp(a, n)              \
    for (ll i = 0; i < n; i++) \
        cout << a[i].ff << " " << a[i].ss << bl;
#define maxa(ar, N) *max_element(ar, ar + N)
#define mina(ar, N) *min_element(ar, ar + N)
#define present(c, x) ((c).find(x) != (c).end())
#define fastio()                  \
    ios_base::sync_with_stdio(0); \
    cin.tie(0);                   \
    cout.tie(0)
typedef pair<int, int> pii;
ll fast_exp(ll base, ll exp)
{
    ll res = 1;
    while (exp > 0)
    {
        if (exp % 2 == 1)
            res = (res * base) % MOD;
        base = (base * base) % MOD;
        exp /= 2;
    }
    return res % MOD;
}
ll fib(ll n)
{
    if (n < 2)
        return n;
    ll a = 0, b = 1, pre;
    int i = 1;
    while (i < n)
    {
        pre = (a + b) % MOD;
        a = b;
        b = pre;
        i++;
    }
    return pre;
}
ll fac(ll n)
{
    if (n % 2 == 0)
        return 2;

    for (ll i = 3; i * i <= n; i += 2)
    {
        if (n % i == 0)
            return i;
    }

    return n;
}
ll fact(ll n)
{
    if (n > 1)
        return n * fact(n - 1);
    else
        return 1;
}
bool sortbysec(const tuple<int, int, int> &a,
               const tuple<int, int, int> &b)
{
    return (get<1>(a) < get<1>(b));
}
int last_digit(int n)
{
    int ans = 0;
    while (n != 0)
    {
        ans = n % 10;
        n /= 10;
    }
    return ans;
}
int main()
{
    int H, A;
    cin >> H >> A;
    if (H % A == 0)
    {
        cout << H / A;
    }
    else
    {
        cout << (H + A) / A;
    }
}