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)

#include <iostream>
#include <string>
#include <algorithm>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <iomanip>
#include <bitset>
#include <set>
#include <map>
#include <stdio.h>
#include <numeric>
#include <cstring>

#define rep(i,n) for (int i = 0; i < (n); i++)
#define rep1(i,n) for (int i = 1; i < (n); i++)
#define FOR(i,a,b) for (int i=(a); i < (b); i++)
#define MOD 1000000007 //10^9+7
using namespace std;
using ll = long long;
using PII = pair<int, int>;
using PLL = pair<ll, ll>;
const int INF = numeric_limits<int>::max();
const ll INFL = numeric_limits<ll>::max();


// 負の数にも対応した % 演算
long long mod(long long val, long long m) {
    long long res = val % m;
    if (res < 0) res += m;
    return res;
}

//greatest common divisor
long long gcd(ll a, ll b) 
{
    if (a % b == 0) {
        return b;
    } else {
        return gcd(b, a % b);
    }
}

//least common multiple
long long lcm(ll a, ll b)
{
    return a / gcd(a, b) * b ;
}

ll factorial(ll n) {
    ll res = 1;
    for (ll i = 1; i <= n; i++) {
        res *= i;
    }
    return res;
}

int main()
{
    double h, a; cin >> h >> a;
    cout << ceil(h/a) << endl;
    return 0;
}