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<bits/stdc++.h>
#define speedio ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
using namespace std;
// #include <ext/pb_ds/assoc_container.hpp>
// using namespace __gnu_pbds;

typedef long long       ll;
typedef long double     ld;

typedef vector<int>     vi;
typedef vector<bool>    vb;
typedef vector<ll>      vll;
typedef vector<vi>      vvi;
typedef vector<vll>     vvll;
typedef vector<string>  vs;

typedef pair<int, int>  pii;
typedef pair<ll, int>   pli;
typedef pair<ll, ll>    pll;
typedef vector< pii >   vpi;
typedef vector< pll >   vpll;

// typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;

#define pb           push_back
#define mod          1000000007
#define INF          1e17
#define mini         INT_MIN
#define maxo         INT_MAX
#define eb           emplace_back
#define ff           first
#define ss           second
#define re           return 
#define endl         "\n" 

#define all(c)       c.begin(),c.end()
#define be           begin()
#define en           end()
#define sz(x)        (ll)x.size()
#define mem(a,val)   memset(a,val,sizeof(a))
#define mk(x,y)      make_pair(x,y)
#define mid(s,e)     ((s)+((e)-(s))/2) 

#define fo(i,a,b)     for(ll i = (ll)(a); i <= (ll)(b); i++)
#define rf(i,a,b)    for(ll i = (ll)(a); i >= (ll)(b); i--)
#define read(a,n)    for(int i=0;i<n;i++)cin>>a[i];
#define show(a,n)    for(int i=0;i<n;i++)cout<<a[i]<<' ';cout<<endl;
#define p2(n)        (1LL << (n))
#define deb(x) cout <<#x << " = " << x << " | " << "LINE: "<<__LINE__<< endl;
template<typename... T> void rd(T&... args) { ((cin >> args), ...); }
template<typename... T> void pp(T... args) { ((cout << args << " "), ...); cout<<"\n"; }

void add(ll &x, ll y){ x += y; if(x>=mod) x %= mod; if(x<0) x = x%mod + mod; }

//============== *** START OF CODE ***




void solve()
{
       ll h,a;
    rd(h,a);
    ll ans = 0;
    if(h%a!=0)ans++;
    ans+=h/a;
    pp(ans);
    
}


void prep(){}

//=============== *** END OF CODE ***

int main(){
    speedio; cout<<fixed<<setprecision(12);

// #ifndef ONLINE_JUDGE
//     freopen("input.txt", "r", stdin);
//     freopen("output.txt", "w", stdout);
// #endif

    prep();
    int t = 1;
    // cin >> t;
    while(t--)
        solve();

    cout<<"\n";
    return 0;
}