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++14 (GCC 5.4.1) (judge verdict: Accepted)

/*
    AUTHOR : TANMAY SINHA
    INORMATION TECHNOLOGY
    BIT MESRA ,RANCHI
*/
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
#define ld long double
#define ull unsigned long long int
#define ll long long int
#define SPEED ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define mod 1000000007
#define rep(i,L,R) for(i=L;i<R;i++)
#define DEBUG
#ifdef DEBUG
#define deb2(args...) {dbg,args; cerr<<endl;}
#else
#define deb2(args...)
#endif
struct debb{template<typename T> debb& operator,(const T &v){cerr<<v<<" ";return *this;}}dbg;
#define deb1(args...) do {cerr << #args << " : "; dbg,args; cerr << "\n";} while(0)
#define printa(a,L,R) rep(i,L,R) cout << a[i] <<' ';cout<<"\n";
typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> indexed_set;
template<typename T>T gcd(T a,T b){if(b==0)return a;else return gcd(b,a%b);}
template<typename T>T lcm(T a,T b){return a/gcd(a,b)*b;}
ll power(ll x,ll y,ll p){ll res=1;x=x%p;while(y>0){if(y&1)res=((res%p)*(x%p))%p;y=y>>1;x=(x*x)%p;}return res;}
ll pow1(ll x,ll y){ll res=1;while(y>0){if(y&1)res=(res*x);y=y>>1;x=x*x;}return res;}
const ll INF = 1e18l+10,N = 1e5+10,MAX = 500500;
#define sq(X) ((X)*(X))

int main()
{
    ll h,a;
    cin>>h>>a;
    if(h%a==0)
    cout<<h/a<<"\n";
    else
    {
        cout<<h/a+1<<"\n";
    }
    
}