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)

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
#pragma GCC optimize("Ofast")
//*********************************************DO IT NOW****************************************************************
#include<bits/stdc++.h> 
#include<random>
#define int long long   
#define pp pair<int,int> 
#define ss second 
#define ff first 
#define pb push_back
#define mod 1000000007
#define pi 3.14159265359
#define mk(arr,n,type)  type *arr=new type[n];
#define sl s1.length();
#define yes cout<< "YES"<<endl
#define no  cout<< "NO"<<endl
#define all(v) (v).begin(),(v).end()
#define s(v)   sort(v,v+n)
#define  mt mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define read(a,n) for(int i=0;i<n;i++)cin>>a[i]
#define print(a,n) for(int i=0;i<n;i++)cout<<a[i]<<" "
#define for1(i,a,n)  for(int i=0;i<n;i++)
using namespace std;
//**********************************************DO IT NOW***************************************************************
void fastio()
{  
  #ifndef ONLINE_JUDGE
  freopen("INPUT.txt","r",stdin);
  freopen("OUTPUT.txt","w",stdout); 
  #endif
  ios_base :: sync_with_stdio(false); 
  cin.tie(NULL) ; 
  cout.tie(NULL);
}
int dp[101][1001],ans1=1e9,dp1[1000001];
int num[10];
int a[1000000];
signed main()
{   
    fastio();
    int t=1;
   // cin>>t;
    while(t--)
    { 
       int n,m;
       cin>>n>>m;
       printf("%d\n",(n+m-1)/m);
    }
}
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