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 5.4.1) (judge verdict: Accepted)

/*  AUTHOR: AKASH JAIN
*   EMAIL:  akash19jain@gmail.com
*   ID:     akash19jain  
*   DATE:   03-02-2020 16:18:58
*/


// #include<algorithm>
// #include <bits/stdc++.h>
// using namespace std;
 
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<stdbool.h>
#include<ctype.h>
#define SC1(x)          scanf("%lld",&x)
#define SC2(x,y)        scanf("%lld%lld",&x,&y)
#define SC3(x,y,z)      scanf("%lld%lld%lld",&x,&y,&z)
#define SCS(x)          scanf("\n%s", x)
#define SCA(a,n)        for(long long i=0;i<n;i++)    scanf("%lld",&a[i]);
#define PF1(x)          printf("%lld\n",x)
#define PF2(x,y)        printf("%lld %lld\n",x,y)
#define PF3(x,y,z)      printf("%lld %lld %lld\n",x,y,z)
#define PFA(a,n)        for(long long i=0;i<n;i++)    printf("%lld ",a[i]); printf("\n");
#define PFN             printf("\n")
#define PFS(x)          printf("%s\n",x)
#define REP(i,n)        for(long long i=0;i<(n);++i)
#define FOR(i,a,b)      for(long long i=(a);i<=(b);++i)
#define FORD(i,a,b)     for(long long i=(a);i>=(b);--i)
#define WHILE(n)        while(n--)
#define MEM(a, b)       memset(a, (b), sizeof(a))
#define ITOC(c)         ((char)(((ll)'0')+c))
#define MID(s,e)        (s+(e-s)/2)
#define SZ(a)           strlen(a)          
#define PI              3.1415926535897932384626433832795
#define DEB(x)          printf("The value of \"%s\" is: %d\n",#x,x)
#define CASES           ll t;SC1(t);while(t--)
#define ABS(a)          ((a>0)?a:-(a))
#define SWAP(a,b)       ll z=a;a=b;b=z
#define SWAPC(a,b)      char z=a;a=b;b=z
#define FLSH            fflush(stdout)
typedef long long ll;
typedef unsigned long long ull;
const ll INF = 1<<29;
const ll MOD = 1000000007;
const ll FMOD = 998244353;
const ll MAX = 10000000005;
const ll MIN = -10000000005;
#define FILEIO(name) \  freopen(name".in", "r", stdin); \   freopen(name".out", "w", stdout);
int cmp(const void * a,const void * b);
long long maxv(long long a,long long b);
long long minv(long long a,long long b);
long long gcd(long long u,long long v);
long long digits(long long n);
bool ispoweroftwo(long long n);
bool isvowel(char x);
ll chartoint(char ch);
ll CEIL(ll x,ll y);
ll FLOOR(ll x,ll y);


int main()
{
    ll h,a;
    SC2(h,a);
    ll ans=CEIL(h,a);
    PF1(ans);
    return 0;
}
 
 
//qsort(arr,n,sizeof(arr[0]),cmp);
int cmp (const void * a, const void * b)
{
    if( *(ll*)a - *(ll*)b < 0 ) return -1;
    if( *(ll*)a - *(ll*)b > 0 ) return 1;
        return 0;
}
long long maxv(long long a,long long b)
{
    if(a>b) return a;
        return b;
}
long long minv(long long a,long long b)
{
    if(a<b) return a;
        return b;
}
long long gcd(long long u,long long v)
{
    if (v == 0) return u; 
       return gcd(v, u%v);
}
long long digits(long long n)    //to calculate no of digits in a number
{
    return floor(log10(n))+1;
}
bool ispoweroftwo(long long x)
{
    return x && (!(x&(x-1)));
}
bool isvowel(char x)
{
    return(x=='a' || x== 'e' ||x=='i' || x== 'o' || x=='u' );
}
ll chartoint(char ch)
{
    if(ch>='A' && ch<='Z') return(ch-'A');
    else if(ch>='0' && ch<='9') return (ch-'0');
    else if(ch>='a' && ch<='z') return (ch-'a');
    else return 0;
}
ll CEIL(ll x,ll y)
{
    if(x%y==0) return(x/y);
    else return(x/y +1);
}
 
ll FLOOR(ll x,ll y)
{
    if(x%y==0) return(x/y);
    else return(x/y -1);
}