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 Python (3.4.3) (judge verdict: Accepted)

# coding:utf-8

import sys
import math
import time
import numpy as np
import collections
from collections import deque
import queue
import copy

# 最大公約数
def gcd(m,n):
  x = max(m,n)
  y = min(m,n)
  if x%y == 0:
    return y
  else:
    while x%y != 0:
      z = x%y
      x = y
      y = z
    else:
      return z

# 幅優先探索(BFS)でスタート座標(x, y)からの最大距離返す
def bfs(SS, x, y, H, W):#入力(座標系(迷路), x, y, xの端, yの端)
  x_next = [0, 0, -1, 1]
  y_next = [1, -1, 0, 0]
  tempx = x
  tempy = y
  mlength = 0
  q = deque()
  xytuple = (x, y, 0)
  q.append(xytuple)


  while q:
    xytuple = q.popleft()
    SS[xytuple[0]][xytuple[1]] = "x" #探索した場所は探索できないようにする
    for i in range(4): #このforで上下左右を調べてる
      tempx = xytuple[0] + x_next[i]
      tempy = xytuple[1] + y_next[i]
      #print(tempx, tempy, xytuple[2])
      #print(S)
      if(0 <= tempx and tempx < H and 0 <= tempy and tempy < W and SS[tempx][tempy] == "."):
        temptuple = (tempx, tempy, xytuple[2]+1)
        SS[temptuple[0]][temptuple[1]] = "x" #探索した場所は探索できないようにする
        mlength = max(mlength, xytuple[2] +1)
        q.append(temptuple)
      #print(q)
  return mlength

def brick(a,n,m,N):#ABC 148
  #n番目以降にmがあるか調べる
  if not m in a[n:]:
    print(N-(m-1))
  else:
    #b = copy.deepcopy(a)
    brick(a,a[n:].index(m),m+1,N)

#X = str(input()).split()
#a = [int(x) for x in input().split()]
HA = str(input()).split()
H = int(HA[0])
A = int(HA[1])
count = 1

while(H>A):
    H=H-A
    count+=1
print(count)