对于任意数量的项,返回最小X(可以包含Y项)

//

number_required = y * z
container_holds = x
reqd_containers = (number_required + container_holds - 1) // container_holds

n=(y*z+(x-1))//x;

n=(y*z+x-1)//x

10 / 3 -> 3 10 / 3 -> 3.3333333333333335

# wrong in Python 3; works with Python 2.3 to 2.7
# int overflow with Pythons up to 2.2
>>> int((100000000000000000 + 2)/3)
33333333333333332 # last digit should be 4

# wrong with Python 2.3 onwards; int overflow with earlier versions
>>> import math
>>> int(math.ceil(float(100000000000000000) / 3))
33333333333333332L

min_containers = y*z/x

min_full_containers = floor(y*z/x)
remaining_items = y*z%x

def f(X, Y, Z):
  d, r = divmod(Y * Z, X)
  return d + bool(r)

#python
import math

int(math.ceil(float(Y) * Z / X))

内德的回答是正确的。通过执行以下操作,还可以避免对math.ceil()的函数调用开销:

minContainers = int((y*z+(x-1))/x);

($people*$breadsticksperson)/$holders不正确?

编辑:抱歉,误读了您的问题,在评论中发布了正确的解决方案