## Task
To charge your mobile phone battery, do you know how much time it takes from 0% to 100%? It depends on your cell phone battery capacity and the power of the charger. A rough calculation method is:

```
0% --> 85%  (fast charge)
(battery capacity(mAh) * 85%) / power of the charger(mA)

85% --> 95% (decreasing charge)
(battery capacity(mAh) * 10%) / (power of the charger(mA) * 50%)

95% --> 100% (trickle charge)
(battery capacity(mAh) * 5%) / (power of the charger(mA) * 20%)
```
For example: Your battery capacity is 1000 mAh and you use a charger 500 mA,  to charge your mobile phone battery from 0% to 100% needs time:

```
0% --> 85%  (fast charge) 1.7 (hour)

85% --> 95% (decreasing charge) 0.4 (hour)

95% --> 100% (trickle charge) 0.5 (hour)

total times = 1.7 + 0.4 + 0.5 = 2.6 (hour)
```

 Complete function `calculateTime` that accepts two arguments `battery` and `charger`, return how many hours can charge the battery from 0% to 100%. The result should be a number, round to 2 decimal places (In Haskell, no need to round it).

Difficulty: introductory

Starter code:

def calculate_time(battery,charger):
	

Solutions:
calculate_time=lambda b,c: round(b/float(c)*1.3+0.0001,2)
    


CHARGE_CONFIG = ((.85, 1), (.1, .5), (.05, .2)) 

def calculate_time(battery, charger):
    return round(1e-4 + sum(battery * pb / (charger * pc) for pb,pc in CHARGE_CONFIG), 2)

def calculate_time(battery, charger):
    return round(battery / charger * 1.3, 2)


def calculate_time(battery,charger):
    return round(1e-10 + 1.3 * battery / charger, 2)

def calculate_time(bat,charger):
    return round(((bat*0.85)/charger)+((bat*0.1)/(charger*0.5))+((bat*0.05)/(charger*0.2)),2)

from decimal import Decimal, ROUND_HALF_UP


def calculate_time(battery,charger):
    n = 2.6 * battery/1000 * 500/charger
    return float(Decimal(str(n)).quantize(Decimal('1.23'), rounding=ROUND_HALF_UP))

#2.55 +  
def calculate_time(battery,charger):
    max = 100
    time = 0
    fast = (battery/charger)*.85
    dcharge = (battery/charger) *  .2
    tcharge = (battery/charger) * .25
    if max == 100:
        time += fast
        max += -85
    if max == 15:
        time += dcharge
        max += -10
    if max == 5:
        time += tcharge
        max += -5
    if time == 1.105:
        return 1.11
    return round(time,2)

calculate_time = lambda b,c: round(b*1.3/c+0.004, 2)

import math

def calculate_time(battery,charger):
    return math.ceil(battery / charger * 130) / 100

def calculate_time(b, c):
    return round(b * (0.85/c + 0.1/(c * 0.5) + 0.05/(c * 0.2)), 2)