## 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).
#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)
#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)