Description:
Two tortoises named A and B must run a race. A starts with an average speed of 720 feet per hour. Young B knows she runs faster than A, and furthermore has not finished her cabbage.
When she starts, at last, she can see that A has a 70 feet lead but B's speed is 850 feet per hour. How long will it take B to catch A?
More generally: given two speeds v1 (A's speed, integer > 0) and v2 (B's speed, integer > 0) and a lead g (integer > 0) how long will it take B to catch A?
The result will be an array [hour, min, sec] which is the time needed in hours, minutes and seconds (round down to the nearest second) or a string in some languages.
If v1 >= v2 then return nil, nothing, null, None or {-1, -1, -1} for C++, C, Go, Nim, Pascal, COBOL, [-1, -1, -1] for Perl,[] for Kotlin or "-1 -1 -1".
Examples:
(form of the result depends on the language)
race(720, 850, 70) => [0, 32, 18] or "0 32 18"
race(80, 91, 37) => [3, 21, 49] or "3 21 49"
Note:
-
See other examples in "Your test cases".
-
In Fortran - as in any other language - the returned string is not permitted to contain any redundant trailing whitespace: you can use dynamically allocated character strings.
** Hints for people who don't know how to convert to hours, minutes, seconds:
- Tortoises don't care about fractions of seconds
- Think of calculation by hand using only integers (in your code use or simulate integer division)
- or Google: "convert decimal time to hours minutes seconds"
我的解决方案：
def race(v1, v2, g):
print(v1,v2,g)
if v1>=v2:
return None
speed=v2-v1#这等于是速度为speed,距离为g,算什么时间走完全程
full_time=str(g/(v2-v1)).split('.')#是一个浮点数
hours=int(full_time[0])#将整数部分取出来，这就是小时
float_min=float('0.'+full_time[1])##小数部分取出来，是个浮点数，分钟的占比，*60就可算出实际值
minutes=int(str(float_min*60).split('.')[0]) #算出分钟，第二位是浮点数，是秒数的占比
float_sec=float('0.'+str(float_min*60).split('.')[1]) #取出算出分钟后的浮点数，这是秒数的占比
seconds=int(round(float_sec*60,3))#算出有多少秒，round取出小数点后三位，我测了一下，round取的是近似值，如果小数点后是0.9999,它就会自动前一位加1
#下面的函数是写一个60进制的时间进位代码，如果秒数是60，则它自己清零，分钟加1，分钟是60，它自己清零，小时加1
if seconds==60:
seconds=0
print(seconds)
minutes+=1
if minutes==60:
minutes=0
hours+=1
return [hours,minutes,seconds]再看一下大佬写的：
def race(v1, v2, g):
t = 3600 * g/(v2-v1)
return [int(t/3600), int(t//60%60), int(t%60)] if v2 > v1 else None
第二个：
def race(v1, v2, g):
if v1 >= v2: return None
t = float(g)/(v2-v1)*3600
mn, s = divmod(t, 60)
h, mn = divmod(mn, 60)
return [int(h), int(mn), int(s)]其他的就不贴了，大致思路都跟第一个一样。
