Tortoise racing---codewar刷题总结_seconds=int(round(-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_41821465/article/details/124058655

本文总结了一道关于乌龟赛跑的编程题，题目中乌龟A和B进行比赛，B需要追赶速度较慢的A。根据两者速度差和距离差，计算B追上A所需的时间。解决方案提供了将时间转换为小时、分钟和秒的算法，并给出了Python实现示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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。