【Kata Daily 190908】How Much?

原题：

I always thought that my old friend John was rather richer than he looked, but I never knew exactly how much money he actually had. One day (as I was plying him with questions) he said:

• "Imagine I have between m and n Zloty..." (or did he say Quetzal? I can't remember!)
• "If I were to buy 9 cars costing c each, I'd only have 1 Zloty (or was it Meticals?) left."
• "And if I were to buy 7 boats at b each, I'd only have 2 Ringglets (or was it Zloty?) left."

Could you tell me in each possible case:

1. how much money f he could possibly have ?
2. the cost c of a car?
3. the cost b of a boat?

So, I will have a better idea about his fortune. Note that if m-n is big enough, you might have a lot of possible answers.

Each answer should be given as ["M: f", "B: b", "C: c"] and all the answers as [ ["M: f", "B: b", "C: c"], ... ]. "M" stands for money, "B" for boats, "C" for cars.

Note: m, n, f, b, c are positive integers, where 0 <= m <= n or m >= n >= 0. m and n are inclusive.

Examples:

howmuch(1, 100)      => [["M: 37", "B: 5", "C: 4"], ["M: 100", "B: 14", "C: 11"]]
howmuch(1000, 1100)  => [["M: 1045", "B: 149", "C: 116"]]
howmuch(10000, 9950) => [["M: 9991", "B: 1427", "C: 1110"]]
howmuch(0, 200)      => [["M: 37", "B: 5", "C: 4"], ["M: 100", "B: 14", "C: 11"], ["M: 163", "B: 23", "C: 18"]]

Explanation of the results for howmuch(1, 100):

In the first answer his possible fortune is 37:

• • so he can buy 7 boats each worth 5: 37 - 7 * 5 = 2
  • or he can buy 9 cars worth 4 each: 37 - 9 * 4 = 1
• The second possible answer is 100:
  • he can buy 7 boats each worth 14: 100 - 7 * 14 = 2
  • or he can buy 9 cars worth 11: 100 - 9 * 11 = 1

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题目大意为给出一个数的范围，在此范围中找出数字，满足公式（这个整数-7*某个整数）=2以及满足（这个整数-9*某个整数）=1，然后返回这个数以及符合条件的“某个整数”的值

解题办法：

看一下我的解题办法，重复代码好多，，，，

def howmuch(m, n):
    # your code
    L = []
    if n >= m:
        for i in range(m, n+1):
            L1 = []
            b = (i-2) % 7
            c = (i-1) % 9
            if b == 0 and c == 0:
                L1 = ["M: "+str(i), "B: "+str(int((i-2)/7)), "C: "+str(int((i-1)/9))]
                L.append(L1)
    else:
        for i in range(n, m+1):
            L1 = []
            b = (i-2) % 7
            c = (i-1) % 9
            if b == 0 and c == 0:
                L1 = ["M: "+str(i), "B: "+str(int((i-2)/7)), "C: "+str(int((i-1)/9))]
                L.append(L1)

    return L

再看看网友优秀的思路：

def howmuch(m, n):
    return [['M: %d'%i, 'B: %d'%(i/7), 'C: %d'%(i/9)] for i in range(min(m,n), max(m,n)+1) if i%7 == 2 and i%9 == 1]

简洁到令人发指，，，羡慕啊！！！

知识点：

1、取两个数的最大值和最小值，可以使用min()和max()函数来取出最大值和最小值；也可以使用n, m = m, n if m>n:

2、i = 1; (["M: %s"])%i -------->这是一种错误的办法