Question

项目：使用输入nickels，pennies，sum创建一个Python代码，如果可以创建sum的{{1}}，则返回nickels }和pennies。

我的代码是：

def corect_change(nickles, pennies, summ):
    result = False
    i = 0
    j = 0
    while i <= nickles:
        if summ == i * 5:
            result = True
        else:
            while j <= pennies:
                if summ == (i * 5) + j:
                    result = True
                j += 1
        i += 1
    return result

print corect_change(5,3,11)

现在这适用于可以被5整除的总和但是当它必须包含便士时会失败。

有人可以告诉我为什么，并给我一些改进它的想法。

Answer 1

您的代码存在一些问题：

它总是假设至少有一个镍可用
总是假设summ＆gt; 5，因为i从1
在用完镍之前用尽了你的便士

这里有几个实现基于首先使用尽可能多的镍，然后是便士。

def correct_change2(nickels, pennies, sum):
    # While you have nickels left and sum is still more than 5 cents...
    while sum > 5 and nickels:
        sum -= 5
        nickels -= 1
    # You've removed as many nickels as you had or sum is now less than 5
    # If you have enough pennies left, return True
    return pennies >= sum

def correct_change(nickels, pennies, sum):
    # Remove as many nickels as you have or as many as sum needs,
    # whichever is less
    sum -= 5 * min(nickels, sum / 5)
    # You're out of nickels or sum is less than 5
    # If you have enough pennies to equal sum, return True
    return pennies >= sum

print correct_change(5,3,11)

此外，您的方法轻轻地重新安排工作：

def corect_change(nickles, pennies, summ):
    i = 0
    while i <= nickles:
        if summ - (i * 5) < 5:
            break
        i += 1
    j = 0
    while j <= pennies:
        if summ == (i * 5) + j:
            break
        j += 1
    return summ == (i * 5) + j

但我们可以通过使用for循环和xrange()：

来实现更多Python惯用语

def corect_change(nickles, pennies, summ):
    for i in xrange(nickles):
        if summ - (i * 5) < 5:
            break
    for j in xrange(pennies):
        if summ == (i * 5) + j:
            break
    return summ == (i * 5) + j

更改计算器镍币便士

1 个答案: