python - 根据另一个列表和附加条件修改列表

Question

我已经删除了之前的问题并重写了它

我有 2 个大小列表 n (假设它是 5):

列表a由非零随机整数组成

列表b它由符合这些标准的随机非负整数组成:

此列表的元素按降序排列。 i第 th 个元素大于或等于 i+1第 1 个元素

这个列表的总和等于一些 f (假设 10 个)

在这里，我要修改列表 b这样除了上面的两个条件，它也符合这个条件:

每i此列表的第 th 个元素小于或等于 i列表的第 th 个元素 a

如果由于某种原因无法满足所有三个标准，则总和的标准等于 f可能会变松变 f-a哪里 a也从列表中减去 b s 最右边的元素和 a是可能的最小整数(见最后一个例子)

一些示例( r 是修改后的列表 b ):

a = [6, 7, 14, 16, 5]
b = [8, 2, 0, 0, 0]

r = [6, 4, 0, 0, 0]


a = [10, 1, 14, 16, 5]
b = [8, 2, 0, 0, 0]

r = [8, 1, 1, 0, 0]


a = [10, 5, 14, 16, 5]
b = [8, 2, 0, 0, 0] # here, b is already complying to all criteria and result should not change

r = [8, 2, 0, 0, 0]

# another example where this list is of size 14 with `f` being also 14
a = [7, 5, 14, 16, 5, 7, 203, 22, 10, 4, 5, 8, 12, 9]
b = [8, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

r = [7, 5, 2,  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

a = [5, 1, 1, 1, 1]
b = [8, 2, 0, 0, 0]

r = [5, 1, 1, 1, 1]

我的 numpy方法适用于 b 的情况不需要任何编辑:

a = [10, 5, 14, 16, 5]

b = [8, 2, 0, 0, 0]

def edit(a, b):
    a_ = np.array(a)
    b_ = np.array(b)
    criteria_check = (a_[:len(b_[b_>0])] >= b_[b_>0])
    if criteria_check.all():
        return b
    else:
        pass


edit(a, b)
[8, 2, 0, 0, 0]

添加另一种方法来解决 numpy 中的最后一个示例案例:

a = [5, 1, 1, 1, 1]
b = [8, 2, 0, 0, 0]

a = np.array(a)
b = np.array(b)

def edit2(f, s):
    diff = f - s
    s[(s + diff).cumsum()<sum(s)] = (s + diff)[(s + diff).cumsum()<sum(s)]
    return s

Answer 1

此解决方案适用于您的所有测试用例:

def solve(a, b):
    f = 0
    for i in range(len(a) - 1):
        if (a[i] < b[i]):
            b[i+1] += (b[i] - a[i])
            b[i] = a[i]
            f += a[i]
        else:
            f += b[i]

    x = 0
    if sum(b) > sum(a):
        x = sum(b) - sum(a)

    b[-1] = sum(b) - f - x

    return b