你正在寻找模运算符:

a % b

例如:

>>> 26 % 7
5

当然，也许他们希望您自己实现它，这也不会太难。

26 % 7(你将得到余数)

26 / 7(你会得到除数，可以是浮点值)

26 // 7(你将得到除数，只有整数值)

除法的余数可以使用运算符%来发现:

>>> 26%7
5

如果你同时需要商和模，有一个内置的divmod函数:

>>> seconds= 137
>>> minutes, seconds= divmod(seconds, 60)

除法时用%代替/。这将为您返回余数。所以在你的案例中

26 % 7 = 5

您可以定义一个函数，并将其称为具有2个值的remainder，如rem(number1,number2)，返回number1%number2 然后创建一个while，并将其设为true然后为持有编号1和2的函数输出两个输入然后输出(rem(number1,number2)

从Python 3.7开始，有一个新的math.remainder()函数:

from math import remainder
print(remainder(26,7))

输出:

-2.0  # not 5

注意，如上所述，它与%不同。

引用文档:

math.remainder(x, y) Return the IEEE 754-style remainder of x with respect to y. For finite x and finite nonzero y, this is the difference x - n*y, where n is the closest integer to the exact value of the quotient x / y. If x / y is exactly halfway between two consecutive integers, the nearest even integer is used for n. The remainder r = remainder(x, y) thus always satisfies abs(r) <= 0.5 * abs(y). Special cases follow IEEE 754: in particular, remainder(x, math.inf) is x for any finite x, and remainder(x, 0) and remainder(math.inf, x) raise ValueError for any non-NaN x. If the result of the remainder operation is zero, that zero will have the same sign as x. On platforms using IEEE 754 binary floating-point, the result of this operation is always exactly representable: no rounding error is introduced.

Issue29962描述了创建新函数的基本原理。