这是Python中余数的整数版本，它的结果应该与C的“%”操作符相同:

def remainder(n, d):
    return (-1 if n < 0 else 1) * (abs(n) % abs(d))

预期结果:

remainder(123, 10)   ==  3
remainder(123, -10)  ==  3
remainder(-123, 10)  == -3
remainder(-123, -10) == -3

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。

取模是正确的答案，但如果你手动做的话，这应该是可行的。

num = input("Enter a number: ")
div = input("Enter a divisor: ")

while num >= div:
    num -= div
print num

你可以用模算子求余数 例子

a=14
b=10
print(a%b)

它会输出4

这是Python中余数的整数版本，它的结果应该与C的“%”操作符相同:

def remainder(n, d):
    return (-1 if n < 0 else 1) * (abs(n) % abs(d))

预期结果:

remainder(123, 10)   ==  3
remainder(123, -10)  ==  3
remainder(-123, 10)  == -3
remainder(-123, -10) == -3

我们可以用模算子(%)来解决这个问题

26%7 = 5;

但 26 / 7 = 3，因为它会给出商，而%运算符会给出余数。