python 中出现溢出错误？

Question

在

处出现 OverflowError: int 太大而无法转换为 float

l = (number // 10)*n

这是什么意思，如何解决？

这就是全部代码

def main():
    count = 0
    n = 10
    t = int(input())
    arr = []
    for i in range(t):
        number = int (input())
        print(number)
        while number != 0:
            l = (number // 10)*n
            print(l)
            if l != 0:
                count += 1
                arr.append(l)
            number = number / 10
            n = n * 10
        print(count)
        print(arr)


if __name__ == '__main__':
    main()

Answer 1

当 while 循环为运行时，数字值的变化不会影响 while 条件。所以你必须使用 if-else.

break 循环

def main():
    count = 0
    n = 10
    t = int(input())
    arr = []
    for i in range(t):
        number = int (input())
        print(number)
        while True:
            l = (number // 10)*n
            print(l)
            if l != 0:
                count += 1
                arr.append(l)
            number = number / 10
            n = n * 10
            if int(number) == 0:
                break
        print(count)
        print(arr)


if __name__ == '__main__':
    main()

输出：

python 中出现溢出错误？

OverflowError in python?

python

overflow