用x乘x米查找米圆的面积,并返回余数

我认为你已经接近解决方案了。 我实施了一个简单但有效的解决方案。

import math

def square(n):
    lis = []
    
    rest = n
    while n > 0:
        rest = math.floor(math.sqrt(n))**2
        lis.append(rest)
        n-= rest
    return(lis)

你可以看到,这行math.floor(math.sqrt(n))^2是计算低于n的最接近的完美平方。然后,我将结果保存在一个列表中,并用n减去其余部分来更新n值。通过while循环重复此过程,我获得了一个包含你指定结果的最终列表。

square(12)
>>> [9, 1, 1, 1]

square(22)
>>> [16, 4, 1, 1]

square(15324)
>>> [15129, 169, 25, 1]

编辑
执行函数的提示命令。

n = float(input(" Please Enter any numeric Value : "))
square(n)

总之,如果你想有一个内部有提示的all-In函数。

import math

def square_prompt():
    n = 1
    while n>0:
        n = float(input(" Please Enter any numeric Value : "))
        if n:
            print(square(n))

square_prompt()

注:。如果您插入一个小于0的数字,您将停止循环(这意味着用户没有任何其他数字可向程序请求)。

您可以使用while循环来完成此操作:

import numpy as np


def func(a):
    lst = []
    while a: # Repeat until a reaches 0
        n = int(np.sqrt(a))**2 # Largest square number < a
        a -= n # Reduce a by n
        lst.append(n) # Append n to a list and repeat
    return lst

print(func(12))
print(func(22))
print(func(15324))

输出:

[9, 1, 1, 1]
[16, 4, 1, 1]
[15129, 169, 25, 1]

from math import sqrt

def square(n):
    lst = [] # Empty List
    while True: 
        for a in range(1,int(n)): 
            if n-a > 0:
                 if round(sqrt(n-a)) == sqrt(n-a):
                    lst.append(str((round(n-a)))) # I am check that if round of square of a number is equal to square of number because sqrt gives square for every not but I need only perfect square. Square Of 3 is 1.732 but this is a perfect square not then round(1.732) returns 2 which is not equal to 1.732.

                    n = a 
        if n==4: # This line is because 4-Anything is not a perfect square therefore I need to add a custom if statement. 
            lst.append(str(4)) # Because 4 is a perfect square of therefore I add 4 to the list.
            n = 0
            break # exit While Loop 
        if n<=4: # Because if n is 3 then I need to add 1 to the list for n times.
            break # exit While loop
    for a in range(int(n)): # add 1 to n times in lst if n is less than 4.
        lst.append('1')
    return ' '.join(lst)

def area(num):
    return num * num

number = float(input(" Please Enter any numeric Value : "))

area= square(number)

print(area)

输出:

IN:22, OUT:16 4 1 1

这是我可以向你解释的。

您可以将此问题转换为 find the closest square number 问题,因此您需要计算输入面积的平方根,并使用 math 库来计算此值,得到最接近的平方数:

import math

area = float(input(" Please Enter any numeric Value : "))
sqrt = area**0.5
closest = math.floor(sqrt)**2
print(closest, area-closest)

输出:

Please Enter any numeric Value : 12
9 3.0