平方根化简功能

Question

我正在尝试创建一个函数，将数字简化为根下的整数。例如：

• √(27) = 3 √(3)
• √(567) = 9 √(7)

我编写了以下函数，但它只能部分工作，因为我认为 while 循环无效。 请问有什么帮助吗？

def sqrt2(num):
    numbers=[]
    roots=[]
     #i=1-11
    while num>1:    
        for i in range(1,num+1):
            if num%i==0:    
                num=num//i  
                if num%i==0: 
                    num=num//i
                    numbers.append(i)
                else:
                    roots.append(i)
        break

    result=1
    for i in numbers:
        result= result*i
    u_root=1
    for j in roots:
        u_root=u_root*j

    print (result,"sqrt (",u_root,")")  

Answer 1

此循环只是尝试除以 2*2、3*3、4*4 等，直到找到除数或直到除数对于除数而言太大。

唯一有趣的部分是递归。如果找到部分结果，我们会尝试简化较小的数字。例如。 567 = (3*3)*63，然后 63 = (3*3)*7。这两个结果相加得出 (9*9)*7。

def my_sqrt(num):
    for i in range(2, num):
        div, mod = divmod(num, i*i)
        if mod == 0:
            sq1, sq2 = my_sqrt(div)
            return (i * sq1, sq2)
        if div == 0:
            break
    return (1, num)

print(my_sqrt(27))
print(my_sqrt(567))

Answer 2

这是我为数学作业编写的小代码

def sqr_sim(und_root):
    und_root = int(und_root)
    und_root0 = round(und_root)
    rt_fc = []
    coef = 1
    if und_root < 0:
        return None
    elif und_root == 0:
        return 0
    else:
        for i in range(2, und_root0):
            if und_root%(i**2) == 0:
                rt_fc.append(i)
                und_root /= i**2

                for i0 in range(2, und_root0):
                    if und_root%(i0**2) == 0:
                        rt_fc.append(i0)
                        und_root /= i0**2

        for ele in rt_fc:
            coef *= ele
        print('the final solution is',
              coef, '√', und_root)

sqr_sim(input())

Answer 3

如果您的缩进实际上如问题中所示，则您的

break

语句需要进一步缩进。当第一个

for

被击中时，它应该脱离

if

循环，而不是脱离

while

循环。

但是代码还有另一个问题。您将开始

for

循环，其中

i

为

1

，但在这种情况下

num % 1

将始终为 true，从而导致无限循环。您应该跳过

i=1

并循环

range(2, num+1)

:

while num>1:    
    for i in range(2,num+1):  # start range with 2
        if num%i==0:    
            num=num//i  
            if num%i==0: 
                num=num//i
                numbers.append(i)
            else:
                roots.append(i)
            break  # increase indent on this line by two levels!

Answer 4

#created 9/24/24
#without spaces the code is 26 lines long

#You enter a number and root you want to do to it
#example (number : 64, root : 2) = 8 sqr(1) 
# while (number : 64, root : 5) = 2 sqr(2)

import math

Factors = [] # holds the factors
originalNumber = int(input("enter a number: ")) # enter the number
rootAmount = int(input("enter the root: ")) # enter the root
calculating = True # used for while loop to get factors

#This program can't do imaginary numbers
if originalNumber <= 0:
    print("Enter a number more than 0")
    exit()


def getFactors(number): #each time this is run it gets one factor of the number
    for i in range(2, number):
        if number % i ** rootAmount == 0:
            Factors.append(i)
            number /= (i ** rootAmount)
            return(number) # gets the new number to factor
        if i ** rootAmount > number: # allows the program to run faster by using roots
            Factors.append(number)
            return(-1) #tells the program the number is fully factored
    Factors.append(number) #adds the last factor
    return(-1) #tells the program the number is fully factored

while calculating: # since getFactors only gets one factor, I have to keep looping it.
    originalNumber = getFactors(int(originalNumber))
    if originalNumber == -1: #tells the program to stop
        root = Factors[-1] #the last factor goes in the root
        del Factors[-1]
        print(math.prod(Factors), u"\u221a", root)
        calculating = False # stops the program

Answer 5

def simplifyRadical(number, root):
    biggestPerfectRoot = None
    originalNumber = number

    for i in range(2, number):
        if (number % i ** root == 0):
            biggestPerfectRoot = i #holds the biggest root that fits into the number
        elif (i ** root > originalNumber):
            if biggestPerfectRoot == None: #tells if the number is prime or not
                biggestPerfectRoot = 1
            break
    return(biggestPerfectRoot, "\u221a", number / biggestPerfectRoot ** root)