使用 python 进行平方根循环时遇到问题

发布于 2024-10-27 16:47:06 字数 398 浏览 9 评论 0原文

我有一个用户输入“n”，我正在寻找平方根。我知道是 math.sqrt(n)，但我需要让程序继续查找平方根，直到它小于 2。还向用户返回程序运行了多少次，使用 a柜台。我正在使用Python。

迄今为止：

import math
root = 0
n = input('Enter a number greater than 2 for its root: ')

square_root = math.sqrt(n)
print 'The square root of', n, 'is', square_root

keep_going = 'y'

while keep_going == 'y':
    while math.sqrt(n) > 2:
        root = root + 1

原文

i have a user input 'n' and i am finding the square root. Which i know is math.sqrt(n), but i need to make the program continue find the square root until it is less than 2. also return to the user how many times the program ran to find the root less than 2 using a counter. I am using python.

So far:

import math
root = 0
n = input('Enter a number greater than 2 for its root: ')

square_root = math.sqrt(n)
print 'The square root of', n, 'is', square_root

keep_going = 'y'

while keep_going == 'y':
    while math.sqrt(n) > 2:
        root = root + 1

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海的爱人是光 2024-11-03 16:47:06

import math
user_in = input()
num = int(user_in)

num = math.sqrt(num)
count = 1
while(num > 2):
  num = math.sqrt(num)
  count += 1

print count

import math
user_in = input()
num = int(user_in)

num = math.sqrt(num)
count = 1
while(num > 2):
  num = math.sqrt(num)
  count += 1

print count

回复收藏 0 原文

北凤男飞 2024-11-03 16:47:06

假设 2.x

count = 0
user_input = int(raw_input("enter:"))
while true:    
    num = math.sqrt( user_input )
    if num < 2: break
    print num
    count+=1
print count

不存在对用户输入的错误检查。

Assuming 2.x

count = 0
user_input = int(raw_input("enter:"))
while true:    
    num = math.sqrt( user_input )
    if num < 2: break
    print num
    count+=1
print count

Error checking for user input not present.

回复收藏 0 原文

哽咽笑 2024-11-03 16:47:06

明显的方式：

def sqrtloop(n):
    x = 0
    while n >= 2:
        n **= 0.5
        x += 1
    return x

没那么多：

def sqrtloop2(n):
    if n < 2:
        return 0
    return math.floor(math.log(math.log(n, 2), 2)) + 1

Obvious way:

def sqrtloop(n):
    x = 0
    while n >= 2:
        n **= 0.5
        x += 1
    return x

Not that much:

def sqrtloop2(n):
    if n < 2:
        return 0
    return math.floor(math.log(math.log(n, 2), 2)) + 1

回复收藏 0 原文

好菇凉咱不稀罕他 2024-11-03 16:47:06

num = 0
while num < 100:
   
   num_sqrt = num ** 0.5
   print('The square root of %0.3f is %0.3f'%(num ,num_sqrt))
   
   num = num + 1
print ("Good bye!")

num = 0
while num < 100:
   
   num_sqrt = num ** 0.5
   print('The square root of %0.3f is %0.3f'%(num ,num_sqrt))
   
   num = num + 1
print ("Good bye!")

回复收藏 0 原文

~没有更多了~