发现任何两个数字之间的三角数

Question

我的代码表明波纹管

import math,sys 

#create a list with numbers 
def create_list(): 
    num_list=[] 
    for num in range(int(input("insert start point: ")),int(input("Insert end point: "))): 
     num_list.append(num) 
    return num_list 

#function to find triangular numbers 
def get_triangles(numlist): 
    triangles = [] 
    for i in numlist: 
     if (check_triangle(i)): 
      triangles.append(i) 
    return triangles 

#function to check number is triangular or not 
def check_triangle(n): 
    return math.sqrt((8*n)+1).is_integer() 

#function main to run the process 
def main(): 
    numlist = create_list() 
    print(get_triangles(numlist)) 

即使它看起来像任务完成它不是。我尝试了范围为0 - 100000000（1 * 10^8）的数字。这是因为我的笔记本电脑卡住了任何可以完成此任务的方法？

Answer 1

看来你只是试图在一个范围内生成所有的三角形数字。如果是这样，直接计算它们比通过平方根检查快得多。

请注意，您可以通过简单地添加连续数字来生成三角形数字。

T_0 = 0 
T_1 = T_0 + 1 = 1 
T_2 = T_1 + 2 = 3 
T_3 = T_2 + 3 = 6 
... 

如果你想保持它的简单，你可以创建一个功能，以保持从n = 0产生这些数字，让他们当他们进入期望的范围，并继续下去，直到它超过了上限。

def generate_triangular_numbers(a, b): 
    """Generates the triangular numbers in [a, b] (inclusive)""" 
    n, t = 0, 0 
    triangles = [] 

    while t < a: 
     n += 1 
     t += n 

    while t <= b: 
     triangles.append(t) 
     n += 1 
     t += n 

    return triangles 

Answer 2

不要打印一个大的表格。而是把它写入文件，这样你可以在后面打开文件。该程序无法高效地将更多的信息写入控制台。我发现在控制台上打印内容会使程序效率降低很多。

此外，我读了一些关于您的代码的评论，他们声明效率不高，我不得不同意。

这是我写的一段代码。这需要一点解释，但我很匆忙。只需回复，如果你需要帮助理解它。

def getTriangles(input1,input2): #input1 is the min value and input2 is the max value 
    li = [] #the list where all of the numbers will go 
    i = 0 #an integer that acts how much another layer of the triangle would have 
    j = 0 #the most current number that it is on 
    while True: #I whipped up this algorithm in a couple minutes, so there is probably a more efficient way than just looping through all of them, but it is faster than the current one being used 
     i += 1 #i has to increment to act as the increase of a side 
     if j > input2: #if the value that could be added is greater than the max number, than just end the function and return the list 
      return li 
     if j >= input1: #if the number qualifies the minimum number requirements, then the program will add it to the list, otherwise it will ignore it and continue on with the function 
      li.append(j) 
     j += i #this simulates adding in another layer of the triangle to the bottom 

这将是一个方式来使用它： 打印（getTriangles（1,45）） 我相信你可以看一下如何将内容写入一个文件。