def getPythagoreanlist (num) : # this function time complexity are not of the two should be run too slow
"" "number 3 is the relationship between the Pythagorean theorem, a2 + b2=c2, find out within 30 (c<=30) all satisfy this relation triples, and as a tuple to join a list,
"" "
Ret=[] # return value definition
If num & gt;=3:
For I in range (3, num + 1) : # I set to triple in maximum number
For j in range (1, I) : # j set to triple the minimum number of
X=(I * I) - (j * j) # with larger squared minus Numbers
Y=int (x 0.5) * * # number of big and minimize the arithmetic square root of the number of integer
If y * y==x and y & gt; J:
# print (j, y, I)
Ret. Append ((j, y, I))
Return ret


If __name__=="__main__" :
R=getPythagoreanlist (30) # you specified 30
Print (r) # a print
M=int (input (" please enter the largest number: "))
R=getPythagoreanlist (m) # read input number
Print (" eligible all triples: ", R)
# truth to you, take no thanks ~

thank you very much

CodePudding user response:

refer to the second floor Hajo_ response:

Quote: refer to the original poster and sank Milky Way you reply:

number 3 is the relationship between the Pythagorean theorem, a2 + b2=c2, find out within 30 (c<=30) all satisfy this relation triples, and as a tuple to join a list,

 
List_a=list () 
For I in range (1, 31) : 
For j in range (1, 31) : 
For k in range (1, 31) : 
If I j * * * * 2 + 2=k * * 2: 
List_a. Append ((I, j, k)) 
Print (' meet the Pythagorean theorem: within 30/n {list_a} ') 

For reference, this also can be optimized, but this is a simple approach

Why is running out to meet within 30 of the Pythagorean theorem:/n {list_a}