How can I optimize Largest Prime factor in given Number? - python-3.x

Question

Constraints
1<=T<=10
10<=N<=10^2
Input
2
10
17
Output
5
17
This is my code
n=int(input())
for f in range(n):
b=[]
a=int(input())
for i in range(1,a+1):
if i>1:
for j in range(2,i):
if (i%j)==0:
break
else:
if a%i==0:
b.append(i)
print(max(b))
Explanation
10 are {2,5}, so answer 5
17 is 17 itself.

Answer 1

In the Constraints,
if T means there are at most 10 test cases,
and N means the range of the given number is between 10 to 100,
One of the most optimized ways is to create an initialized list in the code and print the pre-computed answer for every input read. :)
answer = [0, 1, 2, 3, 2, 5, 3, 7, 2, 3,
5, 11, 3, 13, 7, 5, 2, 17, 3, 19,
5, 7, 11, 23, 3, 5, 13, 3, 7, 29,
5, 31, 2, 11, 17, 7, 3, 37, 19, 13,
5, 41, 7, 43, 11, 5, 23, 47, 3, 7,
5, 17, 13, 53, 3, 11, 7, 19, 29, 59,
5, 61, 31, 7, 2, 13, 11, 67, 17, 23,
7, 71, 3, 73, 37, 5, 19, 11, 13, 79,
5, 3, 41, 83, 7, 17, 43, 29, 11, 89,
5, 13, 23, 31, 47, 19, 3, 97, 7, 11, 5]

Answer 2

Create a list of primes under 100.
Run a loop for each 'i' starting from 'a' to 'a/2-1' only, and check if 'i' divides 'a' completely and is present in 'primes'.
primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
n=int(input())
for f in range(n):
a=int(input())
for i in range(a,int(a/2-1),-1):
if(a%i==0 and i in primes):
print(i)
break
If you dont want to create list of primes manually, you can use this first to create primes list through for loop:
import math
def isPrime(n):
if (n <= 1):
return False
#check from sqrt(n) to 2
for i in range(int(math.sqrt(n)),1,-1):
if (n % i == 0):
return False
return True
primes = []
for i in range(100):
if(isPrime(i)):
primes.append(i)
n=int(input())
for f in range(n):
a=int(input())
for i in range(a,int(a/2-1),-1):
if(a%i==0 and i in primes):
print(i)
break