素数指示器指示巨大的非素数作为素数

Question

def main():
    try:
        a = int(input())
        if isinstance(a, int):
            a = int(a)
            if a ==2:
                print('YES')
            if a > 1:
                for i in range(2, a):
                    if a % i == 0:
                        print('NO')
                        break
                    else:
                        print('YES')
                        break
            else:
                print('NO')
        else:
            print('NO')
    except EOFError:
        print('NO')
    except ValueError:
        print('NO')
main()

This code is working but it's calculating huge numbers as prime number for example 3456734572525. It is successful for 4 test cases out of 5. First i thought it was failing on 0 or decimal numbers but that wasn't the case.

Answer 1

The for loop never runs to the end, since regardless of whether a % i == 0 you will break the loop. You should move your print("YES") outside of the loop and use return enstead of break to ensure that your function doesnt run this code if your loop breaks. Try this:

def main():
    try:
        a = int(input())
        if isinstance(a, int):
            a = int(a)
            if a ==2:
                print('YES')
            if a > 1:
                for i in range(2, a):
                    if a % i == 0:
                        print('NO')
                        return
                print('YES')
                return
            else:
                print('NO')
        else:
            print('NO')
    except EOFError:
        print('NO')
    except ValueError:
        print('NO')
main()

Answer 2

you're breaking out of the loop the on the first run through
say you put 9 in, think about it
it would fail the == 2 condition
then hit a % i ==0 condition and as 9%2 = 1, it would return YES

There are many errors with this program alongside that, but the main issue is breaking after the first run through, that doesnt make any sense, if you removed the break and print on the else for the if a%i==0 it would work, but there are many optimizations you could throw in there.

I reccomend taking a few courses/tutorials on python and not trying to jump in with little prior knowledge
good luck

Answer 3

Your code is actually incorrect to even begin with.
You're using 'break' inside the loop, after one 'if' check, so basically you're only checking once. So all your code does is tell if the number is divisible by 2.

try:
    a = int(input())
    flag = 0
    if isinstance(a, int):
        a = int(a)
        if a==1:
            print('Neither')
        elif a ==2:
            print('YES')
        else:
            for i in range(2, a):
                if a % i == 0:
                    flag=1
                    break
    else:
        print('NO')
except EOFError:
    print('NO')
except ValueError:
    print('NO')
if flag:
    print('NO')
else:
    print('YES')