PROGRAMMING:Judge whether a number is prime or not
The so-called prime number is a positive integer which can not be divided by any other integer except 1 and itself. 2 is the smallest prime number< br>
Now, the user enters an integer to determine whether it is a prime.
###Input format:
An integer n (n is int).
###Output format:
If n is a prime, output n is a prime. Otherwise, output n is not a prime.
###Input sample 1:
```in
seven
```
###Output sample 1:
```out
7 is a prime.
```
###Input sample 2:
```in
twenty-eight
```
###Output sample 2:
```out
28 is not a prime.
```
###Input sample 3:
```in
one
```
###Output sample 3:
```out
1 is not a prime.
```
###Input sample 4:
```in
-11
```
###Output sample 4:
```out
-11 is not a prime.
```
answer：If there is no answer, please comment
Obviously, the simplest way to judge whether a number is a prime number is to use the formula of prime number. First, n should be greater than 1. Secondly, let the variable I start from 2 until n-1. As long as n is divisible by I once in the cycle, it means that n is not prime. At this time, there is no need to continue to try other factors and exit the cycle directly< br>
It can be found that there are two ways to exit the cycle. One is to find the factor I that can divide N and exit the cycle by breaking. The other is to exit the cycle when I have changed to N and do not meet the cycle conditions. The former means that n is not a prime, and because it is exited by break in the middle of the loop, I must be < = n-1 at this time; The latter means that n is a prime, and I is exactly n. Therefore, you can judge how to exit the loop according to I = = n or I < = n-1, and output accordingly< br>
In fact, I only need to try from 2 to sqrt (n), and the number of cycles can be greatly reduced. At this time, the basis for judging which way the program exits from the cycle will also change accordingly.
Now, the user enters an integer to determine whether it is a prime.
###Input format:
An integer n (n is int).
###Output format:
If n is a prime, output n is a prime. Otherwise, output n is not a prime.
###Input sample 1:
```in
seven
```
###Output sample 1:
```out
7 is a prime.
```
###Input sample 2:
```in
twenty-eight
```
###Output sample 2:
```out
28 is not a prime.
```
###Input sample 3:
```in
one
```
###Output sample 3:
```out
1 is not a prime.
```
###Input sample 4:
```in
-11
```
###Output sample 4:
```out
-11 is not a prime.
```
answer：If there is no answer, please comment
Obviously, the simplest way to judge whether a number is a prime number is to use the formula of prime number. First, n should be greater than 1. Secondly, let the variable I start from 2 until n-1. As long as n is divisible by I once in the cycle, it means that n is not prime. At this time, there is no need to continue to try other factors and exit the cycle directly< br>
It can be found that there are two ways to exit the cycle. One is to find the factor I that can divide N and exit the cycle by breaking. The other is to exit the cycle when I have changed to N and do not meet the cycle conditions. The former means that n is not a prime, and because it is exited by break in the middle of the loop, I must be < = n-1 at this time; The latter means that n is a prime, and I is exactly n. Therefore, you can judge how to exit the loop according to I = = n or I < = n-1, and output accordingly< br>
In fact, I only need to try from 2 to sqrt (n), and the number of cycles can be greatly reduced. At this time, the basis for judging which way the program exits from the cycle will also change accordingly.