Efficient Determination of prime numbers and Determination of prime numbers

Efficient Determination of prime numbers and Determination of prime numbers

Twin Prime Number: the so-called twin prime number refers to the adjacent prime number with an interval of 2, and the distance between them is no longer near.

If n is ≥6 and N-1 and n + 1 are twin prime numbers, n must be a multiple of 6. Proof: When N-1 and n + 1 are prime numbers, then there are numbers between them, where ① between N-1 and n + 1 are odd numbers, where n is an even number, that is to say, n is a multiple of 2, where n is not a multiple of 3. n = 3x + 1 or n = 3x + 2. If n = 3x + 1, n-1 = 3x, which is contrary to ①, so n = 3x + 1; if n = 3x + 2, n + 1 = 3 (x + 1), and ①, therefore, n = 3x + 2; n is a multiple of 3, and n is a multiple of 6.
From the above law, we can draw the following conclusion: If x contains 1 and n = 6x-1 or n = 6x + 1 is not a prime number, n must not be a multiple of 2 and 3. Proof: Limit n = 6x-1 or n = 6x + 1, that is, n = 2 (3x)-1 or n = 2 (3x) + 1 or n = 3 (2x) -1 or n = 3 (2x) + 1. Limit n must not be a multiple of 2 and 3.
Regular Expression of prime numbers: WHEN n reaches 5, if n is a prime number, n mod 6 = 1 or n mod 6 = 5, that is, n must appear in 6x (x ≥ 1). Proof: when x is greater than or equal to 1, the following representation is provided: 2017100006x, 6x + 1, 6x + 2, 6x + 3, 6x + 4, 6x + 5, 6 (x + 1), 6 (x + 1) + 1. The number of rows not on both sides of 6x is 6x + 2, 6x + 3, 6x + 4, that is, 2 (3x + 1), 3 (2x + 1), 2 (3x + 2), they must not be prime numbers, so prime numbers must appear on both sides of 6x.

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #define LL long long  6 using namespace std; 7 int n,m,p; 8 bool judge(int num) 9 {10     if(num==2||num==3)11     return true;12     13     if(num%6!=1&&num%6!=5)14     return false;15     16     for(int i=5;i*i<=num;i+=6)17         if(num%i==0||num%(i+2)==0)18         return false;19     20     return true;21 }22 int main()23 {24     scanf("%d%d",&n,&m);25     for(int i=1;i<=m;i++)26     {27         scanf("%d",&p);28         if(judge(p)&&p!=1)printf("Yes\n");29         else printf("No\n");30     }31     return 0;32 }