Ordinary screen prime number and linear screen Prime Number

General screening method:

Normal screening method: const int maxn = 70000; void prime () // screening method for prime number table P [I] = 1 is a prime number, similar to hash ing! {Int I, j; for (I = 0; I <maxn; I ++) prime [I] = 1; prime [0] = prime [1] = 0; for (I = 2; I <maxn; I ++) {If (! Prime [I]) continue; For (j = I * 2; j <maxn; j + = I) prime [J] = 0 ;}}

Linear screening method:

Understanding if (I % prime [J] = 0) is the key.

Principle:
1. Any sum can represent the product of a prime number and a number.
2. Assume that A is a combination of a and a = x * Y. Here, X is also a combination, so there are:
A = x * Y; (assume that Y is a prime number and X is a combination of numbers)
X = a * B; (assume that A is a prime number, and a <X --> A <Y)
-> A = a * B * Y = A * z (Z = B * Y)
That is, a sum (X) and a prime number (Y) Can be expressed as a larger sum (z) and a smaller prime number (A) Product !!!!
This is also the key to understanding if (I % primes [J] = 0) break; in the code.
For example, if I = 8; then because I % 2 = 0; therefore, for I = 8, you only need to check primes [1] = 2, because for prime numbers greater than primes [1], such as 3, there are:
8*3 = 2*4*3 = 12*2
That is to say, 24 (8*3 = 24) does not need to be checked at 8 hours.

Linear screening method: # include <iostream> using namespace STD; const int n = 200000; long prime [N] = {0}, num_prime = 0; // num_pirme records the number of prime numbers int main () {int m; CIN> m; int A [n] = {1, 1}, I, j; for (I = 2; I <m; I ++) {If (! A [I]) prime [num_prime ++] = I; for (j = 0; j <num_prime & I * prime [J] <m; j ++) {A [I * prime [J] = 1; // when the sum mark is 1, prime [J] is the minimum prime factor if (! (I % prime [J]) // obtain the break by multiplying the product of a prime number with a larger sum with a smaller prime number ;}} for (I = 0; I <num_prime; I ++) // output {if (I % 10 = 0) printf ("\ n "); printf ("% 3d", prime [I]);} printf ("\ n"); Return 0 ;}