Sieve method, sieve method

Sieve method, sieve method

2016.1.25

 

We all know that determining whether a number is a prime number can be solved within the time complexity of O (√ n. However, if the number of prime numbers in [1, n] is required, it is obviously a little slow to judge one by one.

But we know an obvious property: all the prime factors of a given number are smaller than this sum, and there is no prime factor smaller than a prime number.

Then, we can use the obtained prime factor to compare the total number of bins to perform sieve division.

The specific operation is as follows: first, we will write down all the numbers in 2 to n. 2 is the decimal number in the table, that is, the prime number, and then all the multiples of 2 are excluded. At this time, the remaining minimum number in the table is 3, that is, 3 is the prime number, and then the multiples of all three in the table are crossed. After this operation is performed, the smallest number in each table is a prime number and its multiples are all crossed.

 

	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20
Multiple of sieve 2	2	3	-	5	-	7	-	9	-	11	-	13	-	15	-	17	-	19	-
Multiple of 3	2	3	-	5	-	7	-	-	-	11	-	13	-	-	-	17	-	19	-

The Code is as follows:

Int n, prime [1000005], e; bool is_prime [1000005]; // If is_prime is 0, void solve () // prime screening {prime [0] = prime [1] = 1; for (int I = 2; I <= n; I ++) {if (! Is_prime [I]) {prime [++ e] = I; if (long) I * I> n) continue; // prevent int for (int j = I * I; j <= n; j + = I) is_prime [j] = 1 ;}}}View Code

The time complexity of this screening method is not obvious. It looks like O (n), and the actual measurement is indeed close to linear under noip data scale, however, his time complexity is actually O (n loglogn), which seems to be a linear constant with a slight increase.

The author does not understand the time complexity of this strange story, but posted it here. If there is a better proof, he can tell me the story.

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