4739. Sieve of Eratosthenes

Given the values of a and b, print all primes in the interval from a to b inclusively.

Input. Two integers a and b (1 ≤ a ≤ b ≤ 100000).

Output. Print in one line all prime numbers in the interval from a to b inclusively.

Sample input	Sample output
1 10	2 3 5 7

SOLUTION

sieve of Eratosthenes

Algorithm analysis

Using the algorithm Sieve of Eratosthenes, fill the primes array, where

· primes[i] = 1, if i is prime;

· primes[i] = 0, if i is composite;

Print all numbers i in the interval from a to b, for which primes[i] = 1.

Example

The filled array primes  has the form:

The prime numbers in the interval [1; 10] are 2, 3, 5, 7.

Algorithm realization

Declare the working array primes.

#define MAX 100001

int primes[MAX];

The gen_primes function implements the Sieve of Eratosthenes and fills the primes array. The numbers 0 and 1 are not prime.

void gen_primes(void)

{

int i, j;

for (i = 0; i < MAX; i++) primes[i] = 1;

primes[0] = primes[1] = 0;

for (i = 2; i * i < MAX; i++)

if (primes[i])

for (j = i * i; j < MAX; j += i) primes[j] = 0;

}

The main part of the program. Fill the primes array.

gen_primes();

Read the input data.

scanf("%d %d", &a, &b);

Print all primes in the interval from a to b.

for (i = a; i <= b; i++)

if (primes[i]) printf("%d ", i);

printf("\n");

Algorithm realization – bitset

Let’s declare a variable primes of type bitset to store information about prime numbers.

#define MAX 100001

bitset<MAX> primes;

The gen_primes function implements the Sieve of Eratosthenes. We fill the primes bitset. The numbers 0 and 1 are not prime.

void gen_primes(void)

{

primes.flip(); // set all to 1

primes.reset(0); primes.reset(1);

for (int i = 2; i <= sqrt(MAX); i++)

if (primes.test(i))

for (int j = i * i; j < MAX; j += i) primes.reset(j);

}

The main part of the program. Fill the primes array.

gen_primes();

Read the input data.

scanf("%d %d", &a, &b);

Print all primes in the interval from a to b.

for (i = a; i <= b; i++)

if (primes.test(i)) printf("%d ", i);

printf("\n");

Java realization

import java.util.*;

public class Main

{

final static int MAX_SIZE = 100001;

static BitSet bits = new BitSet(MAX_SIZE);

public static void Eratosthenes()

{

bits.set(2, MAX_SIZE, true);

for (int i = 2; i < Math.sqrt(MAX_SIZE); i++)

{

if (bits.get(i))

for (int j = i * i; j < MAX_SIZE; j += i)

bits.set(j, false);

}

}

public static void main(String args[])

{

Scanner con = new Scanner(System.in);

Eratosthenes();

int a = con.nextInt();

int b = con.nextInt();

for (int i = a; i <= b; i++)

if (bits.get(i)) System.out.print(i + " ");

System.out.println();

con.close();

}

}

Python realization

The seive_of_eratosthenes function implements the Sieve of Eratosthenes and fills the is_prime list. The numbers 0 and 1 are not prime.

def seive_of_eratosthenes(n):

is_prime = [True] * (n+1)

is_prime[0], is_prime[1] = False, False

for i in range(2, int(n ** 0.5) + 1):

if is_prime[i]:

for j in range(i * i, n + 1, i):

is_prime[j] = False

return is_prime

The main part of the program. Fill the prime list.

prime = seive_of_eratosthenes(100000)

Read the input data.

a, b = map(int, input().split())

Print all primes in the interval from a to b.

for i in range (a, b + 1):

if prime[i]: print(i,end=' ')