Eratosthenes筛

什么是Eratosthenes筛？ (What is Sieve of Eratosthenes?)

Sieve of Eratosthenes is an ancient algorithm of finding prime numbers for any given range. It's actually about maintaining a Boolean table to check for corresponding prime no.

Eratosthenes的筛子是一种古老的算法，可以找到任何给定范围的质数。实际上，这是关于维护布尔表以检查相应的素数。

算法 (Algorithm)

The algorithm is pretty simple.

该算法非常简单。

Say, we need information up to integer N.
假设我们需要最大为N的信息 。
Create a table of size N+1, sieve[N+1]
创建一个大小为N + 1的表，筛子[N + 1]
Assign all the table entries TRUE initially.
最初将所有表条目分配为TRUE 。
Base case:
基本情况：

0 and 1 is not prime
0和1不是素数

Thus
从而

sieve[0]=FALSE=sieve[1]
sieve [0] = FALSE = sieve [1]

For(i=2; i*i<=N; i=i+1)
IF(sieve[i] is TRUE) //i is prime
Make all multiple ofi FALSE since they are not prime
sieve[j]=FALSE where j is multiple of i within range N
END IF
END FOR

Table is built. Now to check any integer K whether prime or not
表已建立。现在检查任何整数K是否为素数

Return
返回

sieve[k]
筛[k]

Example with explanation:

带有说明的示例：

    Let's say our range is up to 20
Thus initially declare sieve[21] and all are true.
sieve[0]=FALSE
sieve[1]=FALSE
------------------------------------------------------------
sieve[2]=true
Thus all multiple of 2, needed to be false
(Of course they are not prime since divisible by 2)
Thus,
sieve[4]=FALSE
sieve[6]=FALSE
sieve[8]=FALSE
sieve[10]=FALSE
sieve[12]=FALSE
sieve[14]=FALSE
sieve[16]=FALSE
sieve[18]=FALSE
sieve[20]=FALSE
------------------------------------------------------------
sieve[3]=true
Thus all multiple of 3, needed to be false
(Of course they are not prime since divisible by 3)
Thus,
sieve[6]=FALSE (it was already false though)
sieve[9]=FALSE
sieve[12]=FALSE(it was already false though)
sieve[15]=FALSE
sieve[18]=FALSE(it was already false though)
------------------------------------------------------------
Sieve[4]=FALSE (Nothing to do more with it)
------------------------------------------------------------
sieve[5]=TRUE
Thus all multiple of 5, needed to be false
(Of course they are not prime since divisible by 5)
Thus,
sieve[10]=FALSE (it was already false though)
sieve[15]=FALSE (it was already false though)
sieve[20]=FALSE (it was already false though)
In this way, after completing all possible entries we can find
Only true entries are
sieve[2]
sieve[3]
sieve[5]
sieve[7]
sieve[11]
sieve[13]
sieve[17]
sieve[19]
Others are false.

So now for any given number with in the N

所以对于N中的任何给定数

We can tell whether the number is prime or not in O(1) time (based on corresponding TRUE/FALSE value).

我们可以在O(1)时间内判断数字是否为质数(基于相应的TRUE / FALSE值)。

何时在编程中使用此算法？ (When to use this algorithm in programming?)

Sieve of Eratosthenes can be very efficient for any program we require to check prime numbers for multiple times (Multiple test cases too).

Eratosthenes筛网对于我们需要多次检查质数的任何程序(也可以是多个测试用例)来说都是非常有效的。

In such cases, we construct the Sieve of Eratosthenes only single time and for all query each only takes O(1) time reducing the time complexity overall.

在这种情况下，我们只构造一次Eratosthenes筛，对于所有查询，每个筛仅花费O(1)时间，从而降低了总体时间复杂度。

C++ implementation:

C ++实现：

#include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cout<<"Enter max range to check for prime numbers\n";
cin>>n;
bool sieve[n+1];
//sieve of eratosthenes
memset(sieve,true,sizeof(sieve));//initially all true
sieve[0]=sieve[1]=false;//0,1 not prime
for(int i=2;i*i<=n;i++)
if(sieve[i])
for(int j=i*2;j<=n;j+=i)//for multiple of i
sieve[j]=false;//can't be prime
cout<<"Enter non-negative integer to check prime or not\n";
cout<<"Negative no to exit\n";
int k;
cin>>k;
while(k>=0){
if(sieve[k])
cout<<k<<" is prime\n";
else
cout<<k<<" is not prime\n";
cout<<"Try more or exit\n";
cin>>k;
}
cout<<"Exited\n";
return 0;
}

Output

输出量

Enter max range to check for prime numbers
1000
Enter non-negative integer to check prime or not
Negative no to exit
997
997 is prime
Try more or exit 
993
993 is not prime 
Try more or exit 
13 
13 is prime 
Try more or exit 
103
103 is prime
Try more or exit 
111
111 is not prime 
Try more or exit 
-1 
Exited