A C language programming every day: finding prime numbers within N (common method + optimization method)

topic description

Find the prime numbers in N.

input format

N

output format

Prime numbers from 0 to N

sample input

100

sample output

2
3
5
7
11
13
17
19
23
29
31
37
41
43
47
53
59
61
67
71
73
79
83
89
97

Ordinary method idea:
Traverse the numbers within 2~N, and then traverse the one-digit number i in 2~i, see if the number in 2~i can be divisible by i, count by count, if count=0, it means that there is no number that can be divided by i i divisible number, then this number is a prime number, the code is as follows:

#include<stdio.h>

int main()

{

//素数就是除一和本身外没有其他的数能与它本身整除 

    int i,j,n,count;//count用来记录可以和i整除的数 

    scanf("%d",&n);

    for(i=2;i<=n;i++)

    {

    count=0;//注意每次都要重新赋零值，不然后面判定就错了 

    for(j=2;j<i;j++)

    {

    if(i%j==0) count++;//如果i是素数的话count应该是零的 

}

if(count==0) printf("%d\n",i);//如果是素数就就输出了记得要加换行符 

}

    return 0;

}

Optimization method idea : A method I saw on the Internet is a new idea for me, and I will share it with you here

• Any natural number N greater than 1, if N is not a prime number, then N can be uniquely decomposed into a product of finite prime numbers

•So if prime[i]==0, then prime[i*j]==1, that is, prime[i*j] is not a prime number

Note:

prime[2]==0 means 2 is a prime number prime[8]==1 means 8 is not a prime number

prime[0]=prime[1]=1; //0 and 1 need special handling 

code show as below

#include<stdio.h>
int main(){
     int prime[10000]={0};
     int i,j;
     int n;
  
     scanf("%d",&n);
  
     prime[0]=prime[1]=1;
  
     for(i=2;i<n;i++)
         if(prime[i]==0)
               for(j=2;i*j<=n;j++)
                    prime[i*j]=1;
  
     for(i=0;i<n;i++)
          if(prime[i]==0)
               printf("%d\n",i);
  
     return 0;
}