c语言求素数以及改进算法

Posted 2022-06-28 passedbylove

代码需要使用c99编译

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
//是否为素数
//从2到x-1测试是否可以整除 
//时间复杂度O(n-2)，n趋向正无穷
int isPrime(int x)

    int ret = 1;
    for(int i = 2; i < x; i++)    
    
        if(x % i == 0)
        
            ret = 0;
            break;
        
    
    return ret;


//除了2以外，所有的偶数都不是素数，从3到x-1，每次加2 
//x为偶数时间复杂度O((n-3)/2+1)
//x很大时时间复杂度接近于(n/2）
int isPrime2(int x)

    int ret = 1;
    if(x == 1 || (x % 2 ==0 && x != 2))
        ret = 0;
        
    for (int i = 3; i < x; i += 2)
    
        if( x % i == 0)
        
            ret = 0;
            break;
        
    
    
    return ret;


//无须到x-1,到sqrt(x) 
//时间复杂度O(Log2n) 即sqrt(x) 
int isPrime3(int x)

    int ret = 1;
    
    if(x == 1 || x % 2 == 0 && x != 2)
        ret = 0;
    for(int i = 3 ; i < sqrt(x); i += 2)
    
        if( x % i == 0)
        
            ret = 0;
            break;
        
    
    
    return ret;


//判断是否能被已知的且<x的素数整除
int isPrime4(int x, int knownPrimes[], int numberOfKnownPrimes)

    int ret = 1;
    int i;
    
    for( i = 0; i < numberOfKnownPrimes; i++)
    
        if (x % knownPrimes[i] == 0 )
        
            ret = 0;
            break;
        
    
    
    return ret;

 
int main()

    const int number = 100;
    int prime[number] = 2;
    int count = 1;
    int i = 3;
    
    while( count < number )
    
        if(isPrime4(i ,prime, count))
        
            prime[count++] = i;
        
        i++;
    
    
    for( i = 0; i < number; i++)
    
        printf("%d",prime[i]);
        if( (i + 1) % 5 ) printf("\t");
        else printf("\n");
    
    return 0;