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  • 质数筛选——素数筛选法

    质数筛选

    最终算法模板,比赛就写这个!!

    int prime[100010];
    
    void Prime(){
        for (int i = 2; i <= 1000000; i++) {
            prime[i] = true;
        }
        for (int i = 1; i * i <= 1000000; i++) {
            if (prime[i]) {
                for (int j = i * i; j <= 1000000; j += i) {
                    prime[j] = false;
                }
            }
        }
    }
    

    或者写下面这个:线性筛

    #define SIZE 1000000
    int main()
    {
        int check[SIZE] = {0};//元素值为0代表是素数
        int prime[SIZE] = {0};
        int pos=0;
        int flag;
        for (int i = 2 ; i < SIZE ; i++)
        {
            if (!check[i])//如果是素数
                prime[pos++] = i;
    
            for (int j = 0 ; j < pos && i*prime[j] < SIZE ; j++)
            {
                check[i*prime[j]] = 1;//筛掉
                //标注一
                if (i % prime[j] == 0)
                    break;
            }
        }
    
        printf("%.2f", (double)clock()/CLOCKS_PER_SEC);
    
        return 0;
    }
    
    下面是具体解释,可以不看,比赛套上面模板即可!!!
    
    ![](https://img2018.cnblogs.com/blog/1454456/201811/1454456-20181119214228044-1809688019.png)
    ###如何判断一个数是不是质数?简单方法一:
    ![](https://img2018.cnblogs.com/blog/1454456/201811/1454456-20181119215344441-1163614886.png)
    
        int is_prime(int n) {
            for (int i = 2; i * i <= n; ++i) {
                if (n % i == 0) {
                    return 0; // 不是质数
                }
            }
            return 1; // 是质数
        }
    
    ##素数筛选算法:
    ![](https://img2018.cnblogs.com/blog/1454456/201811/1454456-20181119220344796-975079526.png)
    
    ####算法一:时间复杂度---- O(nlogn)
    ![](https://img2018.cnblogs.com/blog/1454456/201811/1454456-20181119220352279-1219400795.png)
    
    	for (int i = 2; i <= n; ++i) {
    	    is_prime[i] = 1;
    	}
    	for (int i = 2; i <= n; ++i) {
    	    for (int j = i * 2; j <= n; j += i) {
    	        is_prime[j] = 0;
    	    }
    	}
    
    ####算法二优化:
    ![](https://img2018.cnblogs.com/blog/1454456/201811/1454456-20181119220423117-1867892215.png)
    
    	for (int i = 2; i <= n; ++i) {
        	    is_prime[i] = 1;
    	}
    	for (int i = 2; i * i <= n; ++i) {
        	    if (is_prime[i]) {
            	for (int j = i * i; j <= n; j +=i) {
               	  is_prime[j] = 0;
            	}
       	     }
    	}
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  • 原文地址:https://www.cnblogs.com/fisherss/p/9986126.html
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