题目描述 题目来源于 LeetCode 204.计数质数,简单来讲就是求"不超过整数 n 的所有素数个数". 常规思路 一般来讲,我们会先写一个判断 a 是否为素数的 isPrim(int a) 函数: bool isPrim(int a){ for (int i = 2; i < a; i++) if (a % i == 0)//存在其它整数因子 return false; return true; } 然后我们会写一个 countIsPrim(int n) 来计算不超过 n…
埃拉托色尼筛法(Sieve of Eratosthenes)是一种用来求所有小于N的素数的方法.从建立一个整数2~N的表着手,寻找i? 的整数,编程实现此算法,并讨论运算时间. 由于是通过删除来实现,而1和0则不是素数,所以从2,3,5以及其倍数删除. 用Data[]来储存所有的数,将替换好的数字存在Data[]当中 而只需做出将2,3,5以及能将这些数整除的数字替换为零:if(Data[j] % i == 0 ) Data[j]==0; 实现的代码段为: for (i = 2; i < n;…
Sieve of Eratosthenes (素数筛选算法) Given a number n, print all primes smaller than or equal to n. It is also given that n is a small number. For example, if n is 10, the output should be “2, 3, 5, 7″. If n is 20, the output should be “2, 3, 5, 7, 11, 13,…
Programming 1.3 In this problem, you'll be asked to find all the prime numbers from 1 to 1000. Prime numbers are used in allkinds of circumstances, particularly in fields such as cryptography, hashing among many others. Any method w ill be sufficient…
Given a number N, the output should be the all the prime numbers which is less than N. The solution is calledSieve of Eratosthenes: First of all, we assume all the number from 2 to N are prime number (0 & 1 is not Prime number). According to the Prim…