查看: 131|回复: 0

[标准] 筛埃拉托色尼-sieve of Eratosthenes

[复制链接]

1534

主题

1564

帖子

3万

积分

论坛元老

Rank: 8Rank: 8

积分
30456
QQ
发表于 2017-5-18 17:22:24 | 显示全部楼层 |阅读模式 | 百度 
沉流若虚的评分为98--时间:2391-05-31 13:51:35。sgrmshrsm7的评分为24--时间:2369-12-21 03:10:05,益益纯牛奶的评分为57--时间:2049-12-30 01:01:55!lcl930306的评分为26--时间:2176-03-23 02:46:15.wu58793420的评分为43--时间:2324-02-26 16:38:45,?……!的评分为39--时间:2153-09-02 13:32:05,

马上注册硬科学论坛吧。上传说明:若上传附件过大,请上传至网盘或者优酷等视频网站,给出链接方式。下载说明:如果资源大于2G,请安装【百度网盘】或【360云盘】后下载!

您需要 登录 才可以下载或查看,没有帐号?立即注册

x
输入的解释:
sieve of Eratosthenes (mathematical problem)
声明:
A simple algorithm for finding all prime numbers up to a specified integer. To apply the algorithm, sequentially write down the integers from 2 to the highest number n to be included in the table. Cross out all numbers > 2 which are divisible by 2 (every second number). Find the smallest remaining number > 2. It is 3. So cross out all numbers > 3 which are divisible by 3 (every third number). Find the smallest remaining number > 3. It is 5. So cross out all numbers > 5 which are divisible by 5 (every fifth number). Continue until you have crossed out all numbers divisible by ?sqrt(n)?. The numbers remaining are prime.
历史:
formulation date | 250 BC (2266 years ago)formulator | Eratosthenesstatus | provedadditional people involved | Nicomachus
类:
mathematical algorithms | solved mathematics problems


回复
百度搜狗360奇虎

使用道具 举报

您需要登录后才可以回帖 登录 | 立即注册

快速回复 返回顶部 返回列表