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Description: Eratosthenes筛法求素数 -Eratosthenes'sieve Method Primes
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Size: 1143 |
Author: 朱亮 |
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Description: 一个素数筛法程序
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Size: 6658 |
Author: 郭凯图 |
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Description: 用筛法求素数的小程序。用Visual C++实现
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Size: 239070 |
Author: 求索 |
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Description: 判断质数的改进方法。-judgment few quality improvements.
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Size: 8985 |
Author: 黄应东 |
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Description: 快速质数表算法
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Size: 397 |
Author: blacksapper |
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Description: 快速质数表算法
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Size: 743 |
Author: blacksapper |
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Description: Eratosthenes筛法求素数 -Eratosthenes'sieve Method Primes
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Size: 1024 |
Author: 朱亮 |
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Description: 判断质数的改进方法。-judgment few quality improvements.
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Size: 9216 |
Author: 黄应东 |
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Description: 一个素数筛法程序-A prime number sieve method procedures
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Size: 6144 |
Author: 郭凯图 |
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Description: 用筛法求素数的小程序。用Visual C++实现-Sieve method using prime number of small procedures. Using Visual C++ Realize
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Size: 432128 |
Author: 求索 |
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Description: 筛法筛素数，结构很好
效率也不错！算法思想很好！-Sieve Method sieve prime numbers, the structure of the efficiency of good very good! Algorithm good thinking!
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Size: 32768 |
Author: 爱因斯坦 |
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Description: Euler函数:
m = p1^r1 * p2^r2 * …… * pn^rn ai >= 1 , 1 <= i <= n
Euler函数：
定义：phi(m) 表示小于等于m并且与m互质的正整数的个数。
phi(m) = p1^(r1-1)*(p1-1) * p2^(r2-1)*(p2-1) * …… * pn^(rn-1)*(pn-1)
= m*(1 - 1/p1)*(1 - 1/p2)*……*(1 - 1/pn)
= p1^(r1-1)*p2^(r2-1)* …… * pn^(rn-1)*phi(p1*p2*……*pn)
定理：若(a , m) = 1 则有 a^phi(m) = 1 (mod m) 即a^phi(m) - 1 整出m
在实际代码中可以用类似素数筛法求出
for (i = 1 i < MAXN i++)
phi[i] = i
for (i = 2 i < MAXN i++)
if (phi[i] == i)
{
for (j = i j < MAXN j += i)
{
phi[j] /= i
phi[j] *= i - 1
}
}
容斥原理：定义phi(p) 为比p小的与p互素的数的个数
设n的素因子有p1, p2, p3, … pk
包含p1, p2…的个数为n/p1, n/p2…
包含p1*p2, p2*p3…的个数为n/(p1*p2)…
phi(n) = n - sigm_[i = 1](n/pi) + sigm_[i!=j](n/(pi*pj)) - …… +- n/(p1*p2……pk)
= n*(1 - 1/p1)*(1 - 1/p2)*……*(1 - 1/pk)
-err
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Size: 1024 |
Author: Campbell |
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Description: 计算百亿内素数个数的效率极高的算法，采用动态分配内存和倒筛法，文件夹内附有报告。-Calculating the number of prime numbers in tens of billions of extremely efficient algorithms, dynamic allocation of memory and inverted sieve method, the folder containing the report.
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Size: 81920 |
Author: slh |
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Description: 筛法求素数。
用筛法求给定范围内的素数。
-Prime number sieve method
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Size: 159744 |
Author: sy |
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Description: 一是利用计算机对正整数进行素因子分解认识素数，通过运用筛法与试除法寻找一定范围内的素数，体会到筛法比试除法有效但无法找出所有素数；二是通过若干特殊形式的素数的研究，寻找素数生成公式，发现一些结论；三是借助于非线性拟合技术，观察素数分布规律，利用其研究有关数论问题。-The first computer was used to prime factor decomposition of positive integers understanding of prime numbers, through the use of screening method and trial division to find prime numbers within a certain range, the division experienced a test of an effective screening method but can not identify all the prime numbers second is through a number of special form of prime numbers research, looking for prime number generation formula, found that some of the conclusions third, by means of non-linear fitting technique to observe the distribution of prime numbers, using their research on number theory problems.
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Size: 223232 |
Author: sean |
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Description: 分解质因数。先用筛法筛出素数，然后求解。-Sift the quality factor decomposition
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Size: 6144 |
Author: 张晓媛 |
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Description: 一些数学问题算法模板，稍微做了下分类
目录
1 欧几里得和模线性方程
欧几里德
扩展欧几里德
模线性方程
2 欧拉φ函数
扩展欧拉函数
3 线性时间素数筛法
生成欧拉函数表
4 因式分解
5 Nim游戏和sg函数
合肥B 题 -Some mathematical problems algorithm template, do a little under
Categories
1and mode of linear equations Euclid Euclid
1-mode extended Euclidean Euler linear equation
extended Euler function φ function linear time to generate prime number sieve
factorization of the Euler function table
Nim game and sg function Hefei Problem B
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Size: 20480 |
Author: Leo |
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Description: 计算素数：各种方法计算素数，平方根法，筛法1 筛法2-Calculate prime numbers: various methods to calculate the prime numbers, square root method, method of sieve sieve method
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Size: 1169408 |
Author: jackchinacai |
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Description: 用筛法求素数的基本思想是：把从1开始的、某一范围内的正整数从小到大顺序排列， 1不是素数，首先把它筛掉。剩下的数中选择最小的数是素数，然后去掉它的倍数。依次类推，直到筛子为空时结束。如有：
1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30
1不是素数，去掉。剩下的数中2最小，是素数，去掉2的倍数，余下的数是：
3 5 7 9 11 13 15 17 19 21 23 25 27 29
剩下的数中3最小，是素数，去掉3的倍数，如此下去直到所有的数都被筛完，求出的素数为：
2 3 5 7 11 13 17 19 23 29(The basic idea of using sieve to find prime numbers is to set up the positive integers from 1 to a certain extent, and 1 are not prime numbers. The smallest number in the remaining number is a prime number, and then the multiple of it is removed. The analogy is in turn until the sieve ends when it is empty. If there are:
12345678910
111213141516171819 20
212223242526272829 30
1 is not a prime number, get rid of it. The smallest number in the remaining 2 is the prime number, and the number of 2 is removed, and the remaining number is:
3579111315171921 23252729)
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Size: 493568 |
Author: ImbaPlayer
|
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Description: 使用筛法遍历求取素数，效率较低，有待优化(Searching for prime numbers by traversing the sieve method)
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Size: 1024 |
Author: 一只猿 |
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