搜索资源列表
PollardRho
- 大数分解算法,根据算法设计分析高级教程,随机算法章节里面所讲的,mentocarlo算法改造的大数分解,能将一个大的合数分成几个素数的相乘,可以判断这个和数是不是由素数组成-factoring algorithm, algorithm design and analysis Senior Guide, randomized algorithm inside chapters have said, mentocarlo majority of the transformation algorith
7_Rsa
- RSA公钥加密算法基于大整数因式分解困难这样的事实。 选择两个素数，p，q。（一般p,q选择很大的数） 然后计算 z=p*q f=(p-1)(q-1) 选择一个n，使gcd(n,f)=1（gcd代表greatest common divider,一般n也选择一个素数）， n和z就作为公钥。 选择一个s,0<s<f,满足n*s ％ f=1，s就作为私钥。-RSA public key encryption algorithm based on the integer fa
Soft20060207115054365
- 查找素数 给定一个范围，自动查找并显示该范围内的所有素数。-finding prime numbers to set a range, automatic search and display within the scope of all prime numbers.
sushudui
- 编制函数prime，用来判断整数n是否为素数：bool prime(int n)； 而后编制主函数，任意输入一个大于4的偶数n，找出满足n=i+j的所有数对，其中要求i与j均为素数（通过调用prime来判断素数）。如偶数18可以分解为11+7以及13+5；而偶数80可以分解为：43+37、61+19、67+13、73+7。-establishment prime function, used to judge whether the integer n is prime : bool prime
lisanchuandibibao
- 离散数学中求解二元关系2到100间的所有素数传递闭包的问题.-Discrete Mathematics for binary relations 2-100 all prime numbers transitive closure of the issue.
HugeCalc
- 这是一个应用软件,用于处理大数,里面包含一些小的实用的软件,如生成 素数 ,大数阶乘.-This is a software application, for handling large numbers, which contains a small number of practical software, such as generating prime numbers, The majority of factorial.
simpleMathFunctions
- some simple but time wasting math logic 一些不难但麻烦（适合用电脑作）的数学函数的源码 1.Test for Prime numbers 2.Find the factorial of a number 3.Find the triangle(total sum) of a number 4.Reverse every integer of a number 5.Find any power of any value 6.Fin
manyPrimeThrea
- 使用多线程技术计算100000之间的所有素数的个数，java程序-use multithreading technology between 100,000 calculation of all the prime numbers, java procedures
sushupanduan
- 判断输入的数字是不是素数,输入数字的范围可以再程序中设定-judgment input is the number of prime numbers, enter the number of procedures can be set
OPt_primeB
- 素数筛选b算法源程序!输出一个数内所有的素数,此算法是上一算法的优化~主要用VC实现-prime source selection algorithm b! Output within a few all the prime numbers, This algorithm is an algorithm, the optimal ~ main VC
twice_OPt_prime
- 两次素数筛选算法源程序!先筛选sprt(n)内的素数,再选n内的素数,些算法更快~主要用VC实现-two prime screening algorithm source! Sprt first screening (n) the prime, n re-election of prime numbers, some faster-key algorithm with VC
RsakitT
- RSA算法,大素数生成,加密解密-RSA algorithm, generating large prime numbers, encryption and decryption
DisPlayingPrimes
- 输入一个数,程序将会把小于等于这个数的所有素数显示到屏幕上面.-importation of a number of procedures will be put to less than the number of all prime numbers shows that the screen above.
素数的求法
- 素数的求法Yes-prime numbers Yes Solution
SeekNumber
- C#寻找素数 素数寻找问题由来已久，一直是一些数学家追求的目的。-C#-finding prime numbers to find a few long-standing problem, mathematicians have been some of the objective pursued.
javapro
- 素数的查找，阶乘的计算，对称矩阵的输出-find prime numbers, the factorial, symmetric matrix output
哥德巴赫
- Goldbach s conjecture： Goldbach(哥德巴赫)曾经提出了一个这样的猜想： 对于2000 以内的正偶数都能够分解为两个素数之和。 要求： 该题是一个猜想，我们不去证明它，而是通过编程序来验证 该猜想是否正确。 输入： 用户从键盘任意输入一个自然数。 输出： 给出满足猜想的两个自然数。-Goldbach conjecture : Goldbach (Goldbach) had raised such a guess : 2000 within the can is eve
2262
- 北京大学ACM比赛题目 In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture: Every even number greater than 4 can be written as the sum of two odd prime numbers.
Rsa1Kit_V1.0
- RSA的小程序，源码产生随机素数调用方法：N.GetPrime(bits)返回值：N被赋值为一个bits位（0x100000000进制长度）的素数-RSA small program source random-number call methods : N. GetPrime (bits) Returns : N has been assigned a bits - (0x100000000 band length) of prime numbers
sy5_2
- 按“开始”按钮可以在文本框中显示所有1001到1100之间的素数。-according to the "start" button in the text box can show all from 1001 to 1100 between the prime numbers.