搜索资源列表
计算素数公约数公倍数
- 计算素数，最大公约数，最小公倍数的VB程序编程-calculating prime numbers, the common denominator, implements the VB Programming
歌德巴赫猜想
- 验证：2000以内的正偶数都能够分解为两个素数之和(即验证歌德巴赫猜想对2000以内的正偶数成立)。 *问题分析与算法设计 为了验证歌德巴赫猜想对2000以内的正偶数都是成立的，要将整数分解为两部分，然后判断出分解出的两个整数是否均为素数。若是，则满足题意；否则重新进行分解和判断。 程序中对判断是否为素数的算法进行了改进，对整数判断“用从2开始到该整数的一半”改为“2开始到该整数的平方根”。原因何在请自行分析。 -certification : 2000 within the
blackeye_elgamalkeygenerator10beta3
- 大素数生成器，根据用户的设置自动生成一个符合要求的大素数，计算速度较快。-large prime number generator, according to the automatic generation to meet the requirements of a large prime numbers, the fast calculation.
乘式还原(2)
- 有乘法算式如下： ○○○ × ○○ ------------ ○○○○ ○○○○ ------------ ○○○○○ 18个○的位置上全部是素数(1、3、5或7)，请还原此算式。 *问题分析与算法设计 问题中虽然有18数位，但只要确定乘数和被乘数后经过计算 就可确定其它的数位。-multiplication formula is as follows : 2000 2000 --- --- 2000 2000 --- ---
验证歌德巴赫猜想实验
- 验证歌德巴赫猜想 （1）：任何大于6的偶数都可以表示为两个素数之和：如8=5+3 （2）：任何大于5的奇数都可以表示为三个素数之和：如：11=3+3+5-certification Goldbach conjecture (1) : any of the even greater than 6 can be expressed as two prime numbers and : 800 = 5 3 (2) : No more than five odd numbers can be e
一个偶数总能表示为两个素数之和
- 用于分析偶数的组成原则，一个偶数总能表示为两个素数之和-even for the analysis of the composition, an even always expressed as two prime numbers and
判断素数的个数
- 判断101-200之间有多少个素数，并输出所有素数。 1.程序分析：判断素数的方法：用一个数分别去除2到sqrt(这个数)，如果能被整除， 则表明此数不是素数，反之是素数。 -judgment 101-200 between the number of primes, and the export of all prime numbers. 1. Program Analysis : judgment in several ways : with a few were t
qiusushu
- 求任意自然数中素数的和，可以校验输入的数是否自然数，并显示出所有素数-seeking arbitrary natural number which primes and can check whether the number of imported natural number, and demonstrated that all prime numbers
classicC100
- 经典C程序100例,对于初学者非常值得研究.内有各种排序,找素数问题,字符串处理等.-classic C program hundred cases for beginners is worth studying. There are various sort, finding prime numbers. string processing.
isprime
- 判断是不是大素数,的快速算法,适用于公钥加密算法-judgment is not large prime numbers, the fast algorithm applicable to the public key encryption algorithm
20060624a
- 一个杨辉三角形和判断显示两个数之间的素数-a triangle and judgment shows the number of two prime numbers
Gdbhcx(C)
- 简易哥德巴赫猜想(任何一个大于6的偶数可以分解成两个素数，可以利用枚举法，先把偶数可以分解的所有奇数对列出，再判断是否为素数，直到验证成功为止)-summary Goldbach conjecture (any one of the even greater than 6 can be decomposed into two prime numbers. can enumeration method, the decomposition can even first of all to the
prime
- MATLAB 寻找素数 源程序代码-MATLAB to find prime numbers source code
prime-number
- 求解质数，利用bitset实现，非常适合求解答大质数。-Solving prime number, use bitset implement, ideal for large prime numbers seeking answers.
The-prime-is-stored
- 请编写函数fun,函数的功能是：将大于形参m的且紧靠m的k个素数存入XX所指的数组中。-Please write a function fun, function function is: greater than the parameter m and m close to the k prime numbers into an array XX referred to in.
prime
- 给定一个范围，就可求出该范围内的所有素数，且压缩包中含有普通算法和厄尔多塞筛法，可进行效率比较-Given a range, you can find all the prime numbers within that range, and the general compression algorithm and Eerduose package contains sieve method, can be effective rate
prime-number
- Using this program you can find out the prime numbers between 1 to 100, 100 to 999 etc. You just need to input the range, for e.g. if you want the prime numbers 100 to 999 then enter numbers 100 and 999 when program prompts for input.-Using this prog
prime
- 关于素数的简单介绍，和简单的一些程序，帮助理解素数(About the simple introduction of prime numbers, and some procedures)
prime_fnc
- how to find prime numbers using function
prime_std
- how to find prime numbers using std