搜索资源列表
cPP
- 大整数运算，用C++实现，可以解决各种大整数运算的问题-Large integer arithmetic, with C++ achieve, and can solve the problem of integer arithmetic
calculator
- 大整数运算，有一些整数，其位数可能上百位。在功能上要对这些整数做加、减、乘、整除及取余运算。在程序组织上要求通过大整数类型的设计，来体现更好的可读性和可维护性，并且对错误采用异常方法处理。-Large integer arithmetic, some integer may be hundreds of bits. The function of these integers do add, subtract, multiply, divisible modulo operator. Progr
zc_CBigNum_10071701
- 自己编写的大数库，可实现大整数的加减乘除求余，取模等多种计算，经过实验证明，所有运算都是正确的。-Write your own library of large numbers, large integer addition and subtraction, multiplication and division remainder modulo a wide variety of computing, after the experiment proved, all the operation
zc_RSA_090721
- 基于大整数运算的RSA加密，大数库是自己编写的版本，经过实验所有的运算都是正确的。-Large integer arithmetic RSA encryption, large numbers library is to write your own version of the experiment all arithmetic is correct.
TSH-big-int
- 大整数算法，算法设计与分析课中的实验代码，希望对你有用-Large integer arithmetic, algorithm design and analysis courses in experimental code, hope useful to you
bignum.tar
- 大整数加法与大整数乘法，本程序是综合运算，可以方便的进行修改-Integer addition and large integer multiplication, the program is integrated computing, you can easily modify
big
- 大整数乘法，编译通过，采用的是分治法，绝对正确-Large integer multiplication, compiled by the divide and conquer, absolutely right.
suanfa
- 大整数运算 图论计算算法 加法 经常性的图论知识-Large integer arithmetic
test
- 超大整数乘法，最大可支持32767位整数相乘-Large integer multiplication 32767 integer multiplied
jianzhifa
- 跟据减治法的大整数乘法，此方法是通过对大整数乘法的改进，运行时间更短，跟有效率的算法-According to conquer reduction of large integer multiplication, this method is improved by a large integer multiplication, the running time is shorter, with the efficiency of the algorithm
simple-dazhengshu
- 此为大整数算法，能够将两个位数极大的位数相乘并得出精确的结果-This is a large integer arithmetic, the number of bits in the two-digit greatly multiplied and accurate results
gaijin-fenzhifa
- 此为对分治法大整数相乘的一个改进，能够给对刚开始学习算法的人一个了解的机会-This is an improved large integer multiplication of divide-and-conquer method to the beginning of the learning algorithm an understanding of the opportunity
mofashu-chazhao
- 运用大整数的存储格式，迅速查找任意倍数内的魔法数（即将一个n位整数的最高移到最后成为一个新的n位整数，二者成整数倍关系）-Use large integer storage format, the Find any multiple magic number quickly (about to a maximum of n-bit integers moved and eventually become a new n-bit integers, both as an integer multi
RSA
- 一个RSA算法的实现， 其中有大整数简单实现， 和求大质数的算法，最后综合成RSA加密解密算法。 代码有一定注释，输出详细的中间过程，对C++初学者很有用。-Simple implementation of an RSA algorithm, where there is a large integer, and demand a large prime number algorithm, finally consolidated into the RSA encryption al
zhengshu
- 大整数的运算，可以运算很大的整数，使用起来会十分方便-Large integer arithmetic
infinite
- 计算机中整数都有一定的范围，对一些需要用到任意大的整数的应用，需要一种解决方案。本方案为用单链表实现的任意大的整数，可以实现输入输出操作、加法操作和赋值操作。-Integer in the computer has a certain range, on the need to use any integer applications, the need for a solution. The program for arbitrarily large integer, single linke
FFT
- FFT是离散傅里叶变换的快速算法，也可用于计算离散傅里叶变换的逆变换。快速傅里叶变换有广泛的应用，如数字信号处理、计算大整数乘法、求解偏微分方程-The FFT is a fast algorithm, can also be used to calculate the discrete Fourier transform of the inverse transform of the discrete Fourier transform. Fast Fourier Transform of a
IntegerMultiply
- 算法分析与设计 用递归分治算法解决大整数乘积问题（java源码）-Algorithm Analysis and Design Recursive divide and conquer algorithm to solve the problem of large integer multiplication. (Java source code)
Fermat_RSAkey
- 从RSAkey.txt中读取已生成的RSA密钥，形如n=pq（p和q皆为素数）的大整数，采用费马因子分解法对其分解，输出分解的结果和所费时间（单位为秒）。-Read from the RSAkey.txt RSA key, shaped like a large integer n = pq (p and q are prime numbers), using Fermat factorization method, decomposition, output decomposition res
cata
- ▲问题描述： (1) 设计一个表示高精度整数的模板类，它支持对任意大整数的输入、输出和四则运算(+、-、*、/)，其中除法运算应输出所得的商和余数。 (2) 用上述高精度整数的模板类，计算大整数组合数和Catalan数的精确值。 组合数C(m,n)=m! / (n!*(m-n)!) Catalan数定义为：C(n)=C(2*n,n)/(n+1) ▲编程任务： 给定大整数m和n，计算C(m,n)和C(m)的精确值。 ▲数据输入： 由文件input.txt