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RSA解密和加密算法的实现和应用
- RSA算法 :首先, 找出三个数, p, q, r, 其中 p, q 是两个相异的质数, r 是与 (p-1)(q-1) 互质的数...... p, q, r 这三个数便是 person_key，接著, 找出 m, 使得 r^m == 1 mod (p-1)(q-1)..... 这个 m 一定存在, 因为 r 与 (p-1)(q-1) 互质, 用辗转相除法就可以得到了..... 再来, 计算 n = pq....... m, n 这两个
m序列
- Verilog编写的M序列发生器，希望能对大家带来帮助。 -Verilog prepared by the M-sequence generator, we hope to bring help.
chmmbox_1_2
- CHMMBOX, version 1.2, Iead Rezek, Oxford University, Feb 2001 Matlab toolbox for max. aposteriori estimation of two chain Coupled Hidden Markov Models. -CHMMBOX, version 1.2, Iead Rezek, Oxford University, Feb 20
USB_CDC_ACM_DRIVER
- USB Abstract Control Model driver for USB modems and ISDN adapters(for Linux)-USB Abstract Control Model driver for USB m odems and ISDN adapters (for Linux)
ns2.28MAODV
- 在ns2.28中实现的多路径路由协议M-AODV-ns2.28 achieve in the multi-path routing protocol M-AODV
Euler_fuction
- 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
rsa
- 1) 找出两个相异的大素数P和Q，令N＝P×Q，M＝（P－1）（Q－1）。 2) 找出与M互素的大数E，用欧氏算法计算出大数D，使D×E≡1 MOD M。 3) 丢弃P和Q，公开E，D和N。E和N即加密密钥，D和N即解密密钥。 -1) to identify two different large prime numbers P and Q, so N = P × Q, M = (P-1) (Q-1). 2) to id
mod-mac
- 实现修改网卡地址，简单过程明细。直接修改，快速上网。-Modified to achieve LAN address, details of a simple process. Directly modify, fast Internet access.
0703_dynamic-mod
- embest 可以动态加载的led驱动 编译成.o文件后直接加载就行 -embest can be dynamically loaded into the led driver to compile. o files directly loaded on the line
lps130-m
- Mod. 130 power supply Schematic diagram
ForcedPendulum
- This simulink model simulates the damped driven pendulum, showing it s chaotic motion. theta = angle of pendulum omega = (d/dt)theta = angular velocity Gamma(t) = gcos(phi) = Force omega_d = (d/dt) phi Gam
mcfpfsf2_1
- 固定窗口的模重复平方算法，计算大数a的n次方模m余下的值，可以有效出理计算机计算过程中的大数溢出问题。-figure out the value of a^n(mod m) when a is large.
figout_e
- 求出一个数对于另一个数的指数，例如求a对模m的指数e,使 a^e(mod m)=1-Obtained a number to another number for the index, such as seeking a pair of mode indices m e, so that a ^ e (mod m) = 1
RK
- 实验RK算法，即利用Hash方法和素数理论，首先定义一个Hash函数（hash (r) = r mod q），然后将模式串P和文本串T中长度为m的子串利用Hash函数转换成数值。显然只需比较那些与模式串具有相同Hash函数值的子串。 当然因为Hash冲突的存在，还要进一步进行字符串比较，但只要选择适当的素数q, Hash冲突的概率就会很小 -Experimental RK algorithm, namely the use of
0
- 5、设计一个程序计算 (mod m)。当a=31,n=48413,m=113时，计算其值。-5, to design a program to calculate (mod m). When a = 31, n = 48413, m = 113, the calculated value.
sffhf4
- 设m不整除a，计算一次同余式ax=b(mod m)。当a=987，b=564，m=2005时，求出x。-Let m not divisible by a, calculate a congruence ax = b (mod m). When a = 987, b = 564, m = 2005, the calculated x.
uerybffd5
- 设计一个程序计算 (mod m)。当a=31,n=48413,m=113时，计算其值。-Design a program to calculate (mod m). When a = 31, n = 48413, m = 113, the calculated value.
2-mochongfu
- 通过C语言实现对于b^n (mod m)形式的式子的模重复平方法计算-C language formula for b ^ n (mod m) in the form of mold repeat method basis
diffeehellmanPROGRAM
- Example 1. To find 1537 x 4248 modulo 10, you could multiply out and take the last digit, but a better way would be to replace 1537 by 7 and 4248 by 8 to start, find 7 x 8 = 56, and then take 56 mod 10 to get 6 as the an
RC4-Prog
- Example 1. To find 1537 x 4248 modulo 10, you could multiply out and take the last digit, but a better way would be to replace 1537 by 7 and 4248 by 8 to start, find 7 x 8 = 56, and then take 56 mod 10 to get 6 as the an