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Caesar--java
- 凯撒密码扩展算法，输入加密参数k1,k2， c =(k1*m+k2) mod 26-Caesar expansion password algorithm, the import of encryption parameters k1, k2, c = (k1 * m k2) mod 26
VC_RSA
- 一、RSA基本原理 对明文分组M和密文分组C，加密与解密过程如下： C = POW (M , e) mod n M = POW(C , d) mod n = POW(POW( M ,e), d) mod n=POW( M,e*d) 其中POW是指数函数，mod是求余数函数。 其中收发双方均已知n，发送放已知e，只有接受方已知d，因此公钥加密算法的公钥为 KU={ e , n},私钥为KR={d , n}。该算法要能用做公钥加密，必须满足下列条件： 1. 可以找到e ，
lps130-m
- Mod. 130 power supply Schematic diagram
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 - 1/p2)*……*(1 - 1/pn)
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 identify and M large numbers cop
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
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 Gamma(t) = (d/dt)omega + omega/Q
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 Hash methods and prime number
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 answer. A handy standard no
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 answer. A handy standard no
ModOfPower
- Write an efficient algorithm to calculate R=B^P mod M
modleach.m
- MATLAB Source Code of MOD-LEACH Routing Protocol
数字签名
- 表示整数k关于某个模数的逆元，并非指 k 的倒数。逆元：满足 a*b mod m =1 的a和b互为关于模数 m 的逆元。 2、H( m )：One-Way Hash函数，m为待签署的明文，DSS中选用SHA1( Secure Hash Algorithm )，此时H(m) 的输出为160bits长的数字串（16进制）。(HFTGHDGHXGHRFGFXZGVZSDFdCXVCFHGFGXFXBVXVDZf)