额 第一题就暴力搜索了

 

已知仿射加密变换为c=（11m+8）mod26，试对密文sjoyuxzr解密

 

#include 
     
      int main(void){    int m,c;    while(scanf("%c",&m))    {        c = (11 * (m -97)+ 8)%26;        printf("%c",c+97);    }    return 0;}
     

看了大神的代码 有所醒悟

c= Ek(m)=(k1m+k2) mod n{c-k2 = k1 *m mod n令k3*k1 mod n =1得(c - k2) * k3 = m mod nm = (c - k2) * k3}m=Dk(c)=k3(c- k2) mod n（其中（k3 ×k1）mod26 = 1）

 

 

def extendedGCD1(a, b):    # a*xi + b*yi = ri    if b == 0:        return (1, 0, a)    (x, y, r) = extendedGCD1(b, a%b)    """    gcd(a, b) = a*xi + b*yi    gcd(b, a %  b) = b*xi+1 + (a - [a/b]*b)*yi+1    gcd(a, b) = gcd(b, a %  b)   =>   a*xi + b*yi = a*yi+1 + b*(xi+1 - [a/b]*yi+1)    xi = yi+1    yi = xi+1 - [a/b]*yi+1    """    tmp = x    x = y    y = tmp - (a/b) * y    return (x, y, r)