参加个CTF比赛，发现自己零代码量，什么东西都没有准备，被吊打当然正常，基本的工具常用的代码可以自己提前写好的呀
常用的代码有：
gcd：求两个数的最大公约数
egcd：求满足ax+by=1，当gcd(a，b)=1时，满足式子的x和y
hextoflag：很多题目都是给的字母的ord值：两位两位表示一个字母，转化成chr字符就是flag的
b64：base64解密
b32：base32解密
qp：quickpow，快速幂运算
modinv：求cd=1(mod m)，在已知c，m，且gcd(c，m)=1的时候，求得c的逆元d
get_phi_n：求n的欧拉函数(这个是做RSA题专用的分解n，因为n是两个大素数相乘)
attacksamen：RSA公模攻击
原理：
modequation：求ax=b(mod c)的x，一次同余方程，有解的条件是，b % gcd(a，c)=0
代码如下：
#coding=utf-8
import base64
import sys
sys.setrecursionlimit(10000000)
def hextonumber(x):
#1234567890abcdef
#1234567890ABCDEF
if x>='0' and x<='9':
return int(x)
elif x>='A' and x<='F':
return ord(x)-55
else:
return ord(x)-87
def hextoflag(s):
#word='666c61677b7769656e65725f61747461636b5f61747461636b5f796f757d'
#flag{wiener_attack_attack_you}
flag = ''
i = 0
while (i
flag += chr(hextonumber(s[i])*16+hextonumber(s[i+1]))
i += 2
return flag
def b64(s):
#word = 'Y3RmezY2NjY2Nn0='
#ctf{666666}
return base64.b64decode(s)
def b32(s):
#word = 'GYYWIY3UMZ5UQML6IIYHSLCXMVWEGMDNMUWVI3ZNJAZVEZL5'
#61dctf{H1~B0y,WelC0me-To-H3Re}
return base64.b32decode(s)
def gcd(a, b):
if a < b:
a, b = b, a
while b != 0:
temp = a % b
a = b
b = temp
return a
def egcd(a,b):
if b==0:
return a,1,0
else:
g,x,y=egcd(b,a%b)
return g,y,x-a/b*y
def qp(n,m,p):
ans=1
while(m):
if (m%2==1):
ans=(ans*n)%p
n=(n*n)%p
m=m/2
return ans
def egcd(a, b):
if a == 0:
return (b, 0, 1)
else:
g, y, x = egcd(b % a, a)
return (g, x - (b // a) * y, y)
def modinv(a, m):
g, x, y = egcd(a, m)
if g != 1:
raise Exception('modular inverse does not exist')
else:
return (x+m) % m
def get_phi_n(p,q):
return (p-1)*(q-1)
def attacksamen(n,e1,e2,c1,c2):
s = egcd(e1 , e2)
s1 = s[1]
s2 = s[2]
if s1 < 0:
s1 = - s1
c1 = modinv (c1 , n)
elif s2 < 0:
s2 = - s2
c2 = modinv (c2 , n)
m = ( qp(c1,s1,n) * qp(c2,s2,n) ) % n
return hextoflag(str(hex(m))[2:-1])
def modequation(a,b,c):
#ax==b(mod c)
#ax+cy==b
#b%gcd(a,c)==0
Gcd = gcd(a,c)
if (b % Gcd != 0):
return 'No Solution'
a /= Gcd
c /= Gcd
b /= Gcd
return b*modinv(a,c)%c
if __name__ == "__main__":
#hextoflag =
#b64 = base64.b64decode
#b32 = base32.b32decode
#gcd = greatest common divisor
#egcd =
#qp = (n^m)%p
#modinv = (e*d)==1 % n , we have number e and number n to get number d
#get_phi_n(p,q)
#attacksamen
#n = 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
#e1 = 0x17e1
#e2 = 0x43a5
#c1 = 0xb6e66aa0d4d5ad1460482f45aab87e80a99c1ff3af605fd9cea82d76d464272f3dd2e1797e3fede64cffcd54b2a7a5e21f45574783f62266cebf3cdb9764c6c04b0b30b5d065d5f6142d498506ea1f6449f428253d4d76bd96778d5f58abf313370b980dcb90daf882c5539ac3df81a431bc2c0e0911ecbe5195d94312218b3854ee14f13bd00c81d7ff11c06a9e112940b7377c20e53738a2ebb77b0534d8d9e481e60e9c87693bd9e1fd1e569083479ff8f53e42337a2b799c2325a7e2588fb046cf228d01d8596e7af4570a3cb0635d2524d234e3993d76b7e60f1c478ba45891de5cc0a1fec116f7c0dd9be7aa54226edf0196e37856afca32c69d790e1L
#c2 = 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
#print attacksamen(n,e1,e2,c1,c2)
#print modequation(5,4,12)
#print modequation(15,11,36)
letter = 'abcdefghijklmnopqrstuvwxyz'
word = 'falszztysyjzyjkywjrztyjztyynaryjkyswarztyegyyj'
flag = ''
for i in word:
flag += letter[modequation(7,(ord(i)-97+4)%26,26)]
print flag
