最后通过字频查找找到flag 3.CheckIn 先用base64解码 再用ROT47解码得到flag 4.childRSA 先用factordb分解N 然后套用脚本得到flag from Crypto.Util.number import * import gmpy2 n = c = e= p= q= phi = (p-1)*(q-1) d=gmpy2.invert(e,phi) m=pow(c,d,n) print(long_to_bytes(m)) 5....
print hex(num)[2:-1].decode("hex") 在解题过程中,一般使用PyCrypto库中的long_to_bytes和bytes_to_long函数进行转换 fromCrypto.Util.numberimportbytes_to_long c=flag{123456}printbytes_to_long(c) urlencode 特点:有% 例子:BUUCTF平台 url编码 打开文件%66%6c%61%67%7b%61%6e%64%20%31%3d%31%...
print hex(num)[2:-1].decode("hex") 在解题过程中,一般使用PyCrypto库中的long_to_bytes和bytes_to_long函数进行转换 fromCrypto.Util.numberimportbytes_to_long c=flag{123456}printbytes_to_long(c) urlencode 特点:有% 例子:BUUCTF平台 url编码 打开文件%66%6c%61%67%7b%61%6e%64%20%31%3d%31%...
p1=n1// qp2=n2// q# 两次解密 d1=inverse(e,(q-1)*(p1-1))d2=inverse(e,(q-1)*(p2-1))print(long_to_bytes(pow(pow(c,d2,n2),d1,n1))) 🚀🚀运行结果如下所示: 代码语言:javascript 复制 b'SangFor{qSccmm1WrgvIg2Uq_cZhmqNfEGTz2GV8}' 低加密指数攻击(小明文攻击) 🚀🚀低加...
不过要注意,直接把上面求得的dd作long_to_bytes会得到乱码,这是因为随机点PP不是椭圆曲线的生成元。所以我们还需要加上PP阶的倍数去枚举: n = 1780694557271320552511299360138314441283923223949197 p = 73997272456239171124655017039956026551127725934222347 o = 3083219685676632130193959041477461850061047352503612 from Crypto.Util.numb...
思路:计算欧拉函数phi,使用inverse函数计算出私钥d,使用私钥对密文c进行解密,使用long_to_bytes函数将解密结果转换为字节数组。 from Crypto.Util.number import inverse,long_to_bytesc=872226907597064443425333959275851278816040891270738763259155213017570784395068431508325049401005543539187903628510381026359195143782941443864030756164572134785965...
print(long_to_bytes(ct)) 执行脚本即可得到flag: DrgnS{w3_fiX3d_that_f0r_y0U} Looking glass 题目描述如下 We found some shady Looking Glass service. Pretty sure there is a way to get that delicious /flag. http://lg.hackable.software:8080/ ...
returnlong_to_bytes(int(ret[:ret.rfind('1')+1],2),2).decode() if__name__=='__main__': print(solve()) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 运行后,最终的答案为:ROIS{base_GA_caN_b3_d1ffeR3nT...
fromCrypto.Util.numberimportlong_to_bytes # 用于求解a和b的最大公约数 defget_p(a,b): p=gmpy2.gcd(a,b) returnp if__name__=='__main__': n1=2322061983964262412720880432932907928927349792735156401198529202625491439483369154255289081051175123965636168607362827330939031488160458020442970846158751250063615816130341991625927107817...
m的bytes_to_long(flag)次方mod n = c 这里运用sympy.discrete_log求解离散对数 bytes_to_long(flag) = sympy.discrete_log(n,c,m) 脚本: importrandomimportsympy fromCrypto.Util.numberimport*n=2**512print(n)m=39119070912452742895948966256527403931830595217293685940385507958140277098689030846908473545120788538631898688104...