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  • 利用gmpy2破解rsa

    gmpy2的相关文档:

    https://gmpy2.readthedocs.io/en/latest/

    ================

    题目:

    来自实验吧的rsarsa:http://www.shiyanbar.com/ctf/1979

    题目内容:

    p =  9648423029010515676590551740010426534945737639235739800643989352039852507298491399561035009163427050370107570733633350911691280297777160200625281665378483
    q =  11874843837980297032092405848653656852760910154543380907650040190704283358909208578251063047732443992230647903887510065547947313543299303261986053486569407
    e =  65537
    c =  83208298995174604174773590298203639360540024871256126892889661345742403314929861939100492666605647316646576486526217457006376842280869728581726746401583705899941768214138742259689334840735633553053887641847651173776251820293087212885670180367406807406765923638973161375817392737747832762751690104423869019034
    
    Use RSA to find the secret message

    求明文

    Python代码:

    import gmpy2
    
    p = gmpy2.mpz(9648423029010515676590551740010426534945737639235739800643989352039852507298491399561035009163427050370107570733633350911691280297777160200625281665378483)
    q = gmpy2.mpz(11874843837980297032092405848653656852760910154543380907650040190704283358909208578251063047732443992230647903887510065547947313543299303261986053486569407)
    e = gmpy2.mpz(65537)
    phi_n = (p - 1) * (q - 1)
    d = gmpy2.invert(e, phi_n)
    print ("private key:")
    print (d)
    
    c = gmpy2.mpz(83208298995174604174773590298203639360540024871256126892889661345742403314929861939100492666605647316646576486526217457006376842280869728581726746401583705899941768214138742259689334840735633553053887641847651173776251820293087212885670180367406807406765923638973161375817392737747832762751690104423869019034)
    print ("plaintext:")
    print (pow(c,d,p*q))

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  • 原文地址:https://www.cnblogs.com/ESHLkangi/p/8576222.html
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