实验吧 rsarsa crypto Writeup

FrancisQiu

已于 2025-03-07 22:17:37 修改

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分类专栏： CTFwriteup CTF crypto 密码学实验吧文章标签：算法

于 2019-02-25 11:12:01 首次发布

本文链接：https://blog.youkuaiyun.com/qq_40737798/article/details/87913505

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本文介绍了解决一个典型的RSA算法题目，通过使用gmpy2库处理大数运算，成功解码了秘密信息。文章提供了完整的计算脚本，包括参数设置、求逆元、计算模幂等步骤。

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题目

Math is cool! Use the RSA algorithm to decode the secret message, c, p, q, and e are parameters for the RSA algorithm.
解题链接： http://ctf5.shiyanbar.com/crypto/rsarsa/rsa.txt

思路

打开链接获取得到rsa.txt内容：

p =  9648423029010515676590551740010426534945737639235739800643989352039852507298491399561035009163427050370107570733633350911691280297777160200625281665378483
q =  11874843837980297032092405848653656852760910154543380907650040190704283358909208578251063047732443992230647903887510065547947313543299303261986053486569407
e =  65537
c =  83208298995174604174773590298203639360540024871256126892889661345742403314929861939100492666605647316646576486526217457006376842280869728581726746401583705899941768214138742259689334840735633553053887641847651173776251820293087212885670180367406807406765923638973161375817392737747832762751690104423869019034

Use RSA to find the secret message

显然这是个很中规中矩的rsa习题，p、q、e、c都挺大的，但是针对rsa大数处理我们可以使用gmpy2这个库。
以下是计算脚本：

p =  9648423029010515676590551740010426534945737639235739800643989352039852507298491399561035009163427050370107570733633350911691280297777160200625281665378483
q =  11874843837980297032092405848653656852760910154543380907650040190704283358909208578251063047732443992230647903887510065547947313543299303261986053486569407
e =  65537
c =  83208298995174604174773590298203639360540024871256126892889661345742403314929861939100492666605647316646576486526217457006376842280869728581726746401583705899941768214138742259689334840735633553053887641847651173776251820293087212885670180367406807406765923638973161375817392737747832762751690104423869019034
from gmpy2 import *
d=invert(e,(q-1)*(p-1))
n=mul(p,q)
a=powmod(c,d,n)
print(a)