安全-rsarsa（BUUCTF）

Posted 2021-09-04 小狐狸FM

文章目录

• 前言
• 一、题目
• 二、WriteUp

前言

如果安装gmpy2的时候报错，可以看下方的文章

Python-解决下载gmpy2的报错问题

安全-RSA算法基础

一、题目

原题链接

Math is cool! Use the RSA algorithm to decode the secret message, c, p, q, and e are parameters for the RSA algorithm.


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

Use RSA to find the secret message

二、WriteUp

目前已知的量是p、q、e、c求m
使用gmpy2来求解d

# coding=utf-8
# 作者：小狐狸FM
import gmpy2
p = int(input("p:"))
q = int(input("q:"))
e = int(input("e:"))
c = int(input("c:"))
n = p*q
fn = (p-1) * (q-1) #欧拉函数
d = int(gmpy2.invert(e,fn))
m = pow(c,d,n)
print("m:",m)