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链接：http://pan.baidu.com/s/1eS9JNP4 密码：ltl4
本体分析比较简单，算法是解一个22元一次方程
这里引入了numpy这样一个python库，灰常强大
import numpy NUM=0x16 matrix=[[0 for i in range(NUM)] for j in range(NUM)] strings = [ "ThelightTokeepinmindtheholylight", "Timeismoneymyfriend", "WelcometotheaugerRuiMa", "Areyouheretoplayforthehorde", "ToarmsyeroustaboutsWevegotcompany", "Ahhwelcometomyparlor", "Slaytheminthemastersname", "YesrunItmakesthebloodpumpfaster", "Shhhitwillallbeoversoon", "Kneelbeforemeworm", "Runwhileyoustillcan", "RisemysoldiersRiseandfightoncemore", "LifeismeaningleshThatwearetrulytested", "BowtothemightoftheHighlord", "ThefirstkillgoestomeAnyonecaretowager", "Itisasitshouldbe", "Thedarkvoidawaitsyou", "InordertomoregloryofMichaelessienray", "Rememberthesunthewellofshame", "Maythewindguideyourroad", "StrengthandHonour", "Bloodandthunder" ] verify=[ 0x000373ca, 0x00031bdf, 0x000374f7, 0x00039406, 0x000399c4, 0x00034adc, 0x00038c08, 0x00038b88, 0x00038a60, 0x0002b568, 0x00032471, 0x00037dea, 0x00036f97, 0x000378e4, 0x00038706, 0x00029010, 0x00034c23, 0x00038ef8, 0x00038e29, 0x0003925e, 0x0002b5fc, 0x0002584e ] def gen_matrix(): for i in range(NUM): for j in range(NUM): matrix[i][j]=0 for i in range(NUM): for j in range(len(strings[i])): if j>=NUM: break matrix[i][j]=ord(strings[i][j]) if __name__==‘__main__‘: gen_matrix() verify=numpy.array(verify) matrix=numpy.array(matrix) a=numpy.linalg.solve(matrix,verify) print a
b=[ 104, 99, 116,102,123,76,74,95,121,54, 99, 100, 99,
95, 113, 119, 101, 101, 114, 116, 33, 125]
c=‘‘
for i in b:
c+=(chr(i))
print c
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原文地址：http://www.cnblogs.com/lomooo/p/5866747.html
