利用matlab求解RE中出现的齐次线性方程组
判断输入的flag时，将flag的每一位当成了未知数带入方程进行计算
一共有32位flag，32个32元1次方程
编写脚本将方程转化成矩阵形式
fd=open('bianl.txt')
a=[0,0,0,0,0,0,0,0,0,0,0,0,0,00,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
def getmidstring(html, start_str, end):
start = html.find(start_str)
if start >= 0:
start += len(start_str)
end = html.find(end, start)
if end >= 0:
return str(html[start:end])
def geta(b):
num=0
i=0
if('*a' in b):
i=0
elif('a1[' in b):
i=int(str(getmidstring(b,'[',']')))
if('<<6' in b and '+' in b):
num='64'
elif('<<6' in b and '-' in b):
num='-64'
elif('+' in b):
num=str(getmidstring(b,'+ ',' *'))
elif('- ' in b):
num='-'+str(getmidstring(b,'- ',' *'))
elif('(' in b):
num=str(getmidstring(b,'( ',' *'))
if('!=' in b):
a[32]=int(getmidstring(b,'!=',' )'))
a[i]=num
#print(a[i])
while(1):
b=fd.readline()
#print(a)
if('ret' in b):
final=''
for i in range(0,33):
final=final+' '+str(a[i])
print(final+';')
a=[0,0,0,0,0,0,0,0,0,0,0,0,0,00,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
elif('a1' in b):
geta(b)存在一点bug，把变量里的0和None手动改一下问题不大
得到的矩阵
导入matlab并化简得到方程的解
>> B=[ XXXXX]; %XXXXX矩阵的具体值，元素之间用空格分隔，行之间用分号分隔% >> format rat >> C=rref(B)
打印出C
