*reverse*_练习1__elrond32

今日学习61

1.

 

老样子，拿到题后，找字符串

 

 

 

 

双击进入，f5反编译

 

找到关键函数，这个

 

 

函数就是flag的关键所在，被允许进入后的函数，和判断函数非常关键

 

 

 

 

 

 

 

 

 

 

根据手算，以及a1%8来得知key的大小为8，

Key[105,115,101,110,103,97,114,100]

 

 

 

 

 

 

print(flag)

key=[105,115,101,110,103,97,114,100]

v2=[0x0F,0x1F,0x04,0x09,0x1C,0x12,0x42,0x09,0x0C,0x44,0x0D,0x07,0x09,0x06,0x2D,0x37,0x59,0x1E,0x00,0x59,0x0F,0x08,0x1C,0x23,0x36,0x07,0x55,0x02,0x0C,0x08,0x41,0x0A,0x14]

flag=''

for i in range(33):

    flag+=chr(v2[i]^key[i%8])

print(flag)

得出Flag

flag{s0me7hing_S0me7hinG_t0lki3n}