normal20

无壳

像叠叠乐一般的累加

没什么好说的

创建一个暂存量来模拟累加的数字,减回去就行

因为data长65,可以算出密钥长34

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data=[0x72, 0xE9, 0x4D, 0xAC, 0xC1, 0xD0, 0x24, 0x6B, 0xB2, 0xF5, 0xFD, 0x45, 0x49, 0x94, 0xDC, 0x10,
0x10, 0x6B, 0xA3, 0xFB, 0x5C, 0x13, 0x17, 0xE4, 0x67, 0xFE, 0x72, 0xA1, 0xC7, 0x04, 0x2B, 0xC2,
0x9D, 0x3F, 0xA7, 0x6C, 0xE7, 0xD0, 0x90, 0x71, 0x36, 0xB3, 0xAB, 0x67, 0xBF, 0x60, 0x30, 0x3E,
0x78, 0xCD, 0x6D, 0x35, 0xC8, 0x55, 0xFF, 0xC0, 0x95, 0x62, 0xE6, 0xBB, 0x57, 0x34, 0x29, 0x0E,
0x03]
key=[0x21, 0x22, 0x23, 0x24, 0x25, 0x26, 0x27, 0x28, 0x29, 0x2A, 0x2B, 0x2C, 0x2D, 0x2E, 0x2F, 0x3A,
0x3B, 0x3C, 0x3D, 0x3E, 0x3F, 0x40, 0x5B, 0x5C, 0x5D, 0x5E, 0x5F, 0x60, 0x7B, 0x7C, 0x7D, 0x7E]
temp=[0]*65
flag=[]
for i in range(34):
flag.append((((data[i]-temp[i])%128)^0x21)%128)
for k in range(32):
temp[k+i]+=flag[i]^key[k]
print(''.join(chr(x)for x in flag))
print(len(data))

SYC{4+mile+b3gin+with+sing1e+step}

比赛不会写啊TAT