如其名，这个题确实是easy..通过查找字符串的方法找到的虚假main函数里将`isdebug`段置1然后调用`IsDebuggerPresebt`结束函数真正的主要逻辑是通过crc32校验出来的值解密dll，并`memload`调用dlldll内部的实现了一个aes加密以及一些简单的位运算脚本如下```pythonfrom Crypto.Cipher import AES
enc = [142, 56, 81, 115, 166, 153, 42, 240, 218, 213, 106, 145, 233, 78, 152, 206, 42, 183, 61, 64, 241, 229, 29, 171, 239, 238, 176, 214, 20, 11, 42, 149]
for i in range(len(enc)):    enc[i] ^= 0x55
for i in range(21, 28):    p = (enc[i] & 0xaa)>>1    q = (enc[i]<<1) & 0xaa    enc[i] = (p|q)^0xad
for i in range(14, 21):    p = (enc[i] & 0xcc)>>2    q = (enc[i]<<2) & 0xcc    enc[i] = (p|q)^0xbe
for i in range(7, 14):    p = (enc[i] & 0xf0)>>4    q = (enc[i]<<4) & 0xf0    enc[i] = (p|q)^0xef
#print (enc)cipher = enckey = b"5343544632303230"key = list(key)cipher = bytes(cipher)key = bytes(key)aes = AES.new(key, mode=AES.MODE_ECB)flag = aes.decrypt(cipher)
print (flag)#SCTF{y0u_found_the_true_secret}```