Jarvis OJ （逆向）：FindKey
python文件进行反编译：
# Embedded file name: findkey
import sys
lookup = [196, 153, 149, 206, 17, 221, 10, 217, 167, 18, 36, 135, 103, 61, 111, 31, 92, 152,
21, 228, 105, 191, 173, 41, 2, 245, 23, 144, 1, 246, 89, 178, 182, 119, 38,
85, 48, 226, 165, 241, 166, 214, 71, 90, 151, 3, 109, 169, 150, 224, 69, 156, 158,
57, 181, 29, 200, 37, 51, 252, 227, 93, 65, 82, 66, 80, 170, 77, 49, 177, 81,
94, 202, 107, 25, 73, 148, 98, 129, 231, 212, 14, 84, 121, 174, 171, 64, 180,
233, 74, 140, 242, 75, 104, 253, 44, 39, 87, 86, 27, 68, 22, 55, 76, 35, 248, 96,
5, 56, 20, 161, 213, 238, 220, 72, 100, 247, 8, 63, 249, 145, 243, 155, 222, 122,
32, 43, 186, 0, 102, 216, 126, 15, 42, 115, 138, 240, 147, 229, 204, 117, 223
, 141, 159, 131, 232, 124, 254, 60, 116, 46, 113, 79, 16, 128, 6, 251, 40, 205,
137, 199, 83, 54, 188, 19, 184, 201, 110, 255, 26, 91, 211, 132, 160, 168, 154,
185, 183, 244, 78, 33, 123, 28, 59, 12, 210, 218, 47, 163, 215, 209, 108, 235,
237, 118, 101, 24, 234, 106, 143, 88, 9, 136, 95, 30, 193, 176, 225, 198, 197, 194,
239, 134, 162, 192, 11, 70, 58, 187, 50, 67, 236, 230, 13, 99, 190, 208, 207,
7, 53, 219, 203, 62, 114, 127, 125, 164, 179, 175, 112, 172, 250, 133, 130, 52,
189, 97, 146, 34, 157, 120, 195, 45, 4, 142, 139]
pwda = [188, 155, 11, 58, 251, 208, 204, 202, 150, 120, 206, 237, 114, 92, 126, 6, 42]
pwdb = [53, 222, 230, 35, 67, 248, 226, 216, 17, 209, 32, 2, 181, 200, 171, 60, 108]
flag = raw_input('Input your Key:').strip()
if len(flag) != 17:
print 'Wrong Key!!'
sys.exit(1)
flag = flag[::-1]
for i in range(0, len(flag)):
if ord(flag[i]) + pwda[i] & 255 != lookup[i + pwdb[i]]:
print 'Wrong Key!!'
sys.exit(1)
print 'Congratulations!!'一开始写的脚本：
思想：既然输入为键盘上的字符，即在 33 到 127之间，由于&运算符不可逆。所以，对于lookup[i+pwdb[i]] - pwda[i] 的结果不在可输入字符范围之间的话，即原来的式子左边的 flag[i]) + pwda[i] > 255，由于 & 运算而失去了255（这句话是错的），所以我选择了加255。其实不是失去255，而是失去了256（性质跟%运算有些类似）。
lookup = [196, 153, 149, 206, 17, 221, 10, 217, 167, 18, 36, 135, 103, 61, 111, 31, 92,
152, 21, 228, 105, 191, 173, 41, 2, 245, 23, 144, 1, 246, 89, 178, 182, 119, 38,
85, 48, 226, 165, 241, 166, 214, 71, 90, 151, 3, 109, 169, 150, 224, 69, 156, 158,
57, 181, 29, 200, 37, 51, 252, 227, 93, 65, 82, 66, 80, 170, 77, 49, 177, 81,
94, 202, 107, 25, 73, 148, 98, 129, 231, 212, 14, 84, 121, 174, 171, 64, 180,
233, 74, 140, 242, 75, 104, 253, 44, 39, 87, 86, 27, 68, 22, 55, 76, 35, 248, 96,
5, 56, 20, 161, 213, 238, 220, 72, 100, 247, 8, 63, 249, 145, 243, 155, 222, 122,
32, 43, 186, 0, 102, 216, 126, 15, 42, 115, 138, 240, 147, 229, 204, 117, 223,
141, 159, 131, 232, 124, 254, 60, 116, 46, 113, 79, 16, 128, 6, 251, 40, 205,
137, 199, 83, 54, 188, 19, 184, 201, 110, 255, 26, 91, 211, 132, 160, 168, 154,
185, 183, 244, 78, 33, 123, 28, 59, 12, 210, 218, 47, 163, 215, 209, 108, 235,
237, 118, 101, 24, 234, 106, 143, 88, 9, 136, 95, 30, 193, 176, 225, 198, 197, 194,
239, 134, 162, 192, 11, 70, 58, 187, 50, 67, 236, 230, 13, 99, 190, 208, 207,
7, 53, 219, 203, 62, 114, 127, 125, 164, 179, 175, 112, 172, 250, 133, 130, 52,
189, 97, 146, 34, 157, 120, 195, 45, 4, 142, 139]
pwda = [188, 155, 11, 58, 251, 208, 204, 202, 150, 120, 206, 237, 114, 92, 126, 6, 42]
pwdb = [53, 222, 230, 35, 67, 248, 226, 216, 17, 209, 32, 2, 181, 200, 171, 60, 108]
flag = ""
for i in range(0,17):
if lookup[i+pwdb[i]] - pwda[i] < 33 or lookup[i+pwdb[i]]-pwda[i] > 127:
flag += (chr(lookup[i+pwdb[i]]-pwda[i] + 255))
else:
flag += (chr(lookup[i+pwdb[i]]-pwda[i]))
print(flag[::-1])所以最后的脚本将 +255 改成 +256 ，或者将其改成 & 255，都是对的。
& 255 的效果也可以达到目的的原因是：
a + b & c = d
例如：288 & 255 = 32
假设左边的b为100，那么，带入到这个题的场景 d-b 的结果是 -68 ，之后，-68 & 255 = 188 （性质跟%运算类似），这个思想跟钟表有些类似，限定循环。
flag： PCTF{PyC_Cr4ck3r}