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[原创] KCTF 2022 Win. 第六题 约束与伪随机
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发表于: 2022-11-28 14:50 6950
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 | [ 5 - - 8 ] [ 0 - 4 ]KCTF[ 9 - d] #ida_bytes.get_dword(0x40F000) #0x3bfc #[ida_bytes.get_dword(0x40F000+i*4) for i in range(20)] #ida_bytes.get_dword(0x40F050) #0x2bb6 #[ida_bytes.get_dword(0x40F050+i*4) for i in range(20)] import z3,ida_bytes x = z3. Int ( 'x' ) z3.solve(x> 0 ,x< 100000 ,(x * 0x343FD + 0x269EC3 ) / 0x10000 % 0x8000 = = ida_bytes.get_dword( 0x40F000 )) #[x = 14725] z3.solve(x> 0 ,x< 100000 ,(x * 0x343FD + 0x269EC3 ) / 0x10000 % 0x8000 = = ida_bytes.get_dword( 0x40F050 )) #[x = 83690] 即: 14725KCTF83690 |
最后于 2022-11-28 14:57
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