8C7A:canstants

now you known this:

char [10]={7A D0 07 71 DE 82 5F 40 AA 13}

next is the cacl:

78787879 imul  7A=00000039:696969AA  //get char[0]

39 SAR 3=7

7A ADD 7=81

8C7A ADD 81=8CFB

78787879 imul  8CFB=00004258:00004AA3

4258 SAR 3=084B

084B ADD 8CFB=9546

78787879 imul  D0=00000061:E1E1E250  //get char[1]

61 SAR 3=0C

D0 ADD 0C=DC

9546 ADD DC=9622

78787879 imul  9622=000046A6:9696E612

46A6 SAR 3=08D4

08D4 ADD 9622=9EF6//D0

78787879 imul 07=00000003:4B4B4B4F  //get char[2]

3 SAR 3=0

07 ADD 0=07

9EF6 ADD 7=9EFD

78787879 imul  9EFD=00004AD1:6969BD95

4AD1 SAR 3=095A

095A ADD 9EFD=A857//07

78787879 imul  71=00000035:2D2D2D69  //get char[3]

35 SAR 3=6

71 ADD 6=77

A857 ADD 77=A8CE

78787879 imul  A8CE=00004F70:0000595E

4F70 SAR 3=09EE

09EE ADD A8CE=B2BC//71

78787879 imul  DE=00000068:787878EE  //get char[4]

68 SAR 3=0D

DE ADD 0D=EB

B2BC ADD EB=B3A7

78787879 imul  B3A7=0000548A:D2D331EF

548A SAR 3=0A91

0A91 ADD B3A7=BE38//DE

78787879 imul  82=0000003D:2D2D2D72  //get char[5]

3D SAR 3=07

82 ADD 07=89

BE38 ADD 89=BEC1

78787879 BEC1  BEC1=000059C4:3C3CA139

59C4 SAR 3=0B38

0B38 ADD BEC1=C9F9//82

78787879 imul  5F=0000002C:B4B4B4E7  //get char[6]

2C SAR 3=05

5F ADD 05=64

C9F9 ADD 64=CA5D

78787879 imul  CA5D=00005F3A:D2D33DF5

5F3A SAR 3=0BE7

CA5D ADD 0BE7=D644

78787879 imul  40=0000001E:1E1E1E40  //get char[7]

1E SAR 3=03

40 ADD 03=43

D644 ADD 43=D687

78787879 imul  D687=000064F4:3C3CADCF

64F4 SAR 3=0C9E

0C9E ADD D687=E325//40

78787879 imul  AA=00000050:0000005A  //get char[8]

50 SAR 3=0A

AA ADD 0A=B4

E325 ADD B4=E3D9

78787879 imul  E3D9=00006B38:F0F16991

6B38 SAR 3=0D67

E3D9 ADD 0D67=F140//AA

78787879 imul  13=00000008:F0F0F0FB  //get char[9]

08 SAR 3=01

13 ADD 01=14

F140 ADD 14=F154

78787879 imul  F154=00007190:F0F170B4

7190 SAR 3=0E32

0E32 ADD F154=FF86//13

take the last add:FF86 CACL this:

FF86 sar 8=FF

FF86 sar 3=1FF0

F0 XOR FF=0F

0F XOR 86=89

--------------------------------------------------------------

89^ 13=9A

89^ AA=23

89^ 40=C9

89^ 5F=D6

89^ 82=0B

89^ DE=57

89^ 71=F8

89^ 07=8E

89^ D0=59

89^ 7A=F3

F3598EF8570BD6C9 //get this char[10]

--------------------------------------------------------

now ,i Hypothesis you know char[10]=[F3598EF8570BD6C9],how do you cacl :

char [10]={7A D0 07 71 DE 82 5F 40 AA 13} form C language?i was uesd vc++6.0  .thanks all.please give me some Suggestions ETC.

Edited May 29, 201411 yr by diskgetor

Reverse the code...(add to sub .. etc in reverse.. little brute forcing with the xors though..  

Here is your algo in c 

unsigned int i;

unsigned short k;

unsigned char buf[] = {0x7A,0xD0,0x07,0x71,0xDE,0x82,0x5F,0x40,0xAA,0x13};
k = 0x8C7A;

for (i = 0; i < sizeof(buf); i++) {

    k += buf[i] + buf[i] / 0x11;

    k += k / 0x11;

}

k = (k >> 3) ^ (k >> 8) ^ 0x86;
for (i = 0; i < sizeof(buf); i++) {

    buf[i] ^= (unsigned char)k;

}

ok,this bug was fixed,thanks UniSoft.