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[原创]KCTF Q3 第十三题：大圣归来
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2019-9-24 04:59
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用OD在MessageBoxA下断，跟到校验代码。IDA F5一下。
int __cdecl sub_407F30(HWND hDlg)
{
CHAR *v1; // edi
signed int v2; // esi
signed int v3; // eax
const CHAR *v4; // eax
const char *v5; // eax
CHAR *v6; // ecx
char v7; // al
const CHAR *v8; // eax
_BYTE *v9; // ecx
char v10; // al
char v11; // al
_BYTE *v12; // ecx
char v13; // al
CHAR *v14; // ecx
char pe; // [esp-40h] [ebp-3D8h]
char *v17; // [esp-3Ch] [ebp-3D4h]
int v18; // [esp-38h] [ebp-3D0h]
int v19; // [esp-34h] [ebp-3CCh]
char un; // [esp-30h] [ebp-3C8h]
char *v21; // [esp-2Ch] [ebp-3C4h]
int v22; // [esp-28h] [ebp-3C0h]
int v23; // [esp-24h] [ebp-3BCh]
int kx; // [esp-20h] [ebp-3B8h]
char *v25; // [esp-1Ch] [ebp-3B4h]
int v26; // [esp-18h] [ebp-3B0h]
int v27; // [esp-14h] [ebp-3ACh]
char *input; // [esp-10h] [ebp-3A8h]
char *v29; // [esp-Ch] [ebp-3A4h]
const CHAR *v30; // [esp-8h] [ebp-3A0h]
void (__thiscall *v31)(void *); // [esp-4h] [ebp-39Ch]
char v32; // [esp+13h] [ebp-385h]
char v33; // [esp+14h] [ebp-384h]
LPCSTR v34; // [esp+18h] [ebp-380h]
int v35; // [esp+1Ch] [ebp-37Ch]
int v36; // [esp+20h] [ebp-378h]
char v37; // [esp+24h] [ebp-374h]
int v38; // [esp+28h] [ebp-370h]
unsigned int v39; // [esp+2Ch] [ebp-36Ch]
int v40; // [esp+30h] [ebp-368h]
char v41; // [esp+34h] [ebp-364h]
int v42; // [esp+38h] [ebp-360h]
int v43; // [esp+3Ch] [ebp-35Ch]
int v44; // [esp+40h] [ebp-358h]
char v45; // [esp+44h] [ebp-354h]
const char *v46; // [esp+48h] [ebp-350h]
unsigned int v47; // [esp+4Ch] [ebp-34Ch]
char v48; // [esp+54h] [ebp-344h]
LPCSTR lpText; // [esp+58h] [ebp-340h]
int nIDDlgItem; // [esp+64h] [ebp-334h]
int v51; // [esp+68h] [ebp-330h]
int v52; // [esp+6Ch] [ebp-32Ch]
char *v53; // [esp+70h] [ebp-328h]
char *v54; // [esp+74h] [ebp-324h]
int *v55; // [esp+78h] [ebp-320h]
char **v56; // [esp+7Ch] [ebp-31Ch]
int v57[3]; // [esp+80h] [ebp-318h]
CHAR String; // [esp+8Ch] [ebp-30Ch]
char v59; // [esp+8Dh] [ebp-30Bh]
char v60; // [esp+18Ch] [ebp-20Ch]
char v61; // [esp+28Ch] [ebp-10Ch]
__int16 v62; // [esp+389h] [ebp-Fh]
char v63; // [esp+38Bh] [ebp-Dh]
int v64; // [esp+394h] [ebp-4h]
String = 0;
memset(&v59, 0, 0x2FCu);
v62 = 0;
v63 = 0;
nIDDlgItem = 1005;
v51 = 1000;
v52 = 1001;
v1 = &String;
v2 = 0;
while ( 1 )
{
v3 = GetDlgItemTextA(hDlg, *(int *)((char *)&nIDDlgItem + v2 * 4), v1, 255);
v57[v2] = v3;
if ( v3 < 1 || v3 > 100 )
return 0;
++v2;
v1 += 256;
if ( v2 >= 3 )
{
`eh vector constructor iterator'(&v45, 0x10u, 2, sub_408360, sub_40B410);
v64 = 0;
v33 = v32;
newstring(&v33, 0);
strcpy(&v33, &String, strlen(&String));
v37 = v32;
LOBYTE(v64) = 1;
newstring(&v37, 0);
strcpy(&v37, &v60, strlen(&v60));
LOBYTE(v64) = 2;
v41 = v32;
newstring(&v41, 0);
strcpy(&v41, aKxctf19q3, strlen(aKxctf19q3));
LOBYTE(v64) = 3;
if ( !check1(&v61, (int)&v45, (int)&v48) )// 以KXCTFXXXX分割字符串并判断长度，第一部分长度需大于等于4，第二部分长度需大于等于8
{
v4 = lpText;
if ( !lpText )
v4 = (const CHAR *)&unk_419144;
v31 = 0;
v30 = aBadCode;
goto LABEL_15;
}
v5 = v46;
if ( !v46 )
v5 = (const char *)&unk_419144;
if ( !strcmp(&v37, 0, v39, v5, v47) ) // 前8位是否为username
{
v56 = &input;
LOBYTE(input) = v48;
newstring(&input, 0);
substr_0(&input, &v48, 0, 0xFFFFFFFF);
v55 = &kx;
LOBYTE(v64) = 4;
strmove(&kx, &v41); // KXCTF19Q3
v53 = &un;
LOBYTE(v64) = 5;
strmove(&un, &v37); // username
v54 = &pe;
LOBYTE(v64) = 6;
strmove(&pe, &v33); // PEDIY_CTF2019_Q3_完璧归赵_Crackme_Readyu_
LOBYTE(v64) = 3;
if ( !check2((int)hDlg, pe, v17, v18, v19, un, v21, v22, v23, kx, v25, v26, v27, (char)input, v29, (int)v30) )
{
v8 = v34;
if ( !v34 )
v8 = (const CHAR *)&unk_419144;
MessageBoxA(hDlg, v8, aBadCheck, 0);
if ( v42 )
{
v9 = (_BYTE *)(v42 - 1);
v10 = *(_BYTE *)(v42 - 1);
if ( v10 && v10 != -1 )
*v9 = v10 - 1;
else
sub_40DF24(v9);
}
v42 = 0;
v43 = 0;
v44 = 0;
if ( v38 )
{
v11 = *(_BYTE *)(v38 - 1);
v12 = (_BYTE *)(v38 - 1);
if ( v11 && v11 != -1 )
*v12 = v11 - 1;
else
sub_40DF24(v12);
}
v38 = 0;
v39 = 0;
v40 = 0;
if ( v34 )
{
v13 = *(v34 - 1);
v14 = (CHAR *)(v34 - 1);
if ( v13 && v13 != -1 )
*v14 = v13 - 1;
else
sub_40DF24(v14);
}
v34 = 0;
v35 = 0;
v36 = 0;
v64 = -1;
goto LABEL_41;
}
success(hDlg);
LOBYTE(v64) = 2;
newstring(&v41, 1);
LOBYTE(v64) = 1;
newstring(&v37, 1);
if ( v34 )
{
v6 = (CHAR *)(v34 - 1);
v7 = *(v34 - 1);
if ( v7 && v7 != -1 )
*v6 = v7 - 1;
else
sub_40DF24(v6);
}
v31 = sub_40B410;
v30 = (const CHAR *)2;
v29 = (char *)16;
v34 = 0;
v35 = 0;
v36 = 0;
v64 = -1;
input = &v45;
}
else
{
v4 = v46;
if ( !v46 )
v4 = (const CHAR *)&unk_419144;
v31 = 0;
v30 = aBadName;
LABEL_15:
MessageBoxA(hDlg, v4, v30, (UINT)v31);
LOBYTE(v64) = 2;
newstring(&v41, 1);
LOBYTE(v64) = 1;
newstring(&v37, 1);
LOBYTE(v64) = 0;
newstring(&v33, 1);
v64 = -1;
LABEL_41:
v31 = sub_40B410;
v30 = (const CHAR *)2;
v29 = (char *)16;
input = &v45;
}
`eh vector destructor iterator'(input, (unsigned int)v29, (int)v30, v31);
return 0;
}
}
}
主要是计算一些字符串的hash作为大数，用LUC算法解密两次被KXCTFXXXX分割的第二部分input转成的大数，比较结果是否为0xC4B73CB0DD1D750C69E1755B06BBAFFFF44D2600
总结一下就是
N=0x3FBDAD083DBC11A52FA2AF1A0829C522C1492907F1B9523A17B7A8E65679BB01 message=0x9071232E6B170092668255303E5D824F2879AD56+inputnum message2=LUCdecrypt(message,0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF4B5843544631395133,N) message2+=0x249BA36000029BBE97499C03DB5A9001F6B734EC (LUCdecrypt(message2,N,N)+0x6DC844C73B34D6E6B8DE48DA64EF92AB2B11F461) % N==0xC4B73CB0DD1D750C69E1755B06BBAFFFF44D2600 //(LUCdecrypt(message2,N,N)% N==0x56EEF7E9A1E89E25B1032C80A1CC1D54C93B319F
怎么得到0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF4B5843544631395133的？
LUC( v206, // input+... v207, big_Nadd1, v209, big_Index, // 第一轮为：hash+KXCTF19Q3 // 第二轮为：~hash v211, big_Ncp, v213, (int *)v214, //U (int *)v215, //V (int *)v216);
第一次解密，CM是将其分成两步来算的，第一步取得Lucas序列第 hash+KXCTF19Q3项的值，第二轮取第~hash项的值。
之后将第二轮的U,V值做72次快速递推，换算到项数上就是 ~hash * 2 ^ 0x72 =
~hash * 0x1000000000000000000。
然后再进行一次递推
v169 = big_mul((int)&v263, (int *)v234, (int *)&v235); LOBYTE(v362) = 122; v170 = (int *)big_mul((int)&v258, (int *)&v243, (int *)v239);// v243=(input+...)^2-4 LOBYTE(v362) = 123; v171 = big_mul((int)&v256, v170, (int *)&v240); LOBYTE(v362) = 124; v172 = (int *)big_add((int)&v246, v171, v169); LOBYTE(v362) = 125; v173 = (int *)big_mod((int)&v237, v172, big_hash64); LOBYTE(v362) = 126; big_cpy((int *)&v251, v173); LOBYTE(v362) = 125; free(&v237); LOBYTE(v362) = 124; free(&v246); LOBYTE(v362) = 123; free(&v256); LOBYTE(v362) = 122; free(&v258); LOBYTE(v362) = 18; free(&v263); big_num = (char *)&v215; strmove_0((int *)&v215, big_hash64); big_shr1((int *)&v251); v174 = (int *)big_add((int)&v237, (int)&v251, (int)&v271);// username
精简一下
message2=((v243*v239*v240+v234*235 mod 0x3FBDAD083DBC11A52FA2AF1A0829C522C1492907F1B9523A17B7A8E65679BB01
)>>1)+ 0x249BA36000029BBE97499C03DB5A9001F6B734EC //username
再加上一条性质
将第二轮的项数和第一轮的项数相加（hash是20字节的），就得到
0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF4B5843544631395133了。
之后，用LUC算法把0x56EEF7E9A1E89E25B1032C80A1CC1D54C93B319F
加密回去就行，具体步骤如下：
1.分解N（N=0x3FBDAD083DBC11A52FA2AF1A0829C522C1492907F1B9523A17B7A8E65679BB01）
用ppsiqs或者在线分解都可以。分解得到
128959*277879388132275220363*804543020813130593505380215293268595059305396789149 = N
2.计算它们的平方-1相乘的结果，也就是
(128959^2-1)*(
277879388132275220363 ^2-1)*(
804543020813130593505380215293268595059305396789149 ^2-1) = 0xFDEE7B2FD689345E05543994931E971838A18B641E848A1EEAA9D93353AF6595CFAC531583E0694776464966283075356167F8C8240C43533BCB281DE24E000（M）
3.求逆元
分别求出
d1 * 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF4B5843544631395133
mod M
= 1
d2 * 0x3FBDAD083DBC11A52FA2AF1A0829C522C1492907F1B9523A17B7A8E65679BB01 mod M = 1
求得
d1 = 0xAC79CC5A4E51DFDF32C3234C5ABE56E90DF14AF9F0F659515C21C969F2B293EB7ED8C13072B634372A7EE3428573CA6655F11A369B1BCCE3FF76D4FE29191FB
d2 =
0xA8656E606751D95C207210B59A1ECB880B59F633BA64BF15367614212FBF642D0310EF5C6B1114B688B1E24EB03BA2DDF8725E63887AA9DA9555FDE6D9E0501
4.用d1，d2进行LUC加密（加密和解密其实用的同一个函数）
message2 = 0x56EEF7E9A1E89E25B1032C80A1CC1D54C93B319F
message = LUCencrypt(message2,d2,N) - 0x249BA36000029BBE97499C03DB5A9001F6B734EC
result = LUCencrypt(message,d1,N) - 0x9071232E6B170092668255303E5D824F2879AD56
解得
message = 0xD9C518135ED505550BE71C2430C33DFA16D8F8CFDF7F1FDD4C5471EAC2089D5
result
= 0x753350C5A24300015025083AFBAE5D206A7D83E2C7F98D09ADB6EF51ACFDC83
故flag为
usernameKXCTFXXXX753350C5A24300015025083AFBAE5D206A7D83E2C7F98D09ADB6EF51ACFDC83
参考资料
[培训]内核驱动高级班，冲击BAT一流互联网大厂工 作，每周日13:00-18:00直播授课
最后于 2019-9-25 19:51
被梦游枪手编辑
，原因：
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