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[原创] 看雪 2025 KCTF 签到题 废材少年觉醒
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发表于: 2025-8-15 13:20 87
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通过二进制里的字符串得知是rust程序
从入口点依次找到rust的主函数:
(sub_140001400 -> sub_140001010 -> sub_140001AF0 -> sub_140001620)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 | __int64 __fastcall sub_140001620(){ __int64 v0; // rax char **v1; // rdx char *v2; // r9 unsigned __int8 *v3; // rdx unsigned __int8 *v4; // r10 char v5; // bl unsigned __int8 *v6; // rbx unsigned __int8 *v7; // rax unsigned int v8; // edi int v9; // edx int v10; // r14d int v11; // ebp unsigned int v12; // ebx char v13; // bl char *v14; // rdi unsigned int v15; // ebx char v16; // bp char v17; // r14 int v18; // r14d int v19; // ebp unsigned int v20; // ebp __int64 result; // rax char **v22; // [rsp+28h] [rbp-B0h] BYREF __int64 v23; // [rsp+30h] [rbp-A8h] __int64 v24; // [rsp+38h] [rbp-A0h] __int128 v25; // [rsp+40h] [rbp-98h] __int64 v26; // [rsp+58h] [rbp-80h] BYREF LPVOID lpMem; // [rsp+60h] [rbp-78h] __int64 v28; // [rsp+68h] [rbp-70h] _QWORD v29[2]; // [rsp+70h] [rbp-68h] BYREF _QWORD v30[2]; // [rsp+80h] [rbp-58h] BYREF v22 = &off_1400A10A0; v23 = 1; v24 = 8; v25 = 0; sub_140031F30(&v22); v22 = &off_1400A10C8; v23 = 1; v24 = 8; v25 = 0; sub_140031F30(&v22); v26 = sub_14002FCF0(); v0 = sub_14002FDA0((__int64)&v26); if ( v0 ) { v22 = (char **)v0; sub_140087C50( (unsigned int)aCalledResultUn, 43, (unsigned int)&v22, (unsigned int)&off_1400A1030, (__int64)&off_1400A10D8); } v26 = 0; lpMem = (LPVOID)1; v28 = 0; v29[0] = sub_14002E310(); if ( (sub_14002E3A0(v29, &v26) & 1) != 0 ) { v22 = v1; sub_140087C50( (unsigned int)aCalledResultUn, 43, (unsigned int)&v22, (unsigned int)&off_1400A1030, (__int64)&off_1400A10F0); } v2 = (char *)lpMem + v28; if ( v28 ) { v3 = 0; v4 = (unsigned __int8 *)lpMem; while ( 1 ) { v6 = v4; v7 = v3; v8 = *v4; if ( (v8 & 0x80u) != 0 ) { v9 = v8 & 0x1F; v10 = v4[1] & 0x3F; if ( (unsigned __int8)v8 <= 0xDFu ) { v4 += 2; v8 = v10 | (v9 << 6); } else { v11 = (v10 << 6) | v4[2] & 0x3F; if ( (unsigned __int8)v8 < 0xF0u ) { v4 += 3; v8 = (v9 << 12) | v11; } else { v4 += 4; v8 = ((v8 & 7) << 18) | (v11 << 6) | v6[3] & 0x3F; } } } else { ++v4; } v3 = &v7[v4 - v6]; if ( v8 - 9 >= 5 && v8 != 32 ) { if ( v8 < 0x80 ) goto LABEL_27; v12 = v8 >> 8; if ( v8 >> 8 > 0x1F ) { if ( v12 == 32 ) { v5 = *((_BYTE *)off_1400A0AC0 + (unsigned __int8)v8) >> 1; } else { if ( v12 != 48 ) goto LABEL_27; v5 = v8 == 12288; } } else if ( v12 ) { if ( v12 != 22 ) goto LABEL_27; v5 = v8 == 5760; } else { v5 = *((_BYTE *)off_1400A0AC0 + (unsigned __int8)v8); } if ( (v5 & 1) == 0 ) goto LABEL_27; } if ( v4 == (unsigned __int8 *)v2 ) goto LABEL_53; } } v7 = 0; v3 = 0; v4 = (unsigned __int8 *)lpMem;LABEL_27: if ( v4 == (unsigned __int8 *)v2 ) {LABEL_52: if ( v28 ) goto LABEL_54;LABEL_53: v7 = 0; v3 = 0; goto LABEL_54; } while ( 1 ) { v14 = v2; v15 = *(v2 - 1); if ( (v15 & 0x80000000) != 0 ) { v16 = *(v2 - 2); if ( v16 >= -64 ) { v2 -= 2; v19 = v16 & 0x1F; } else { v17 = *(v2 - 3); if ( v17 >= -64 ) { v2 -= 3; v18 = v17 & 0xF; } else { v2 -= 4; v18 = ((*(v14 - 4) & 7) << 6) | v17 & 0x3F; } v19 = (v18 << 6) | v16 & 0x3F; } v15 = (v19 << 6) | v15 & 0x3F; if ( v15 - 9 < 5 ) goto LABEL_31; } else { --v2; if ( v15 - 9 < 5 ) goto LABEL_31; } if ( v15 == 32 ) goto LABEL_31; if ( v15 < 0x80 ) break; v20 = v15 >> 8; if ( v15 >> 8 <= 0x1F ) { if ( v20 ) { if ( v20 != 22 ) break; v13 = v15 == 0x1680; } else { v13 = *((_BYTE *)off_1400A0AC0 + (unsigned __int8)v15); } goto LABEL_30; } if ( v20 == 32 ) { v13 = *((_BYTE *)off_1400A0AC0 + (unsigned __int8)v15) >> 1; goto LABEL_30; } if ( v20 != 48 ) break; v13 = v15 == 12288;LABEL_30: if ( (v13 & 1) == 0 ) break;LABEL_31: if ( v4 == (unsigned __int8 *)v2 ) goto LABEL_52; } v3 = (unsigned __int8 *)&v14[v3 - v4];LABEL_54: v30[0] = (char *)lpMem + (_QWORD)v7; v30[1] = v3 - v7; if ( sub_140001560((_BYTE *)lpMem + (_QWORD)v7, v3 - v7) ) { v29[0] = v30; v29[1] = sub_140001480; v22 = &off_1400A1140; v23 = 2; v24 = (__int64)v29; v25 = 1u; } else { v22 = &off_1400A1118; v23 = 1; v24 = 8; v25 = 0; } result = sub_140031F30(&v22); if ( v26 ) return sub_140001B50(lpMem); return result;} |
末尾的 sub_140001560 的返回结果经过 if 判断,两个分支的 off_1400A1140 和 off_1400A1118 分别指向成功和失败
打开 sub_140001560 ,把 ascii 数字转换为字符,直接是flag的明文
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 | bool __fastcall sub_140001560(_BYTE *a1, __int64 a2){ return a2 == 25 && *a1 == 'f' && a1[1] == 'l' && a1[2] == 'a' && a1[3] == 'g' && a1[4] == '{' && a1[5] == 'k' && a1[6] == 'c' && a1[7] == 't' && a1[8] == 'f' && a1[9] == '_' && a1[10] == 't' && a1[11] == 'i' && a1[12] == 'm' && a1[13] == 'e' && a1[14] == '_' && a1[15] == 'l' && a1[16] == 'e' && a1[17] == 'a' && a1[18] == 'p' && a1[19] == '_' && a1[20] == '2' && a1[21] == '0' && a1[22] == '2' && a1[23] == '5' && a1[24] == '}';} |
最终flag:flag{kctf_time_leap_2025}
至于主函数的逻辑,AI直接识别出来是UTF-8字符处理,并给出结论其作用为去除输入前后的空白字符,对应到rust则是标准库str::trim的内联
[培训]Windows内核深度攻防:从Hook技术到Rootkit实战!
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