安洵杯——game（混淆控制流平坦化）
发布日期：2021-05-07 12:08:50
浏览次数：20
分类：技术文章
本文共 35396 字，大约阅读时间需要 117 分钟。
文章目录
查壳
拖进ida
main函数
printf("input your flag:", argv, envp); gets(&v8); v10 = general_inspection((int (*)[9])sudoku); v7 = -1804515313; while ( 1 ) { while ( 1 ) { while ( v7 == -2071121728 ) { v4 = blank_num((int (*)[9])sudoku); v5 = mem_alloc(v4); trace(sudoku, v5, v4); check((int (*)[9])sudoku); check1(&v8); check3(&v8); v9 = 0; v7 = -303742386; } if ( v7 != -1804515313 ) break; v3 = -2071121728; if ( v10 ) v3 = 664169471; v7 = v3; } if ( v7 == -303742386 ) break; if ( v7 == 664169471 ) { printf("error"); check((int (*)[9])sudoku); v9 = 0; v7 = -303742386; } } 分析
主函数中，需要分析的函数有general_inspection((int (*)[9])sudoku),blank_num((int (*)[9])sudoku)，trace(sudoku, v5, v4);，check((int (*)[9])sudoku);，check1(&v8);，check3(&v8);check3函数中还有个check2(a1)需要分析
general_inspection((int (*)[9])sudoku)
v10一直是0，sudoku在前后也并非发生什么改变。。。忽略不计 blank_num((int (*)[9])sudoku)
代码
__int64 __fastcall blank_num(int (*a1)[9]){ signed int v1; // eax signed int v2; // eax signed int v3; // eax signed int v5; // [rsp+2Ch] [rbp-18h] unsigned int v6; // [rsp+30h] [rbp-14h] signed int v7; // [rsp+34h] [rbp-10h] signed int v8; // [rsp+38h] [rbp-Ch] v6 = 0; v8 = 0; v5 = 1046773218; while ( 1 ) { while ( v5 == -1892951115 ) { v7 = 0; v5 = -1048142948; } if ( v5 == -1585203536 ) break; switch ( v5 ) { case -1237447983: v5 = 1058605341; break; case -1048142948: v2 = -1237447983; if ( v7 < 9 ) v2 = 1501457574; v5 = v2; break; case -1026222996: ++v7; v5 = -1048142948; break; case -516195663: ++v6; v5 = 710936108; break; case 710936108: v5 = -1026222996; break; case 1046773218: v1 = -1585203536; if ( v8 < 9 ) v1 = -1892951115; v5 = v1; break; case 1058605341: ++v8; v5 = 1046773218; break; case 1501457574: v3 = 710936108; if ( !(*a1)[9 * v8 + v7] ) v3 = -516195663; v5 = v3; break; } } return v6;} 第一步
case 1046773218: v1 = -1585203536; if ( v8 < 9 ) v1 = -1892951115; v5 = v1; break;
v8 < 9的话，就会进入-1892951115，否则进入-1585203536；
v8>9的情况： if ( v5 == -1585203536 ) break;
直接退出。。。所以v8必须<9才出现下一步。
第二步
while ( v5 == -1892951115 ) { v7 = 0; // 初始化v7 v5 = -1048142948; } 这里初始化了一个v7，留心即可。
第三步
case -1048142948: v2 = -1237447983; if ( v7 < 9 ) v2 = 1501457574; v5 = v2; break;
这里判断了v7的大小，利用v7来判断程序的执行流程，感觉应该是个for循环，
老样子，看看v7>9的情况case -1237447983: v5 = 1058605341; break;
继续往下：
case 1058605341: ++v8; v5 = 1046773218; break;
这里进行了v8自增，还没看出有什么猫腻，继续往下：
v5 = 1046773218;
这不就是回到了第一步嘛？？？？？嗯哼，回去了。.。
但是v7 < 9的话，流程进到1501457574;
第四步
case 1501457574: v3 = 710936108; if ( !(*sudoku)[9 * v8 + v7] ) v3 = -516195663; v5 = v3; break;
这里为什么是v8*9+v7呢？我个人认为它把这个数组分成一个二维数组，每行是九个，v8也就代表行数，v7代表列数，当v7大于9时，那么只需要把行数+1（即v8自增，然后v7清零即可）。这个二维数组有9行9列，当行数大于9时（即v8>9时，那么就遍历完了，然后直接退出即可）
- 当数组中的元素值非0时
v3 = 710936108;
流程代码不用管
case 710936108: v5 = -1026222996; break;
依然是流程代码：
case -1026222996: ++v7; v5 = -1048142948; break;
v7自增，然后进行流程代码，往下进行。
v5 = -1048142948;
这个流程-1048142948;直接回到第三步。。灭了
v3 = -516195663;
流程进到了下一步：
第五步
case -516195663: ++v6; v5 = 710936108; break;
这里v6++，也就是统计0的个数，
case 710936108: v5 = -1026222996; break;
这里全是流程代码，没必要分开写了
case -1026222996: ++v7; v5 = -1048142948; break;
紧接着流程回到第三步v5 = -1048142948;
总结
它把这个数组分成一个二维数组，每行是九个，v8也就代表行数，v7代表列数，当v7大于9时，那么只需要把行数+1（即v8自增，然后v7清零即可）。这个二维数组有9行9列，当行数大于9时（即v8>9时，那么就遍历完了，然后直接退出即可）
也就是一个遍历，然后把0的个数找出来，也就是v6自增。（v6就代表0的个数，最后返回是v6，也就是这个数组中0的个数）trace(sudoku, v5, v4);
代码
void __fastcall trace(__int64 a1, signed int *a2, int a3){ signed int v3; // eax signed int v4; // eax signed int v5; // eax signed int v6; // eax int v7; // eax signed int v8; // er8 signed int v9; // eax signed int v10; // eax signed int v11; // eax signed int v12; // eax signed int v13; // [rsp+78h] [rbp-28h] int v14; // [rsp+7Ch] [rbp-24h] signed int v15; // [rsp+80h] [rbp-20h] signed int v16; // [rsp+84h] [rbp-1Ch] signed int v17; // [rsp+88h] [rbp-18h] int v18; // [rsp+8Ch] [rbp-14h] v18 = a3; v14 = 0; v13 = 671940414; while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( v13 == -2124394493 ) { v4 = 338033522; if ( v17 < 9 ) v4 = -1264962160; v13 = v4; } if ( v13 != -2084617164 ) break; ++v18; v17 = a2[12 * v14]; v16 = a2[12 * v14 + 1]; v13 = 295419890; } if ( v13 != -2069701336 ) break; v5 = 942378879; if ( v16 < 9 ) v5 = 1672958513; v13 = v5; } if ( v13 != -1561315505 ) break; v13 = 2016120547; } if ( v13 != -1361654796 ) break; ++v16; v13 = -2069701336; } if ( v13 != -1289862082 ) break; v13 = -1361654796; } if ( v13 != -1264962160 ) break; v16 = 0; v13 = -2069701336; } if ( v13 == -1246113443 ) break; if ( v13 == -446534017 ) { v9 = 1764791757; if ( !a2[12 * v14 + 2] ) v9 = 1923573299; v13 = v9; } else if ( v13 == -264375465 ) { *(_DWORD *)(36LL * a2[12 * v14] + a1 + 4LL * a2[12 * v14 + 1]) = 0; ++v18; --v14; v13 = -446534017; } else if ( v13 == -127108152 ) { a2[12 * v14] = v17; a2[12 * v14 + 1] = v16; v7 = findvalue(a1, &a2[12 * v14]); v8 = 295419890; *(_DWORD *)(36LL * v17 + a1 + 4LL * v16) = v7; if ( *(_DWORD *)(36LL * v17 + a1 + 4LL * v16) == -1 ) v8 = 1601744610; v13 = v8; } else if ( v13 == 67917660 ) { *(_DWORD *)(36LL * a2[12 * v14] + a1 + 4LL * a2[12 * v14 + 1]) = v15; a2[12 * v14 + 2 + v15] = 1; --a2[12 * v14 + 2]; v13 = -2084617164; } else if ( v13 == 295419890 ) { ++v14; v18 = v18 - 1146223301 + 1146223300; v13 = -1289862082; } else if ( v13 == 338033522 ) { v13 = 671940414; } else if ( v13 == 376448068 ) { v17 = 0; v13 = -2124394493; } else if ( v13 == 599244415 ) { v11 = -2084617164; if ( v15 < 10 ) v11 = 1332608024; v13 = v11; } else if ( v13 == 671940414 ) { v3 = -1246113443; if ( v18 ) v3 = 376448068; v13 = v3; } else if ( v13 == 942378879 ) { v13 = 1396614849; } else if ( v13 == 1332608024 ) { v12 = -1561315505; if ( !a2[12 * v14 + 2 + v15] ) v12 = 67917660; v13 = v12; } else if ( v13 == 1396614849 ) { ++v17; v13 = -2124394493; } else if ( v13 == 1601744610 ) { *(_DWORD *)(36LL * v17 + a1 + 4LL * v16) = 0; --v14; v13 = -446534017; } else if ( v13 == 1672958513 ) { v6 = -1289862082; if ( !*(_DWORD *)(36LL * v17 + a1 + 4LL * v16) ) v6 = -127108152; v13 = v6; } else { if ( v13 == 1751405620 ) { printf(aGameOver); exit(1); } switch ( v13 ) { case 1764791757: v15 = 1; v13 = 599244415; break; case 1923573299: v10 = -264375465; if ( !v14 ) v10 = 1751405620; v13 = v10; break; case 2016120547: ++v15; v13 = 599244415; break; } } } free(a2);} 第一步
else if ( v13 == 671940414 ) { v3 = -1246113443; if ( v18 ) v3 = 376448068; v13 = v3; } 这里利用v18来控制流程，先假设反面，v18为0,v3 = -1246113443;,
if ( v13 == -1246113443 ) break;
好了，直接退出。，所以v18必不可能为0.
第二步
else if ( v13 == 376448068 ) { i = 0; v13 = -2124394493; } i赋值为0，然后紧接流程代码
while ( v13 == -2124394493 ) { v4 = 338033522; if ( i < 9 ) v4 = -1264962160; v13 = v4; } 判断了i值和9，先假设坏情况，i>9,v4 = 338033522;,
else if ( v13 == 338033522 ) { v13 = 671940414; } 流程代码，直接进行下一个（好家伙，回到第一步了。回到第一步呢，紧接着就会执行第二步，也就是把i置零。）：
else if ( v13 == 671940414 ) { v3 = -1246113443; if ( v18 ) v3 = 376448068; v13 = v3; } 接下来来分析一下i<9的情况v4 = -1264962160;
第三步：
if ( v13 != -1264962160 ) break; j = 0; v13 = -2069701336; }
这里给j赋值了，然后接下来是流程代码
if ( v13 != -2069701336 ) break; v5 = 942378879; if ( j < 9 ) v5 = 1672958513; v13 = v5; }
这里判断j值，然后再来进行程序的执行，反向，j>9时v5 = 942378879;，
else if ( v13 == 942378879 ) { v13 = 1396614849; } 流程代码，继续往下：
else if ( v13 == 1396614849 ) { ++i; v13 = -2124394493; } 好家伙，i自增，那么这个i就是行数，大于9然后++，盲猜下一步要把j置零。
while ( v13 == -2124394493 ) { v4 = 338033522; if ( i < 9 ) v4 = -1264962160; v13 = v4; } 这里判断了行数是否大于9，那么不大于9的话，下一步就是列数置零喽
if ( v13 != -1264962160 ) break; j = 0; v13 = -2069701336; }
好了，我们继续分析，列数小于9，
if ( j < 9 ) v5 = 1672958513; v13 = v5; }
流程代码，继续分析：
第四步
else if ( v13 == 1672958513 ) { v6 = -1289862082; if ( !*(_DWORD *)(36LL * i + sudoku + 4LL * j) ) v6 = -127108152; v13 = v6; } 这里又来判断元素是否为0喽。。。。假设非0，v6 = -1289862082;
if ( v13 != -1289862082 ) break; v13 = -1361654796; }
流程代码：
if ( v13 != -1361654796 ) break; ++j; v13 = -2069701336; }
也就是继续遍历下一个元素喽。直到遇到0，好了，我们假设0吧：
第五步
else if ( v13 == -127108152 ) // step5 { v5_1[12 * v14] = i; v5_1[12 * v14 + 1] = j; v7 = findvalue(sudoku, &v5_1[12 * v14]); v8 = 295419890; *(_DWORD *)(36LL * i + sudoku + 4LL * j) = v7; if ( *(_DWORD *)(36LL * i + sudoku + 4LL * j) == -1 ) v8 = 1601744610; v13 = v8; } 这里把0位置的行数和列数记在一个新数组中，然后，先往下吧，v8 = 295419890;
else if ( v13 == 295419890 ) { ++v14; v18 = v18 - 1146223301 + 1146223300; v13 = -1289862082; } v14自增，然后v18自减，接着流程代码
if ( v13 != -1289862082 ) break; v13 = -1361654796; }
流程代码，不用管，接着走。
if ( v13 != -1361654796 ) break; ++j; v13 = -2069701336; }
继续遍历下一个元素，那没事了。。。。
第六步
继续下一种情况：v8 = 1601744610;
else if ( v13 == 1601744610 ) { *(_DWORD *)(36LL * i + sudoku + 4LL * j) = 0; --v14; v13 = -446534017; } 把数组相应位置赋0，然后v14自减。接着流程代码：
if ( v13 == -446534017 ) { v9 = 1764791757; if ( !v5_1[12 * v14 + 2] ) v9 = 1923573299; v13 = v9; } 此处判断所处的数组元素是否为0，为0的话v9 = 1764791757;，
case 1764791757: v15 = 1; v13 = 599244415; break;
v15赋值1，然后流程代码
v13 = 599244415; break;
流程代码，
else if ( v13 == 599244415 ) { v11 = -2084617164; if ( v15 < 10 ) v11 = 1332608024; v13 = v11; } 再来判断是否小于10，首先假设大于10，v11 = -2084617164;
if ( v13 != -2084617164 ) break; ++v18; i = v5_1[12 * v14]; j = v5_1[12 * v14 + 1]; v13 = 295419890; }
i ，j被利用数组中的值赋上，然后v18自增，紧接着流程代码：
绕回来了：else if ( v13 == 295419890 ) { ++v14; v18 = v18 - 1146223301 + 1146223300; v13 = -1289862082; } 然后假设小于10，v11 = 1332608024;
else if ( v13 == 1332608024 ) { v12 = -1561315505; if ( !v5_1[12 * v14 + 2 + v15] ) v12 = 67917660; v13 = v12; } 然后判断数组元素是否为0，非零情况下：v12 = -1561315505;
if ( v13 != -1561315505 ) break; v13 = 2016120547; }
流程代码，继续，
case 2016120547: ++v15; v13 = 599244415; break;
v15自增，流程代码：
else if ( v13 == 599244415 ) { v11 = -2084617164; if ( v15 < 10 ) v11 = 1332608024; v13 = v11; } 为0情况下，v12 = 67917660;：
else if ( v13 == 67917660 ) { *(_DWORD *)(36LL * v5_1[12 * v14] + sudoku + 4LL * v5_1[12 * v14 + 1]) = v15; v5_1[12 * v14 + 2 + v15] = 1; --v5_1[12 * v14 + 2]; v13 = -2084617164; } 改变了数组的内容，紧接着流程代码，然后回到上级
if ( v13 != -2084617164 ) break; ++v18; i = v5_1[12 * v14]; j = v5_1[12 * v14 + 1]; v13 = 295419890; }
第七步（第五步的第二种情况）
v8 = 1601744610;这时流程代码如下：
else if ( v13 == 1601744610 ) { *(_DWORD *)(36LL * i + sudoku + 4LL * j) = 0; --v14; v13 = -446534017; } 数组元素被赋值0，v14自减，紧接着流程代码：
if ( v13 == -446534017 ) { v9 = 1764791757; if ( !v5_1[12 * v14 + 2] ) v9 = 1923573299; v13 = v9; } 然后又返回去了。
总结
check(int (*a1)[9])
代码
__int64 __fastcall check(int (*a1)[9]){ __int64 result; // rax signed int v2; // eax signed int v3; // eax signed int v4; // [rsp+24h] [rbp-18h] signed int v5; // [rsp+2Ch] [rbp-10h] signed int v6; // [rsp+30h] [rbp-Ch] v6 = 0; v4 = 1715923540; while ( 1 ) { while ( 1 ) { while ( v4 == -1955732718 ) v4 = 927738670; if ( v4 != -1010822917 ) break; v5 = 0; v4 = -135298689; } result = (unsigned int)(v4 + 849866751); if ( v4 == -849866751 ) break; switch ( v4 ) { case -135298689: v3 = 623468669; if ( v5 < 9 ) v3 = -1955732718; v4 = v3; break; case 623468669: v4 = 2055416056; break; case 927738670: ++v5; v4 = -135298689; break; case 1715923540: v2 = -849866751; if ( v6 < 9 ) v2 = -1010822917; v4 = v2; break; case 2055416056: ++v6; v4 = 1715923540; break; } } return result;} __int64 __fastcall findvalue(__int64 a1, int *a2)
代码
__int64 __fastcall findvalue(__int64 a1, int *a2){ signed int v2; // eax signed int v3; // eax signed int v4; // eax signed int v5; // eax signed int v6; // eax signed int v7; // eax signed int v8; // eax signed int v9; // eax signed int v10; // eax signed int v12; // [rsp+8Ch] [rbp-2Ch] int v13; // [rsp+90h] [rbp-28h] int v14; // [rsp+94h] [rbp-24h] signed int v15; // [rsp+98h] [rbp-20h] signed int v16; // [rsp+9Ch] [rbp-1Ch] unsigned int v17; // [rsp+B4h] [rbp-4h] v14 = *a2; v13 = a2[1]; v16 = 0; v12 = -791724132; while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( v12 == -1704444406 ) { v17 = v16; v12 = 981711445; } if ( v12 != -1635553414 ) break; ++v16; v12 = 1796887582; } if ( v12 != -1431966323 ) break; ++v15; v12 = -53637786; } if ( v12 != -1235584672 ) break; ++v16; v12 = 156864788; } if ( v12 != -1228737224 ) break; v8 = -1104998472; if ( v16 < 10 ) v8 = -890529193; v12 = v8; } if ( v12 != -1144155234 ) break; ++v16; v12 = -1228737224; } if ( v12 != -1104998472 ) break; v10 = -1704444406; if ( v16 == 10 ) v10 = 2089198738; v12 = v10; } if ( v12 != -890529193 ) break; v9 = 862823000; if ( !a2[v16 + 2] ) v9 = 771361621; v12 = v9; } if ( v12 != -791724132 ) break; v2 = -611155481; if ( v16 < 10 ) v2 = 1792341902; v12 = v2; } if ( v12 != -611155481 ) break; v16 = 1; v12 = 522488884; } if ( v12 != -248881259 ) break; ++a2[2]; v12 = 1986744843; } if ( v12 != -53637786 ) break; v5 = -32951658; if ( v15 < 3 ) v5 = 432619495; v12 = v5; } if ( v12 != -32951658 ) break; v12 = -1635553414; } if ( v12 != 156864788 ) break; v6 = 507587421; if ( v16 < 10 ) v6 = 810472009; v12 = v6; } if ( v12 != 432619495 ) break; a2[*(signed int *)(36LL * (v16 - -3 * (v14 / 3)) + a1 + 4LL * (v15 - -3 * (v13 / 3))) + 2] = 1; v12 = -1431966323; } if ( v12 != 507587421 ) break; v16 = 1; v12 = -1228737224; } if ( v12 != 522488884 ) break; v3 = 1921980522; if ( v16 < 10 ) v3 = 1524139080; v12 = v3; } if ( v12 != 771361621 ) break; a2[v16 + 2] = 1; --a2[2]; v12 = -1104998472; } if ( v12 != 810472009 ) break; v7 = 1986744843; if ( !a2[v16 + 2] ) v7 = -248881259; v12 = v7; } if ( v12 != 862823000 ) break; v12 = -1144155234; } if ( v12 == 981711445 ) break; switch ( v12 ) { case 1044110222: ++v16; v12 = -791724132; break; case 1275000702: ++v16; v12 = 522488884; break; case 1353319228: a2[2] = 0; v16 = 1; v12 = 156864788; break; case 1404549511: v15 = 0; v12 = -53637786; break; case 1524139080: a2[*(signed int *)(36LL * v14 + a1 + 4LL * (v16 - 1)) + 2] = 1; a2[*(signed int *)(36LL * (v16 - 832240230 + 832240229) + a1 + 4LL * v13) + 2] = 1; v12 = 1275000702; break; case 1792341902: a2[v16 + 2] = 0; v12 = 1044110222; break; case 1796887582: v4 = 1353319228; if ( v16 < 3 ) v4 = 1404549511; v12 = v4; break; case 1921980522: v16 = 0; v12 = 1796887582; break; case 1986744843: v12 = -1235584672; break; case 2089198738: v17 = -1; v12 = 981711445; break; } } return v17;} 第一步
if ( v12 != -791724132 ) break; v2 = -611155481;
控制流代码，没什么好说的，v2 = -611155481;
if ( v12 != -611155481 ) break; v16 = 1; v12 = 522488884;
v16赋值1，然后控制流代码，v12 = 522488884;
if ( v12 != 522488884 ) break; v3 = 1921980522;
控制流代码v3 = 1921980522;
case 1921980522: v16 = 0; v12 = 1796887582; break;
v16被赋值0，然后控制流代码v12 = 1796887582;
case 1796887582: v4 = 1353319228; if ( v16 < 3 ) v4 = 1404549511; v12 = v4; break;
这里利用v16来判断接下来的流程，先看v16>3的情况吧，
case 1353319228: a2[2] = 0; v16 = 1; v12 = 156864788; break;
数组的第三个元素被赋值为0，然后v16被赋值为1，然后流程代码，v12 = 156864788;
if ( v12 != 156864788 ) break; v6 = 507587421; if ( v16 < 10 ) v6 = 810472009; v12 = v6;
这里又来进行第二波判断，v16的值，假设大于10，流程代码，v6 = 507587421;
if ( v12 != 507587421 ) break; v16 = 1; v12 = -1228737224; }
v16被赋值为1，然后流程代码，
if ( v12 != -1228737224 ) break; v8 = -1104998472; if ( v16 < 10 ) v8 = -890529193; v12 = v8; }
v16>10的话，v8 = -1104998472
if ( v12 != -1104998472 ) break; v10 = -1704444406; if ( v16 == 10 ) v10 = 2089198738; v12 = v10;
- 如果v16等于10，
v10 = 2089198738;
case 2089198738: v17 = -1; v12 = 981711445; break;
返回-1，然后退出，if ( v12 == 981711445 ) break;
- 如果不等于10的话，
while ( v12 == -1704444406 ) { v17 = v16; v12 = 981711445; } v16赋给v17，然后退出
if ( v12 == 981711445 ) break;
然后来看看v16<3的情况
if ( v12 != 432619495 ) break; a2[*(signed int *)(36LL * (v16 - -3 * (v14 / 3)) + a1 + 4LL * (v15 - -3 * (v13 / 3))) + 2] = 1; v12 = -1431966323;
把你赋值给a2数组的某个位置，然后流程控制代码
if ( v12 != -1431966323 ) break; ++v15; v12 = -53637786;
v15自增，紧接着流程控制代码，然后执行回去，近一步探索
if ( v12 != -53637786 ) break; v5 = -32951658; if ( v15 < 3 ) v5 = 432619495; v12 = v5;
最上面的v16的值，小于10时，流程代码，v6 = 810472009;
if ( v12 != 810472009 ) break; v7 = 1986744843; if ( !a2[v16 + 2] ) v7 = -248881259; v12 = v7;
判断数组元素是否为0，
- 非0的情况
v7 = 1986744843;；
case 1986744843: v12 = -1235584672; break;
流程代码：
if ( v12 != -1235584672 ) break; ++v16; v12 = 156864788; }
v16自增，然后流程代码，
if ( v12 != 156864788 ) break; v6 = 507587421; if ( v16 < 10 ) v6 = 810472009; v12 = v6;
- 为0情况
v7 = -248881259;
if ( v12 != -248881259 ) break; ++a2[2]; v12 = 1986744843; }
case 1986744843: v12 = -1235584672; break;
if ( v12 != -1235584672 ) break; ++v16; v12 = 156864788; }
if ( v12 != 156864788 ) break; v6 = 507587421; if ( v16 < 10 ) v6 = 810472009; v12 = v6;
第二步
v16<10的话，v8 = -890529193
if ( v12 != -890529193 ) break; v9 = 862823000; if ( !a2[v16 + 2] ) v9 = 771361621; v12 = v9;
判断数组元素是否为0，
- 非0的情况
v9 = 862823000;；
if ( v12 != 862823000 ) break; v12 = -1144155234;
流程代码：
if ( v12 != -1144155234 ) break; ++v16; v12 = -1228737224;
v16自增，然后流程代码，
if ( v12 != -1228737224 ) break; v8 = -1104998472; if ( v16 < 10 ) v8 = -890529193; v12 = v8;
- 为0情况
v9 = 771361621;
if ( v12 != 771361621 ) break; a2[v16 + 2] = 1; --a2[2]; v12 = -1104998472;
if ( v12 != -1104998472 ) break; v10 = -1704444406; if ( v16 == 10 ) v10 = 2089198738; v12 = v10;
check1(char *a1)
代码
__int64 __fastcall check1(char *a1){ __int64 result; // rax size_t v2; // rax signed int v3; // ecx char v4; // ST6F_1 size_t v5; // rax signed int v6; // ecx char v7; // ST6E_1 size_t v8; // rax signed int v9; // ecx signed int v10; // [rsp+68h] [rbp-18h] int v11; // [rsp+70h] [rbp-10h] int v12; // [rsp+74h] [rbp-Ch] v12 = strlen(a1) >> 1; v11 = 0; v10 = 1519002972; while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( v10 == -2084833488 ) { v8 = strlen(a1); v9 = -67245798; if ( v12 < v8 ) v9 = 1974939745; v10 = v9; } if ( v10 != -1988665894 ) break; v12 = 0; v10 = -2084833488; } if ( v10 != -1393133668 ) break; v5 = strlen(a1); v6 = -1988665894; if ( v12 < v5 ) v6 = -1018472136; v10 = v6; } if ( v10 != -1018472136 ) break; v7 = a1[v12]; a1[v12] = a1[v12 + 1]; a1[v12 + 1] = v7; v10 = -146751883; } if ( v10 != -831482631 ) break; ++v12; v10 = -2084833488; } if ( v10 != -291294424 ) break; ++v11; ++v12; v10 = 1519002972; } if ( v10 != -146751883 ) break; v12 += 2; v10 = -1393133668; } result = (unsigned int)(v10 + 67245798); if ( v10 == -67245798 ) break; switch ( v10 ) { case 75381312: v4 = a1[v12]; a1[v12] = a1[v11]; a1[v11] = v4; v10 = -291294424; break; case 1519002972: v2 = strlen(a1); v3 = 1555725255; if ( v11 < v2 >> 1 ) v3 = 75381312; v10 = v3; break; case 1555725255: v12 = 0; v10 = -1393133668; break; case 1974939745: a1[v12] = (a1[v12] & 0xF3 | ~a1[v12] & 0xC) - 20; v10 = -831482631; break; } } return result;} check3(char *a1)
代码
__int64 __fastcall check3(char *a1){ __int64 result; // rax signed int v2; // eax signed int v3; // [rsp+28h] [rbp-18h] int v4; // [rsp+3Ch] [rbp-4h] v4 = check2(a1); v3 = 16123822; while ( 1 ) { while ( v3 == 16123822 ) { v2 = 1478060410; if ( !v4 ) v2 = 1274132590; v3 = v2; } result = (unsigned int)(v3 - 824643665); if ( v3 == 824643665 ) break; if ( v3 == 1274132590 ) { v3 = 824643665; printf("error!\n"); } else if ( v3 == 1478060410 ) { v3 = 824643665; printf("you get it!\n"); } } return result;} check2(char *a1)
代码
__int64 __fastcall check2(char *a1){ size_t v1; // rax signed int v2; // ecx signed int v3; // eax signed int v4; // eax signed int v5; // eax signed int v6; // eax signed int v7; // eax signed int v8; // eax signed int v9; // eax signed int v11; // [rsp+8Ch] [rbp-C4h] unsigned int v12; // [rsp+90h] [rbp-C0h] int v13; // [rsp+94h] [rbp-BCh] signed int v14; // [rsp+98h] [rbp-B8h] signed int v15; // [rsp+9Ch] [rbp-B4h] int v16[42]; // [rsp+A0h] [rbp-B0h] char *s; // [rsp+148h] [rbp-8h] s = a1; v13 = 0; v12 = 1; v15 = 0; v11 = -2671583; while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( 1 ) { while ( v11 == -2119125118 ) { ++v15; v11 = -94879051; } if ( v11 != -1998111552 ) break; v6 = 396170963; if ( v15 < 9 ) v6 = -512482015; v11 = v6; } if ( v11 != -1695072055 ) break; ++v15; v11 = -1998111552; } if ( v11 != -1658909923 ) break; v8 = -1129833658; if ( D0g3[9 * v15 + v14] != sudoku[9 * v15 + v14] ) v8 = -528396247; v11 = v8; } if ( v11 != -1613667829 ) break; v11 = -2119125118; } if ( v11 != -1369143226 ) break; v14 = 0; v11 = -740861019; } if ( v11 != -1244045086 ) break; D0g3[9 * v15 + v14] = v16[v13++]; v11 = 1611237474; } if ( v11 != -1129833658 ) break; v11 = -90011013; } if ( v11 != -740861019 ) break; v4 = -1613667829; if ( v14 < 9 ) v4 = 705300330; v11 = v4; } if ( v11 != -528396247 ) break; v12 = 0; v11 = 1954800504; } if ( v11 != -512482015 ) break; v14 = 0; v11 = 564268595; } if ( v11 != -334121999 ) break; v15 = 0; v11 = -1998111552; } if ( v11 != -94879051 ) break; v3 = -334121999; if ( v15 < 9 ) v3 = -1369143226; v11 = v3; } if ( v11 != -90011013 ) break; ++v14; v11 = 564268595; } if ( v11 != -2671583 ) break; v1 = strlen(s); v2 = 2101131376; if ( v15 < v1 ) v2 = 441246003; v11 = v2; } if ( v11 == 396170963 ) break; switch ( v11 ) { case 430996436: ++v15; v11 = -2671583; break; case 441246003: v16[v15] = s[v15] - 232084296 + 232084248; v11 = 430996436; break; case 564268595: v7 = 1954800504; if ( v14 < 9 ) v7 = -1658909923; v11 = v7; break; case 705300330: v5 = 1611237474; if ( !D0g3[9 * v15 + v14] ) v5 = -1244045086; v11 = v5; break; case 1611237474: v11 = 2119231421; break; case 1908623879: v11 = -1695072055; break; case 1954800504: v9 = 1908623879; if ( !v12 ) v9 = 2014359934; v11 = v9; break; case 2014359934: v11 = 396170963; break; case 2101131376: v15 = 0; v11 = -94879051; printf("\n"); break; case 2119231421: ++v14; v11 = -740861019; break; } } return v12;} 发表评论
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初次前来，多多关照！
[***.217.46.12]2025年04月05日 14时57分42秒
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