查殼後發現有個upx殼，那我們先脫殼去

之後他還會提示你用32位的ida開啟，開啟之後在main主函式裡面找到了關鍵函式

 1  __main();
 2   qmemcpy(v4, "*F'\"N,\"(I?+@", sizeof(v4));
 3   printf("Please input:");
 4   scanf("%s", v6);
 5   if ( v6[0] != 65 || v6[1] != 67 || v6[2] != 84 || v6[3] != 70 || v6[4] != 123 || v10 != 125 )
 6     return 0;
 7   v5[0] = v7;
 
 8   v5[1] = v8;
 9   v5[2] = v9;
10   for ( i = 0; i <= 11; ++i )
11   {
12     if ( v4[i] != _data_start__[*((char *)v5 + i) - 1] )
13       return 0;
14   }
15   printf("You are correct!");
16   return 0;
17 }

意思判斷v4v5v6....v15這麼多的東西是否都等於上面這串東西相對應減1的數。

如果正確的的話關鍵語句就可以理解成這種  v4[i]!=byte_402000[flag[i]-1]

 1 byte_402000 = '~}|{zyxwvutsrqponmlkjihgfedcba`_^]\[ZYXWVUTSRQPONMLKJIHGFEDCBA@?>=<;:9876543210/.-,+*)(\'&%$# !"'
 2 
 3 v4 = [42,70,39,34,78,44,34,40,73,63,43,64]
 4 
 5 flag = ''
 6 
 7 for i in v4:
 8     flag += chr(byte_402000.find(chr(i)) + 1)
 9 
10 print(flag)

程式碼就單的就這樣就行，

套個flag就是答案