BUUCTF[羊城杯 2020]easyre

64位无壳程序

IDA64打开

main函数:

int __cdecl main(int argc, const char **argv, const char **envp)
{int v3; // eaxint v4; // eaxint v5; // eaxchar Str[48]; // [rsp+20h] [rbp-60h] BYREFchar Str1[64]; // [rsp+50h] [rbp-30h] BYREFchar v9[64]; // [rsp+90h] [rbp+10h] BYREFchar v10[64]; // [rsp+D0h] [rbp+50h] BYREFchar Str2[60]; // [rsp+110h] [rbp+90h] BYREFint v12; // [rsp+14Ch] [rbp+CCh] BYREF
_main();strcpy(Str2, "EmBmP5Pmn7QcPU4gLYKv5QcMmB3PWHcP5YkPq3=cT6QckkPckoRG");puts("Hello, please input your flag and I will tell you whether it is right or not.");scanf("%38s", Str);if ( strlen(Str) != 38|| (v3 = strlen(Str), (unsigned int)encode_one(Str, v3, v10, &v12))|| (v4 = strlen(v10), (unsigned int)encode_two(v10, v4, v9, &v12))|| (v5 = strlen(v9), (unsigned int)encode_three(v9, v5, Str1, &v12))|| strcmp(Str1, Str2) ){printf("Something wrong. Keep going.");return 0;}else{puts("you are right!");return 0;}
}

只要输入的Str不满足if的条件，就能得到正确的flag

首先是第一个条件是Str长度为38，然后经过三个编码函数处理，最后条件是Str1和Str2 = EmBmP5Pmn7QcPU4gLYKv5QcMmB3PWHcP5YkPq3=cT6QckkPckoRG两个字符串要相同

然后逐次分析endocde_one, encode_two, encode_three三个函数

encode_one:

__int64 __fastcall encode_one(const char *a1, int a2, char *a3, int *a4)
{int v5; // esiint v6; // esiint v7; // esiint v8; // [rsp+34h] [rbp-1Ch]int v9; // [rsp+38h] [rbp-18h]int v11; // [rsp+48h] [rbp-8h]int i; // [rsp+4Ch] [rbp-4h]unsigned __int8 *v13; // [rsp+70h] [rbp+20h]
v13 = (unsigned __int8 *)a1;if ( !a1 || !a2 )return 0xFFFFFFFFi64;v11 = 0;if ( a2 % 3 )v11 = 3 - a2 % 3;v9 = a2 + v11;v8 = 8 * (a2 + v11) / 6;for ( i = 0; i < v9; i += 3 ){*a3 = alphabet[(char)*v13 >> 2];if ( a2 + v11 - 3 == i && v11 ){if ( v11 == 1 ){v5 = (char)cmove_bits(*v13, 6u, 2u);a3[1] = alphabet[v5 + (char)cmove_bits(v13[1], 0, 4u)];a3[2] = alphabet[(char)cmove_bits(v13[1], 4u, 2u)];a3[3] = 61;}else if ( v11 == 2 ){a3[1] = alphabet[(char)cmove_bits(*v13, 6u, 2u)];a3[2] = 61;a3[3] = 61;}}else{v6 = (char)cmove_bits(*v13, 6u, 2u);a3[1] = alphabet[v6 + (char)cmove_bits(v13[1], 0, 4u)];v7 = (char)cmove_bits(v13[1], 4u, 2u);a3[2] = alphabet[v7 + (char)cmove_bits(v13[2], 0, 6u)];a3[3] = alphabet[v13[2] & 0x3F];}a3 += 4;v13 += 3;}if ( a4 )*a4 = v8;return 0i64;
}

双击查看alphabet

可以看到这个函数实际上就是base64加密

encode_two:

__int64 __fastcall encode_two(const char *a1, int a2, char *a3, int *a4)
{if ( !a1 || !a2 )return 0xFFFFFFFFi64;strncpy(a3, a1 + 26, 0xDui64);strncpy(a3 + 13, a1, 0xDui64);strncpy(a3 + 26, a1 + 39, 0xDui64);strncpy(a3 + 39, a1 + 13, 0xDui64);return 0i64;
}

函数进行了四次字符串复制操作：

a1: 1~13|14~26|27~39|40~52

最后a3：

a3: 26~38|1~13|40~53|14~26

encode_three

将其中的ascii码的数字转换成对应的字符

__int64 __fastcall encode_three(const char *a1, int a2, char *a3, int *a4)
{char v5; // [rsp+Fh] [rbp-11h]int i; // [rsp+14h] [rbp-Ch]const char *v8; // [rsp+30h] [rbp+10h]
v8 = a1;if ( !a1 || !a2 )return 0xFFFFFFFFi64;for ( i = 0; i < a2; ++i ){v5 = *v8;if ( *v8 <= '@' || v5 > 'Z' ){if ( v5 <= '`' || v5 > 'z' ){if ( v5 <= '/' || v5 > '9' )*a3 = v5;else*a3 = (v5 - '0' + 3) % 10 + '0';}else{*a3 = (v5 - 'a' + 3) % 26 + 'a';}}else{*a3 = (v5 - 'A' + 3) % 26 + 'A';}++a3;++v8;}return 0i64;
}

可以看的出是凯撒密码，而且偏移量为3

第二个函数处理后的结果：

str = BjYjM5Mjk7NzMR4dIVHs5NzJjY3MTEzM5VhMn3=zQ6NzhhMzhlOD

接下来写代码出结果：

import base64
tmp = list('BjYjM5Mjk7NzMR4dIVHs5NzJjY3MTEzM5VhMn3=zQ6NzhhMzhlOD')
for i in range(len(tmp)):if tmp[i] >= '0' and tmp[i] <= '9':tmp[i] = str(int(tmp[i]) - 3)
flag = str(tmp[13:26] + tmp[39:] + tmp[:13] + tmp[26:39])
print(base64.b64decode(flag))

带数字偏移的凯撒，用在线解密比手撸代码简单一些。

后面写代码的时候就直接base64解码了，忘记了数字的偏移，导致一直得不到正确的flag

最终结果：

GWHT{672cc4778a38e80cb362987341133ea2}

在buu里面结果就是

flag{672cc4778a38e80cb362987341133ea2}

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