9.5 BUUCTF reverse3
重点在于
注意到这个函数 是对str进行了加密处理的
查看这个函数:
void *__cdecl sub_411AB0(char *a1, unsigned int a2, int *a3)
{
int v4; // [esp+D4h] [ebp-38h]
int v5; // [esp+D4h] [ebp-38h]
int v6; // [esp+D4h] [ebp-38h]
int v7; // [esp+D4h] [ebp-38h]
int i; // [esp+E0h] [ebp-2Ch]
unsigned int v9; // [esp+ECh] [ebp-20h]
int v10; // [esp+ECh] [ebp-20h]
int v11; // [esp+ECh] [ebp-20h]
void *v12; // [esp+F8h] [ebp-14h]
char *v13; // [esp+104h] [ebp-8h]
if ( !a1 || !a2 )
return 0;
v9 = a2 / 3;
if ( (int)(a2 / 3) % 3 )
++v9;
v10 = 4 * v9;
*a3 = v10;
v12 = malloc(v10 + 1);
if ( !v12 )
return 0;
j_memset(v12, 0, v10 + 1);
v13 = a1;
v11 = a2;
v4 = 0;
while ( v11 > 0 )
{
byte_41A144[2] = 0;
byte_41A144[1] = 0;
byte_41A144[0] = 0;
for ( i = 0; i < 3 && v11 >= 1; ++i )
{
byte_41A144[i] = *v13;
--v11;
++v13;
}
if ( !i )
break;
switch ( i )
{
case 1:
*((_BYTE *)v12 + v4) = aAbcdefghijklmn[(int)(unsigned __int8)byte_41A144[0] >> 2];
v5 = v4 + 1;
*((_BYTE *)v12 + v5) = aAbcdefghijklmn[((byte_41A144[1] & 0xF0) >> 4) | (16 * (byte_41A144[0] & 3))];
*((_BYTE *)v12 + ++v5) = aAbcdefghijklmn[64];
*((_BYTE *)v12 + ++v5) = aAbcdefghijklmn[64];
v4 = v5 + 1;
break;
case 2:
*((_BYTE *)v12 + v4) = aAbcdefghijklmn[(int)(unsigned __int8)byte_41A144[0] >> 2];
v6 = v4 + 1;
*((_BYTE *)v12 + v6) = aAbcdefghijklmn[((byte_41A144[1] & 0xF0) >> 4) | (16 * (byte_41A144[0] & 3))];
*((_BYTE *)v12 + ++v6) = aAbcdefghijklmn[((byte_41A144[2] & 0xC0) >> 6) | (4 * (byte_41A144[1] & 0xF))];
*((_BYTE *)v12 + ++v6) = aAbcdefghijklmn[64];
v4 = v6 + 1;
break;
case 3:
*((_BYTE *)v12 + v4) = aAbcdefghijklmn[(int)(unsigned __int8)byte_41A144[0] >> 2];
v7 = v4 + 1;
*((_BYTE *)v12 + v7) = aAbcdefghijklmn[((byte_41A144[1] & 0xF0) >> 4) | (16 * (byte_41A144[0] & 3))];
*((_BYTE *)v12 + ++v7) = aAbcdefghijklmn[((byte_41A144[2] & 0xC0) >> 6) | (4 * (byte_41A144[1] & 0xF))];
*((_BYTE *)v12 + ++v7) = aAbcdefghijklmn[byte_41A144[2] & 0x3F];
v4 = v7 + 1;
break;
}
}
*((_BYTE *)v12 + v4) = 0;
return v12;
}
结合 aAbcdefghijklmn 的值:
可以判断出是base64加密
写个脚本反向解密即可:
if __name__ == "__main__":
str2 = "e3nifIH9b_C@n@dH"
des = ""
for j in range(0,len(str2)):
des += chr(ord(str2[j])-j)
print(des)
然后base64解密即可得到flag{}:flag{i_l0ve_you}