逆向攻防世界CTF系列50-testre

定位

image-20241208144009004

跟进400D00

image-20241208144029535

逻辑:接收输入,长度17,最后附上一个0符号(可能代表结束符)

跟进sub_400700

复杂

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__int64 __fastcall sub_400700(void *a1, _QWORD *a2, __int64 a3, size_t a4){
s = a1;
v30 = a2;
v29 = a3;
v28 = a4;
v27 = -559038737;
v26 = malloc(0x100uLL);
v25 = v29;
v24 = v6;
v22 = 0LL;
v17 = 0LL;
for ( i = 0LL; i < v28; ++i ){
v13 = *(unsigned __int8 *)(v25 + i);
*((_BYTE *)v26 + i) = byte_400E90[i % 0x1D] ^ v13;
*((_BYTE *)v26 + i) += *(_BYTE *)(v25 + i);
}
while ( 1 ){
v12 = 0;
if ( v17 < v28 )
v12 = ~(*(_BYTE *)(v25 + v17) != 0);
if ( (v12 & 1) == 0 )
break;
++v17;
}
n = 138 * (v28 - v17) / 0x64 + 1;
v23 = ((v17 + v28) << 6) / 0x30 - 1;
v11 = (unsigned __int8 *)v6 - ((138 * (v28 - v17) / 0x64 + 16) & 0xFFFFFFFFFFFFFFF0LL);
memset(v11, 0, n);
v20 = v17;
v18 = n - 1;
while ( v20 < v28 ){
v21 = *(unsigned __int8 *)(v25 + v20);
for ( j = n - 1; ; --j ){
v10 = 1;
if ( j <= v18 )
v10 = v21 != 0;
if ( !v10 )
break;
v22 = v11[j] << 6;
v21 += v11[j] << 8;
v9 = 64;
v11[j] = v21 % 58;
*((_BYTE *)v26 + j) = v22 & 0x3F;
v22 >>= 6;
v21 /= 58;
v27 /= v9;
if ( !j )
break;
}
++v20;
v18 = j;
}
for ( j = 0LL; ; ++j ){
v8 = 0;
if ( j < n )
v8 = ~(v11[j] != 0);
if ( (v8 & 1) == 0 )
break;
}
if ( *v30 > n + v17 - j ){
if ( v17 ){
c = 61;
memset(s, 49, v17);
memset(v26, c, v17);
}
v20 = v17;
while ( j < n ){
v4 = v11;
*((_BYTE *)s + v20) = byte_400EB0[v11[j]];
*((_BYTE *)v26 + v20++) = byte_400EF0[v4[j++]];
}
*((_BYTE *)s + v20) = 0;
*v30 = v20 + 1;
if ( !strncmp((const char *)s, "D9", 2uLL)
&& !strncmp((const char *)s + 20, "Mp", 2uLL)
&& !strncmp((const char *)s + 18, "MR", 2uLL)
&& !strncmp((const char *)s + 2, "cS9N", 4uLL)
&& !strncmp((const char *)s + 6, "9iHjM", 5uLL)
&& !strncmp((const char *)s + 11, "LTdA8YS", 7uLL)){
v6[1] = puts("correct!");
}
v32 = 1;
v14 = 1;
}
else{
*v30 = n + v17 - j + 1;
v32 = 0;
v14 = 1;
}
return v32 & 1;
}

关键是最后得是s=D9cS9N9iHjMLTdA8YSMRMp

代码很长,硬着头看,第一部分:

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for ( i = 0LL; i < v28; ++i )
{
v13 = *(unsigned __int8 *)(v25 + i);
*((_BYTE *)v26 + i) = byte_400E90[i % 0x1D] ^ v13;
*((_BYTE *)v26 + i) += *(_BYTE *)(v25 + i);
}

v28是长度也就是a4 ,v13是输入,看了下后面v26好像没有怎么用到,并且这里的逻辑好像是v13=input,v26=input^a,v26=input,其实v26改成了input而已,跟踪也知道这暗示是个假的

image-20241208144718592

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while ( 1 ){
v12 = 0;
if ( v17 < v28 ) v12 = ~(*(_BYTE *)(v25 + v17) != 0);
if ((v12 & 1) == 0 ) break;
++v17;
}

如果想break,v12必须为0,~(*(_BYTE )(v25 + v17) != 0)为1,((_BYTE )(v25 + v17) != 0)为0,((_BYTE *)(v25 + v17) ≠0

总而言之没啥用,你会发现这里只是做了判断,本身并不会改变值,程序总能break。没啥用

下面代码我去除了无用变量v9,v27,v22 >>= 6;

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while ( v20 < v28 ){
v21 = *(unsigned __int8 *)(v25 + v20);
for ( j = n - 1; ; --j ){
v10 = 1;
if ( j <= v18 ) v10 = v21 != 0;
if ( !v10 ) break;
v22 = v11[j] << 6;
v21 += v11[j] << 8;
v11[j] = v21 % 58;
*((_BYTE *)v26 + j) = v22 & 0x3F;
v21 /= 58;
if ( !j ) break;
}
++v20;
v18 = j;
}

image-20241208151623430

看这里,之前有一个fake,然后下面有一个58长度的字符串,下面有个64长度的(无用),结合v11[j] = v21 % 58;,猜测可能是个Base58编码表

具体逻辑有点复杂,其实不用深究,看看下面,同样用不闪婚,直接过

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for ( j = 0LL; ; ++j ){
v8 = 0;
if ( j < n ) v8 = ~(v11[j] != 0);
if ( (v8 & 1) == 0 ) break;
}
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if ( *v30 > n + v17 - j ){
if ( v17 ){
c = 61;
memset(s, 49, v17);
memset(v26, c, v17);
}
v20 = v17;
while ( j < n ){
v4 = v11;
*((_BYTE *)s + v20) = byte_400EB0[v11[j]];
*((_BYTE *)v26 + v20++) = byte_400EF0[v4[j++]];
}
*((_BYTE *)s + v20) = 0;
*v30 = v20 + 1;
if ( !strncmp((const char *)s, "D9", 2uLL)
&& !strncmp((const char *)s + 20, "Mp", 2uLL)
&& !strncmp((const char *)s + 18, "MR", 2uLL)
&& !strncmp((const char *)s + 2, "cS9N", 4uLL)
&& !strncmp((const char *)s + 6, "9iHjM", 5uLL)
&& !strncmp((const char *)s + 11, "LTdA8YS", 7uLL)){
v6[1] = puts("correct!");
}
v32 = 1;
v14 = 1;
}

先开辟空间,再在while循环里赋值,v26用不上

image-20241208152841631

得flag,flag{base58_is_boring}