强网拟态线上mobile的两道wp
VSole2021-12-22 16:08:56
找到关键代码在mainactivity中。
输入处理：
byte[] v7_1 = a.c(v1_1[v6]);.... while(v9 < v7_1.length) { v10 = v10 << 1 | (v7_1[v9] & 0x80) >>> v2; v7_1[v9] = ((byte)(v7_1[v9] & 0x7F)); ++v9; }
关键比较：
String v9_1 = a.a(v10, v7_1); if(v9_1 != null && (v9_1.equals(a.a(a.b[v6], v7_1)))) { if(!v9_1.equals(a.c[v6])) { } else { ++v6; continue; } }
需要a类中的字符串数组C要等同于a方法对v7_1处理的结果。
查看A类a方法，可以看到是根据a.b[v6]进行hash加密，任选了一个进行爆破，爆不出来。仔细翻了翻了，前面的welcomeactivity类更改了a密钥byte数组和b数组。
for(v0 = 0; true; ++v0) { int[] v1 = a.b; if(v0 >= v1.length) { break; } v1[v0] ^= 0xAB; }....while(v4 < a.a.length) { try { v0_1 = MessageDigest.getInstance("MD5"); } catch(NoSuchAlgorithmException v1_1) { v1_1.printStackTrace(); } v0_1.update(a.a[v4]); a.a[v4] = v0_1.digest(); ++v4; }
对密钥进行MD5。
当b数组的值<8时，不使用密钥，hash加密的方式不同，b数组>=8时，相同的hash加密HmacSha512，不同密钥。
这样就直接开始爆破，8组数据，范围为4个0-0x7f字节。直接抄程序反编译代码去爆破。
HmacSha512: 将密钥和密文对应就行
import java.nio.charset.StandardCharsets;import java.math.BigInteger;import java.security.GeneralSecurityException;import java.security.InvalidKeyException;import java.security.Key;import java.security.MessageDigest;import java.security.NoSuchAlgorithmException;import javax.crypto.Mac;import javax.crypto.spec.SecretKeySpec; public class demo { public static final String v0 = "HmacSha512"; public static void main(String[] args) { MessageDigest v0_1 = null; StringBuilder v1_2; byte[] miwen = new byte[4]; String v6_1 = "%02x"; try { for (int i=0x0; i<=0x7e; i++) { miwen[0] = (byte)i; for (int j = 0x0; j <= 0x7e; j++) { miwen[1] = (byte)j; for (int k = 0x0; k <= 0x7e; k++) { miwen[2] = (byte)k; for (int l = 0x0; l <= 0x7e; l++) { miwen[3] = (byte)l; try { v0_1 = MessageDigest.getInstance("MD5"); } catch(NoSuchAlgorithmException v1_1) { v1_1.printStackTrace(); } byte[] tmp = "ALFjcgztxnUaC89v".getBytes(); v0_1.update(tmp); byte[] miwen2 = v0_1.digest(); SecretKeySpec v1 = new SecretKeySpec(miwen2, v0); Mac v3 = Mac.getInstance(v0); v3.init(((Key)v1)); v3.update(miwen); String v2_2; String s = "78b0be39e63b6837"; for(v2_2 = new BigInteger(1, v3.doFinal()).toString(16); v2_2.length() < 0x20; v2_2 = "0" + v2_2) { } if (v2_2.substring(0,16).equals(s)) { System.out.printf("%x ", i); System.out.printf("%x ", j); System.out.printf("%x ", k); System.out.printf("%x ", l); System.out.printf("%n"); return; } } } } } } catch(InvalidKeyException | NoSuchAlgorithmException v2) { return; } }}
无密钥hash: 将hash方法和密文对应
import java.nio.charset.StandardCharsets;import java.math.BigInteger;import java.security.GeneralSecurityException;import java.security.InvalidKeyException;import java.security.Key;import java.security.MessageDigest;import java.security.NoSuchAlgorithmException;import javax.crypto.Mac;import javax.crypto.spec.SecretKeySpec; public class demo5 { public static final String v0 = "HmacSha512"; public static void main(String[] args) { MessageDigest v1_11; byte[] miwen2; byte[] miwen = new byte[4]; String v6_1 = "%02x"; StringBuilder v1_2; for (int i=0x0; i<=0x7e; i++) { miwen[0] = (byte)i; for (int j = 0x0; j <= 0x7e; j++) { miwen[1] = (byte)j; for (int k = 0x0; k <= 0x7e; k++) { miwen[2] = (byte)k; for (int l = 0x0; l <= 0x7e; l++) { miwen[3] = (byte)l; String s = "f2dda5fc021fe2bf"; try { v1_11 = MessageDigest.getInstance("SHA-384" ); v1_11.update(miwen); miwen2 = v1_11.digest(); v1_2 = new StringBuilder(); for(int v4 = 0; v4 < miwen2.length; ++v4) { v1_2.append(String.format(v6_1, Byte.valueOf(miwen2[v4]))); } String tmp = v1_2.toString().substring(0,16); if (tmp.equals(s)){ System.out.printf("%x ", i); System.out.printf("%x ", j); System.out.printf("%x ", k); System.out.printf("%x ", l); System.out.printf("%n"); return; } } catch(NoSuchAlgorithmException v0_2) { v0_2.printStackTrace(); } } } } } }}
这里我没有再去考虑大于7F的输入，我直接分析flag的输出：
if((v10 >>> 3 & 1) == 0) { v6_1[v10 & 7] = new String(v8_1);}else { for(v9 = 0; v9 < v8_1.length / v5; ++v9) { byte v11 = v8_1[v9]; v8_1[v9] = v8_1[v8_1.length - 1 - v9]; v8_1[v8_1.length - 1 - v9] = v11; } v6_1[v10 & 7] = new String(v8_1); // 输出顺序和加密方式有关}
这里可以看到，如果使用了HmacSha512，需要反序。
最后爆破了半天得到flag。
FLAG{H@5H_15_7H3_8@PP1N355_C11P}
Studydesk
输入处理：
byte[] v0 = a.a;.....if(v6 < v5) { v9 = in.indexOf(v0[v6]); if(v9 == -1) { v0 = ((byte[])v7); } else { v4.append(Integer.toBinaryString((v9 & 0x3F) + v2).substring(1)); ++v6; continue; } }
输入为32字节，然后依次查询a类的a数组再输入中的位置，将位置+32后转化为2进制去掉首位保存字符串。
保存的范围就是00000 - 11111，一共是16个。
查看a类的a数组，可以编写脚本处理发现只有32种。要求的是每个数都再输入中能找到，不能重复。
之后就是2进制连起来，依次取8位，转化为byte数组。
密文的生成直接抄反编译代码，然后调试可以发现，是圆周率。长度是360位。
校验：
while(true) { if(v3_3 < v2_2.length()) { v4_1 = v3_3 + 2; if((((byte)(Integer.parseInt(v2_2.substring(v3_3, v4_1), 10) & 0xFF))) != v0[v3_3 / 2]) { v0_1 = 0; } else { v3_3 = v4_1; continue; } } else { break; } goto label_120; }
要求的是依次取圆周率字符串2位，视为10进制数和5位2进制数生成的byte数组一一比较，一共180次。
这里不妨算一下，a数组的长度为288位，转化出来的2进制长度为288 5 == 1440，密文长度是180 8 == 1440，刚好对应。
c = [0x73, 0x6F, 43, 0x72, 0x74, 45, 0x30, 36, 84, 98, 89, 36, 38, 66, 38, 43, 84, 0x79, 50, 101, 101, 43, 100, 87, 69, 0x6F, 51, 66, 89, 49, 69, 51, 101, 51, 53, 0x74, 45, 98, 98, 0x72, 50, 36, 98, 50, 85, 85, 85, 107, 66, 36, 53, 51, 0x6F, 0x72, 89, 89, 66, 50, 33, 66, 0x5F, 66, 101, 0x79, 0x5F, 0x40, 33, 66, 50, 0x40, 85, 85, 45, 43, 36, 50, 0x74, 0x30, 85, 0x73, 0x5F, 0x40, 49, 0x72, 50, 101, 101, 51, 51, 43, 53, 51, 53, 51, 85, 50, 0x40, 0x79, 53, 36, 0x40, 69, 89, 98, 45, 0x6F, 101, 36, 97, 66, 100, 0x30, 0x73, 97, 0x30, 36, 0x6F, 101, 50, 0x5F, 49, 0x30, 0x40, 89, 0x74, 85, 0x30, 85, 0x73, 89, 43, 89, 97, 0x30, 89, 0x72, 97, 100, 38, 50, 0x74, 51, 98, 0x75, 0x5F, 50, 0x74, 0x73, 0x6F, 84, 98, 89, 69, 0x6F, 100, 0x30, 0x6F, 98, 89, 0x72, 0x40, 50, 36, 66, 89, 101, 0x72, 51, 84, 51, 50, 36, 38, 0x40, 0x30, 53, 51, 0x30, 49, 97, 0x74, 89, 101, 85, 97, 66, 84, 97, 45, 43, 100, 89, 45, 0x30, 0x73, 0x30, 0x40, 97, 100, 98, 51, 100, 0x6F, 0x73, 50, 53, 101, 66, 101, 0x6F, 0x75, 50, 45, 0x5F, 51, 82, 50, 89, 87, 101, 50, 89, 0x30, 89, 101, 43, 89, 36, 38, 61, 101, 0x40, 84, 89, 0x5F, 66, 0x74, 49, 0x40, 87, 97, 43, 0x5F, 0x73, 43, 0x30, 89, 45, 84, 89, 33, 89, 107, 53, 85, 0x30, 98, 98, 0x5F, 50, 107, 66, 101, 0x6F, 51, 97, 33, 66, 97, 0x75, 51, 0x74, 51, 97, 0x40, 89, 107, 98, 51, 69, 0x40, 73, 0x5F, 0x30, 85, 0x74, 0x30, 97]b = []for i in range(0x20): b.append("{0:b}".format(i).zfill(5))a = [31,41,59,26,53,58,97,93,23,84,62,64,33,83,27,95,2,88,41,97,16,93,99,37,51,5,82,9,74,94,45,92,30,78,16,40,62,86,20,89,98,62,80,34,82,53,42,11,70,67,98,21,48, 8,65,13,28,23, 6,64,70,93,84,46, 9,55, 5,82,23,17,25,35,94, 8,12,84,81,11,74,50,28,41, 2,70,19,38,52,11, 5,55,96,44,62,29,48,95,49,30,38,19,64,42,88,10,97,56,65,93,34,46,12,84,75,64,82,33,78,67,83,16,52,71,20,19, 9,14,56,48,56,69,23,46, 3,48,61, 4,54,32,66,48,21,33,93,60,72,60,24,91,41,27,37,24,58,70, 6,60,63,15,58,81,74,88,15,20,92, 9,62,82,92,54, 9,17,15,36,43,67,89,25,90,36]d = ""for i in a: d += "{0:b}".format(i).zfill(8)e = [0] * 32k = 0i = 0while i < 8 * 180: for j in range(len(b)): if d[i:i + 5] == b[j]: e[j] = c[k] k += 1 i += 5 for i in e: print(chr(i), end="")
加上flag{}就ok。
附件中的HAHAHAHA我重新打包过。
VSole
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