强网拟态

Posted by marginal on 2021-10-25
Estimated Reading Time 9 Minutes
Words 1.8k In Total
Viewed Times

HaHaHaHa

找到关键代码在mainactivity中.

输入处理:

1
2
3
4
5
6
7
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:

1
2
3
4
5
6
7
8
9
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.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
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

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
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

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
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:

1
2
3
4
5
6
7
8
9
10
11
12
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.

1
FLAG{H@5H_15_7H3_8@PP1N355_C11P}

Studydesk

输入处理:

1
2
3
4
5
6
7
8
9
10
11
12
13
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:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
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.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
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] * 32
k = 0
i = 0
while 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.

如果您喜欢此博客或发现它对您有用，则欢迎对此发表评论。 也欢迎您共享此博客，以便更多人可以参与。 如果博客中使用的图像侵犯了您的版权，请与作者联系以将其删除。 谢谢 ！

特色标签
链友
ESC