大家好關于這次比賽密碼學和部分移動安全如下

沒錯在這里我們來解題吧（不要吐槽我，嘴下饒人）

1

某加密演算法實作如下
import java.nio.charset.Charset;
public class DeEnCode {
    private static final String key0 = "2021.2.26";
    private static final Charset charset = Charset.forName("UTF-8");
    private static byte[] keyBytes = key0.getBytes(charset);
    public static String encode(String enc){
        byte[] b = enc.getBytes(charset);
        for(int i=0,size=b.length;i<size;i++){
            for(byte keyBytes0:keyBytes){
                b[i] = (byte) (b[i]^keyBytes0);
            }
        }
        return new String(b);
    }
加密flag后為：Q[VPLDRTwQBF^YJ
寫出解密演算法求出flag

解：

mport java.nio.charset.Charset;
public class DeEnCode {

    private static final String key0 = "2021.2.26";
    private static final Charset charset = Charset.forName("UTF-8");
    private static byte[] keyBytes = key0.getBytes(charset);


    public static String decode(String dec){
        byte[] e = dec.getBytes(charset);
        byte[] dee = e;
        for(int i=0,size=e.length;i<size;i++){
            for(byte keyBytes0:keyBytes){
                e[i] = (byte) (dee[i]^keyBytes0);
            }
        }
        return new String(e);
    }

    public static void main(String[] args) {
         String dec = decode("Q[VPLDRTwQBF^YJ")
;
        System.out.println(enc);
        System.out.println(dec);
    }
}

Flag:flag{sec@fuqin}

2、凱撒大帝用MD5三步跨欄套娃

R000VEdOUlRHTTNETU5SV0dNMlRHTlJUR1VaVENOUlVHTTRER05aV0dRM0RNTVpTR00zREdOSldHNFpUSU1aWEdNMkRNTlpUSEFaVEtOUlhHTTRUTU9KVEdRWlRFTVpWR000VEdNST0=

解：

打開題目
R000VEdOUlRHTTNETU5SV0dNMlRHTlJUR1VaVENOUlVHTTRER05aV0dRM0RNTVpTR00zREdOSldHNFpUSU1aWEdNMkRNTlpUSEFaVEtOUlhHTTRUTU9KVEdRWlRFTVpWR000VEdNST0=
題目為套娃猜想可能與base家族有關
https://www.qqxiuzi.cn/bianma/base64.htm

GM4TGNRTGM3DMNRWGM2TGNRTGUZTCNRUGM4DGNZWGQ3DMMZSGM3DGNJWG4ZTIMZXGM2DMNZTHAZTKNRXGM4TMOJTGQZTEMZVGM4TGMI=

3936336666353635316438376466323635673437346738356739693432353931

963ff5651d87df265g474g85g9i42591
題目提示凱撒三步則推測凱撒加密位移為3
https://www.qqxiuzi.cn/bianma/kaisamima.php
963cc5651a87ac265d474d85d9f42591
題目提示有MD5則在線解密
https://www.cmd5.com/

Flag：flag{sec2021}

3、另類rsa

在RSA體制中，某給定用戶的公鑰e=31，n=3599，那么給該用戶的私鑰等于多少？
解法如下：
1.根據n=3599可得p，q分別為59和61
2.根據pq的值可推出φ(n)=5961=3480
3.ed≡1(modφ(n))等價于ed–kφ(n)=1
因此d31-3480*y=1

根據擴展歐幾里得演算法（輾轉相除法）求解：
1式：3480=31*112+8 => 8=3480+(-112)31
2式：31=83+7 => 7=31+(-3)8
3式：8=71+1 => 1=8+(-1)*7

將2代入3：1=8+31+8*(-3)
1=31(-1)+84
將1代入： 1=31(-1)+[3480+31*(-112)]4
1=31(-449)+3480*4
由此得d=-449，y=4

由于d一般取正整數，所以d=d+kφ(n)=-449+1*3480=3031
所以給用戶的私鑰為（3599,3031）
Flag為：flag{3031}

4、移動安全

在這里插入圖片描述

package com.example.myapplication;

import android.os.Bundle;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.Toast;
import androidx.appcompat.app.AppCompatActivity;

public class MainActivity extends AppCompatActivity {
    private Button button;
    private EditText et1;
    private EditText et2;

    /* access modifiers changed from: protected */
    @Override // androidx.activity.ComponentActivity, androidx.core.app.ComponentActivity, androidx.appcompat.app.AppCompatActivity, androidx.fragment.app.FragmentActivity
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        this.button = (Button) findViewById(R.id.bu1);
        this.et1 = (EditText) findViewById(R.id.et1);
        this.et2 = (EditText) findViewById(R.id.et2);
        this.button.setOnClickListener(new View.OnClickListener() {
            /* class com.example.myapplication.MainActivity.AnonymousClass1 */

            public void onClick(View v) {
                String s1 = MainActivity.this.et1.getText().toString();
                String s2 = MainActivity.this.et2.getText().toString();
                new DeEnCode();
                String a = DeEnCode.encode("flag{fuqin}");
                if (!s1.equals(a) || !s2.equals(a)) {
                    Toast.makeText(MainActivity.this, "登陸失敗", 0).show();
                } else {
                    Toast.makeText(MainActivity.this, "解密：^TY_C^MIQVK][E", 0).show();
                }
            }
        });
    }
}

可以看到有個flag提交了是錯的再看用了DeEnCode類里面的方法，我們查看這個源代碼
在這里插入圖片描述

package com.example.myapplication;

import java.nio.charset.Charset;

public class DeEnCode {
    private static final Charset charset;
    private static final String key0 = "2021.1.19";
    private static byte[] keyBytes;

    static {
        Charset forName = Charset.forName("UTF-8");
        charset = forName;
        keyBytes = key0.getBytes(forName);
    }

    public static String encode(String enc) {
        byte[] b = enc.getBytes(charset);
        int size = b.length;
        for (int i = 0; i < size; i++) {
            for (byte keyBytes0 : keyBytes) {
                b[i] = (byte) (b[i] ^ keyBytes0);
            }
        }
        return new String(b);
    }
}

可以看到這是加密演算法分析一下啊將引數轉換為位元組存在陣列b里面然后遍歷這個陣列并且和keyBytes進行異或應該需要對這個flag{fuqin}進行加密試著去加密

import java.nio.charset.Charset;


public class DeEnCode {

    private static final String key0 = "2021.1.19";
    private static final Charset charset = Charset.forName("UTF-8");
    private static byte[] keyBytes = key0.getBytes(charset);

    public static String encode(String enc){
        byte[] b = enc.getBytes(charset);
        for(int i=0,size=b.length;i<size;i++){
            for(byte keyBytes0:keyBytes){
                b[i] = (byte) (b[i]^keyBytes0);
            }
        }
        return new String(b);
    }
 public static void main(String[] args) {
        String s="flag{fuqin}";
        String enc = encode(s);
        System.out.println(enc);
    }
}

運行一下
在這里插入圖片描述

彈出來一串字串讓我們解密這個東西回想當時的加密演算法寫出解密演算法解這個字串

import java.nio.charset.Charset;


public class DeEnCode {

    private static final String key0 = "2021.1.19";
    private static final Charset charset = Charset.forName("UTF-8");
    private static byte[] keyBytes = key0.getBytes(charset);
public static String decode(String dec){
    byte[] e = dec.getBytes(charset);
    byte[] dee = e;
    for(int i=0,size=e.length;i<size;i++){
        for(byte keyBytes0:keyBytes){
            e[i] = (byte) (dee[i]^keyBytes0);
        }
    }
    return new String(e);
}
 public static void main(String[] args) {
        String dec = decode("^TY_C^MIQVK][E");
        System.out.println(dec);
    }
}

運行一下：
在這里插入圖片描述
flag{fuqinsec}
5、安卓1
不要脫殼

6、rsa

腳本我放這里

import binascii
import sys 
sys.setrecursionlimit(1000000)
def ByteToHex(bins):
    return ''.join(["%02X" % x for x in bins]).strip()
def n2s(num):
    t = hex(num)[2:-1]  # python
    if len(t) % 2 == 1:
        t = '0' + t
    #print(t)
    return(binascii.a2b_hex(t).decode('latin1'))
def egcd(a, b):
    if a == 0:
        return (b, 0, 1)
    else:
        g, y, x = egcd(b % a, a)
        return (g, x - (b // a) * y, y)
def modinv(a, m):
    g, x, y = egcd(a, m)
    if g != 1:
        print('modular inverse does not exist')
        return 'null'
    else:
        return x % m
c = 38230991316229399651823567590692301060044620412191737764632384680546256228451518238842965221394711848337832459443844446889468362154188214840736744657885858943810177675871991111466653158257191139605699916347308294995664530280816850482740530602254559123759121106338359220242637775919026933563326069449424391192
p = 28805791771260259486856902729020438686670354441296247148207862836064657849735343618207098163901787287368569768472521344635567334299356760080507454640207003
q = 15991846970993213322072626901560749932686325766403404864023341810735319249066370916090640926219079368845510444031400322229147771682961132420481897362843199
e = 354611102441307572056572181827925899198345350228753730931089393275463916544456626894245415096107834465778409532373187125318554614722599301791528916212839368121066035541008808261534500586023652767712271625785204280964688004680328300124849680477105302519377370092578107827116821391826210972320377614967547827619
n = p * q
d = modinv(e, (p - 1) * (q - 1))
m = pow(c, d, n)
print (m)

這次出題主要是引領小白來學習安全的，各位大佬不要吐槽題目簡單，后面有機會會出點難題

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標籤：java

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大家好關于這次比賽密碼學和部分移動安全如下

1

2、凱撒大帝用MD5三步跨欄套娃

3、另類rsa

4、移動安全

6、rsa