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2024-02-14 21:59:16 723 阅读 AI 写同款 · GEO 优化AI 选题数据监测一键分发 ›

逆向实战：从x-mini签名到RPC调用，Frida动态分析小红书加密协议本文通过实战案例，详细拆解了小红书App的加密协议逆向分析过程。从识别x-mini签名等关键参数入手，结合静态分析与Frida动态Hook技术，追踪到核心的RPC调用入口，并最终封装成可独立调用的RPC函数，为安全研究和协议理解提供了完整的技术路径。阅读详情

2.14打卡

1030 完美数列

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main()
{
    long long int N,p;
    cin>>N>>p;
    vector<long long int> v1(N);
    for(int i=0;i<N;i++)
    {
        cin>>v1[i];
    }
    sort(v1.begin(),v1.end());
    int temp=0;
    int ans=0;
    int j=0;
    for(int i=0;i<N;i++)
    {
        long long int min=v1[i];
        while(min*p>=v1[j]&&j<N)
        {
            j++;
        }
        ans=j-i;
        if(ans>temp)
        {
            temp=ans;
        }
    }
    cout<<temp<<endl;
    system("pause");
    return 0;
}

1031 查验身份证

#include<iostream>
#include<algorithm>
#include<string>
#include<map>
#include<cctype>
using namespace std;
int main()
{
    int N;
    cin>>N;
    int count=0;
    int jiaquan[17]={7,9,1,5,8,4,2,1,6,3,7,9,10,5,8,4,2};
    map<int,char> m1={{0,'1'},{1,'0'},{2,'X'},{3,'9'},{4,'8'},{5,'7'},{6,'6'},{7,'5'},{8,'4'},{9,'3'},{10,'2'}};
    for(int i=0;i<N;i++)
    {
        string s;
        cin>>s;
        int d1=0;
        int temp=0;
        for(int j=0;j<s.size()-1;j++)
        {
            if(isalpha(s[j]))
            {
                cout<<s<<endl;
                temp=1;
                count+=1;
                break;
            }
            d1+=(s[j]-'0')*jiaquan[j];
        }
        if(temp==0)
        {
            d1=d1%11;
            if(m1[d1]!=s[17])
            {
                cout<<s<<endl;
                count+=1;
            }
        }
    }
    if(count==0)
    {
        cout<<"All passed"<<endl;
    }
    system("pause");
    return 0;
}

1032 挖掘机技术哪家强

#include<iostream>
#include<set>
#include<map>
#include<vector>
#include<algorithm>
using namespace std;
int cmp(pair<int,int> a, pair<int,int> b) {
    return a.second > b.second;
}
int main()
{
    int N;
    cin>>N;
    set<int> s1;
    map<int,int> m1;
    for(int i=0;i<N;i++)
    {
        int num,score;
        cin>>num>>score;
        if(s1.count(num)==0)
        {
            m1[num]=score;
            s1.insert(num);
        }
        else if(s1.count(num)==1)
        {
            m1[num]=m1[num]+score;
        }
    }
    vector<pair<int,int>> vec;
    map<int,int>::iterator it;
    for(it = m1.begin();it != m1.end(); it++)
    {
        vec.push_back(pair<int,int>(it->first,it->second));
    }
    sort(vec.begin(),vec.end(),cmp);
    cout<<vec[0].first<<" "<<vec[0].second<<endl;
    system("pause");
    return 0;
}

1033 旧键盘打字

#include<iostream>
#include<string>
#include<set>
#include<cctype>
using namespace std;
int main()
{
    string s1,s2;
    getline(cin,s1);
    getline(cin,s2);
    set<char> set1;
    int num=0;
    for(int i=0;i<s1.size();i++)
    {
        set1.insert(s1[i]);   
    }
    for(int i=0;i<s2.size();i++)
    {
        if(islower(s2[i])&&set1.count(toupper(s2[i]))==1)
        {
            continue;
        }
        else if(isupper(s2[i]))
        {
            if(set1.count('+')==1||set1.count(s2[i])==1)
            {
                continue;
            }
            else
            {
                num=1;
                cout<<s2[i];
                continue;
            }
        }
        else if(set1.count(s2[i])==1)
        {
            continue;
        }
        num=1;
        cout<<s2[i];
    }
    if(num==0)
    {
        cout<<endl;
    }
    system("pause");
    return 0;
}