CCF-CSP 2023 第三题 解压缩(50％)

Question

以下代码在ccf-csp官网运行通过50％，本人实在找不到bug了，想请教一下大家看看哪里错啦。
题目：http://118.190.20.162/view.page?gpid=T168

#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <math.h>

using namespace std;

//const int N = 2000010;

string str;
string result;
int n;
int len = 0;  //数据长度
int cnt = 0; 


int byte[8];

void change(char c, char c2)    //将一个字节 转为 二进制 放到 byte
{
    int num;
    memset(byte, 0, sizeof byte);
    if(c >= '0' && c <= '9') num = c - '0';
    else if(c >= 'a' && c <= 'f') num = c-'a' + 10;

    int x = 4;
    while(x)
    {
        byte[8-x] = num & 1;
        x--;
        num >>= 1;
    }

    if(c2 >= '0' && c2 <= '9') num = c2 - '0';
    else if(c2 >= 'a' && c2 <= 'f') num = c2-'a' + 10;

    x = 4;
    while(x)
    {
        byte[4-x] = num & 1;
        x--;
        num >>= 1;
    }

}

int GetLength(int idx)
{
    while(1)
    {
        char h = str[idx];
        char l = str[idx + 1];
        
        change(h, l);

        if(byte[7] == 1)
        {
            int sum = 0;
            for(int i = 6; i >= 0; --i) sum = sum * 2  + byte[i];
            //len += sum * pow(128, cnt);
            len += sum * (1 << (7 * cnt));  // 相当于 128 的 cnt 次幂
            cnt++;
        }
        else if(byte[7] == 0)
        {
            int sum = 0;
            for(int i = 6; i >= 0; --i) sum = sum * 2  + byte[i];
            //len += sum * pow(128, cnt);
            len += sum * (1 << (7 * cnt));  // 相当于 128 的 cnt 次幂
            cnt++;

            return idx + 2;
        }

        idx += 2;
    }
    
}

int Check(int idx)
{
    while(1)
    {
        char h = str[idx];
        char l = str[idx + 1];

        change(h, l);

        int sum = 0;

        if(byte[0] == 0 && byte[1] == 0)
        {
            for(int i = 7; i >= 2; --i) sum = sum * 2 + byte[i];
            int all = 0;
            if(sum <= 59)   //小于60 高六位表示 l-1 
            {
                for(int i = idx + 2; i <= idx + 2 * (sum + 1) + 1; i++) result += str[i];   //右边界再加一 为了完整输出最后一个字节
                idx += 2 * (sum + 2);
                return idx;
            }
            else if(sum >= 60)
            {
                int j = sum - 59;
                for(int i = 0; i < j; ++i)
                {
                    idx += 2;
                    h = str[idx];
                    l = str[idx + 1];

                    change(h, l);

                    sum = 0;
                    

                    for(int i = 7; i >= 0; --i) sum = sum * 2 + byte[i];

                    
                    all += sum * pow(128, i);
                }
                for(int i = idx + 2; i <= idx + 2 * (all + 1) + 1; i++) result += str[i];
                idx += 2 * (all + 2);
                return idx;
                
            }
  
        }
        else if(byte[0] == 1 && byte[1] == 0)
        {
            int o = 0, lsb = 0;
            for(int i = 4; i >= 2; --i) sum = sum * 2 + byte[i];

            lsb = sum + 4;              //得到lsb
            for(int i = 7; i >= 5; --i) o = o * 2 + byte[i];

            idx += 2;
            h = str[idx];
            l = str[idx + 1];

            change(h, l);

            for(int i = 7; i >= 0; --i) o = o * 2 + byte[i]; //得到o

            int temidx = result.length() - 1 - 2 * (o - 1) - 1;

            if(o >= lsb)
            {
                for(int i = temidx; i < temidx + 2 * lsb ; ++i) result += result[i];
            }
            else
            {
                int py = 0;
                while(lsb--)
                {
                    result += result[temidx + py];
                    result += result[temidx + py + 1];
                    py+=2;
                    py %= (o*2);
                }
            }

            return idx + 2;

        }
        else if(byte[0] == 0 && byte[1] == 1)
        {
            int o = 0, lsb = 0;
            for(int i = 7; i >= 2; --i) sum = sum * 2 + byte[i];

            lsb = sum + 1;              //得到l


            idx += 2;
            h = str[idx];
            l = str[idx + 1];

            change(h, l);

            sum = 0;
            for(int i = 7; i >= 0; --i) sum = sum * 2 + byte[i];
            o += sum;     

            idx += 2;
            h = str[idx];
            l = str[idx + 1];

            change(h, l);

            sum = 0;
            for(int i = 7; i >= 0; --i) sum = sum * 2 + byte[i];
            o += sum * 128;     //得到o

            int temidx = result.length() - 1 - 2 * (o - 1) - 1;

            if(o >= lsb)
            {
                for(int i = temidx; i < temidx + 2 * lsb ; ++i) result += result[i];
            }
            else
            {
                int py = 0;
                while(lsb--)
                {
                    result += result[temidx + py];
                    result += result[temidx + py + 1];
                    py+=2;
                    py %= (o*2);
                }
            }

            return idx + 2;
        }
        else return -1;

    }



}

int main()
{
    cin >> n;
    str.clear();
    result.clear();
    for(int i = 0; i < n/8 + 1; ++i)
    {
        string ss;
        cin >> ss;
        str += ss;
    }

    int idx = 0;
    
    idx = GetLength(idx);
    n -= idx / 2;

    while(n)
    {
        int ttem = idx;
        idx = Check(idx);
        n -= (idx - ttem) / 2;
        if(result.length() > 65535)
        {
            // 计算需要删除的字符数量
            int excess = result.length() - 65535;
            excess -= excess % 16;
            for(int i = 0; i < excess; ++i)
            {
                cout << result[i];
                if((i + 1) % 16 == 0) cout << endl;
            }
            // 从字符串的开头清空字符
            result.erase(0, excess);
        }

    }    
    for(int i = 0; i < result.length(); ++i)
    {
        cout << result[i];
        if((i + 1) % 16 == 0) cout << endl;
    } 

    return 0;
}

Answer 1

过了，私我给代码：