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做法： 用字典樹存好譯碼詞，然後模擬即可

AC程式碼：

#include <bits/stdc++.h>
using namespace std;

#define IO          ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define pb(x)       push_back(x)
#define sz(x)       (int)(x).size()
#define sc(x)       scanf("%d",&x)
#define abs(x)      ((x)<0 ? -(x) : x)
 

#define all(x)      x.begin(),x.end()
#define mk(x,y)     make_pair(x,y)
#define fin         freopen("in.txt","r",stdin)
#define fout        freopen("out.txt","w",stdout)

typedef long long ll;
typedef pair<int,int> P;
const int mod = 1e9+7;
const int maxm = 1e8+5;
const int maxn = 1e6+5;
const
 
 int INF = 0x3f3f3f3f;
const ll LINF = 1ll<<62;

int trie[maxn][2],cnt,tot;
int val[maxn],num[maxn],ans[maxn];
char s[maxn];
inline void add(char *s,int k)
{
    int rt = 0;
    for(int i=0;s[i];i++)
    {
        int x = s[i]-'0';
        if(trie[rt][x] == 0) trie[rt][x] = ++cnt;
        rt = trie[
 
rt][x];
    }
    val[rt] = k;
}
int main()
{
//    fin;
    IO;
    int t;
    cin>>t;
    while(t--)
    {
        memset(val,0,sizeof(val));
        memset(trie,0,sizeof(trie));
        tot = 0;
        cnt = 0;
        int n,m,x;
        cin>>n>>m;
        char str[15];
        for(int i=0;i<m;i++)
        {
            cin>>x>>str;
            add(str,x);
        }
        cin>>s;
        for(int i=0;s[i];i++)
        {
            int now;
            if(s[i]>='0' &&s[i]<='9') now = s[i]-'0';
            else if(s[i]>='a' && s[i]<='z') now = s[i]-'a'+10;
            else if(s[i]>='A' && s[i]<='Z') now = s[i]-'A'+10;
            num[++tot] = (now>>3)&1;
            num[++tot] = (now>>2)&1;
            num[++tot] = (now>>1)&1;
            num[++tot] = now&1;
        }
        int cnt2 = 0;
        for(int i=1;i<=tot;i+=9)
        {
            int k = 0;
            for(int j=i;j<=i+7;j++) if(num[j]) k++;
            if((k&1) && num[i+8] == 0 || (k&1) == 0 && num[i+8] == 1)
            {
                for(int j=i;j<=i+7;j++) ans[++cnt2] = num[j];
            }
        }
        int tot2 = 0,rt = 0;
        for(int i=1;i<=cnt2;i++)
        {
            rt = trie[rt][ans[i]];
            if(val[rt]){
                cout<<(char)val[rt];
                tot2++;
                rt = 0;
                if(tot2 == n) break;
            }
        }
        cout<<endl;
    }
    return 0;
}