感觉这场好简单，去年场上做的CCPC都好难，嘤嘤嘤

这场的题目都挺卡的。

L. LRU Algorithm

只能说这个时间非常的卡。

写了哈希和散列表T掉了，然后改成暴力for询问判断，这下时间正常了，然后一直wa。最后发现询问全\(0\)的时候也要输出\(\text{Yes}\)，是不是有一点点坑。

哈希需要注意：

• 哈希是将字符串看成\(n\)进制数，选取的\(base\)必须\(\geq\)字符集大小。

• 散列表这东西就是看上去很快，真正卡常的时候是不是应该看作一个\(\log\)。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef double db;
typedef pair <int, int> pin;
typedef pair <ll, ll> pll;

#ifndef ONLINE_JUDGE
bool MEMORY_ST;
#endif

const int N = 5005;

int T, n, qn, a[N], s[N], hq1[N], hq2[N], pos[N];
bool ans[N];
vector <int> q[N];

inline void inc(int &x, int y, int P) {
    if (!(y >= 0 && y < P)) y = (y % P + P) % P;
    x += y;
    if (x >= P) x -= P;
}

namespace HashTable {
    int b1 = 8209, b2 = 7307, P1 = 999749, P2 = 10001659;
}
using HashTable::b1;
using HashTable::b2;
using HashTable::P1;
using HashTable::P2;

namespace Fread {
    const int L = 1 << 15;
    
    char buffer[L], *S, *T;
    
    inline char Getchar() {
        if(S == T) {
            T = (S = buffer) + fread(buffer, 1, L, stdin);
            if(S == T) return EOF;
        }
        return *S++;
    }
    
    template <class T> 
    inline void read(T &X) {
        char ch; T op = 1;
        for(ch = Getchar(); ch > '9' || ch < '0'; ch = Getchar())
            if(ch == '-') op = -1;
        for(X = 0; ch >= '0' && ch <= '9'; ch = Getchar()) 
            X = (X << 1) + (X << 3) + ch - '0'; 
        X *= op;
    }
    
} using Fread::read;   

#ifndef ONLINE_JUDGE
bool MEMORY_ED;
#endif

int main() {
#ifndef ONLINE_JUDGE
    freopen("sample.in", "r", stdin);
    freopen("test.out", "w", stdout);
    clock_t st_clock = clock();
#endif

    read(T);
    for (; T--; ) {
        read(n), read(qn);
        for (int i = 1; i <= n; i++) {
            read(a[i]);
            s[i] = pos[i] = 0;
        } 
        for (int i = 1; i <= qn; i++) ans[i] = 0;
        for (int m, i = 1; i <= qn; i++) {
            read(m);
            int cur1 = 0, cur2 = 0;
            for (int ix, j = 1; j <= m; j++) {
                read(ix);
                cur1 = 1LL * cur1 * b1 % P1;
                inc(cur1, ix, P1);
                cur2 = 1LL * cur2 * b2 % P2;
                inc(cur2, ix, P2);
            } 
            q[m].emplace_back(i);
            hq1[i] = cur1, hq2[i] = cur2;
            if (cur1 == 0 && cur2 == 0) ans[i] = 1;
        }
        for (int i = 1; i <= n; i++) {
            int p = pos[a[i]];
            if (!p) {
                for (int j = n - 1; j >= 1; j--) s[j + 1] = s[j];
            } else {
                for (int j = p - 1; j >= 1; j--) s[j + 1] = s[j];
            }
            s[1] = a[i];
            int cur1 = 0, cur2 = 0;
            for (int j = 1; j <= n; j++) {
                if (s[j] != 0) pos[s[j]] = j;
                cur1 = 1LL * cur1 * b1 % P1;
                inc(cur1, s[j], P1);
                cur2 = 1LL * cur2 * b2 % P2;
                inc(cur2, s[j], P2);
                for (int k = 0; k < q[j].size(); k++) {
                    int id = q[j][k];
                    if (hq1[id] == cur1 && hq2[id] == cur2) {
                        ans[id] = 1;
                    }
                }
            }
            // for (int j = 1; j <= n; j++)
            //     printf("%d%c", s[j], " \n"[j == n]);
        }
        for (int i = 1; i <= qn; i++) puts(ans[i] ? "Yes" : "No");
        for (int i = 1; i <= n; i++) q[i].clear();
    }

#ifndef ONLINE_JUDGE
    clock_t ed_clock = clock();
    printf("time = %f ms\n", (double)(ed_clock - st_clock) / CLOCKS_PER_SEC * 1000);
    printf("memory = %.2f MB\n", (&MEMORY_ED - &MEMORY_ST) / 1024.0 / 1024.0);
#endif
    return 0;
}