11572 Unique Snowflakes - bitset

#include <bits/stdc++.h>
using namespace std;
#define PB push_back
#define ZERO (1e-10)
#define INF (1<<29)
#define CL(A,I) (memset(A,I,sizeof(A)))
#define DEB printf("DEB!\n");
#define D(X) cout<<"  "<<#X": "<<X<<endl;
#define EQ(A,B) (A+ZERO>B&&A-ZERO<B)
typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef pair<int,int> ii;
typedef vector<ii> vii;
#define IN(n) int n;scanf("%d",&n);
#define FOR(i, m, n) for (int i(m); i < n; i++)
#define REP(i, n) FOR(i, 0, n)
#define F(n) REP(i, n)
#define FF(n) REP(j, n)
#define FT(m, n) FOR(k, m, n)
#define aa first
#define bb second
void ga(int N,int *A){F(N)scanf("%d",A+i);}
#define MX (1<<20)
#define MD (MX-1)
#define SQ (1<<14)
struct bs{
    ll A[SQ];
    void clr(){CL(A,0);}
    void tr(int H){F(H>>6)A[i]=-1;FT(H-(H&63),H)ad(k);}
    bool is(bs&r){F(SQ)if(A[i]&r.A[i])return 1;return 0;}
    void operator|=(bs&r){F(SQ)A[i]|=r.A[i];}
    void operator&=(bs&r){F(SQ)A[i]&=r.A[i];}
    void operator^=(bs&r){F(SQ)A[i]^=r.A[i];}
    void ad(int u){A[u>>6]|=1ull<<(u&63);}
    void cg(int u){A[u>>6]^=1ull<<(u&63);}
    int xt(int u){
        while(++u&63)if((A[u>>6]>>(u&63))&1)return u;
        while(u<SQ<<6&&!A[u>>6])++u;
        if(u==SQ<<6)return -1;
        while(!ct(u))++u;
        return u;
    }
    bool ct(int v){return (A[v>>6]>>(v&63))&1;}
    int bt(){int l(0);F(SQ)l+=__builtin_popcountll(A[i]);return l;}
}S;
int N,a,A[MX],B,E,X;
void er(int a){
    while(A[B]!=a)S.cg(A[B++]),B&=MD;
    ++B,B&=MD;
}
int main(void){
    IN(tt)F(tt){
        S.clr(),B=E=X=0;
        scanf("%d",&N);
        F(N){
            scanf("%d",&a);
            if(S.ct(a))er(a);
            A[E++]=a,E&=MD,S.ad(a);
            if(E>B)X=max(X,E-B);
            else X=max(X,E+MX-B);
        }
        printf("%d\n",X);
    }
    return 0;
}