luogu_P2757 [National Team] arithmetic sequence

https://www.luogu.org/problem/P2757

Title Description

Arrangement to a 1 to N {Ai}, asking whether there

$\le p_1<p_2<p_3<p_4<p_5<…<p_{Len} \le N (Len \ge 3)1≤p1<p2<p3<p4<p5<…<pLen≤N(Len≥3)$

Such $AP_1, AP_2,, AP_3, \ cdots, ap_ Len} { A P . 1 , A P 2 , A P . 3 , ⋯ , A P L E n- is an arithmetic sequence.$

Input Format

The first line of the input contains an integer T, it denotes the number of groups.

T then the set of data, each line of a first integer N, each arranged in a second line 1 to N, the number separated by a space between any two.

Output Format

For each test, if an arithmetic sequence, it outputs a line "Y", and otherwise outputs a row "N".

Sample input and output

Input # 1

Output # 1

N 
Y

Description / Tips

For 5% of data, N <= 100

For 30% of the data, N <= 1000

To 100% of the data, N <= 10000, T <= 7

Arithmetic sequence x, y, z, x and z are symmetric about the y by comparison bitset

The first to write bitset, a cancer too

#include<iostream>
#include<cstdio>

#define ri register int
#define u int

namespace opt {

    inline u in() {
        u x(0),f(1);
        char s(getchar());
        while(s<'0'||s>'9') {
            if(s=='-') f=-1;
            s=getchar();
        }
        while(s>='0'&&s<='9') {
            x=(x<<1)+(x<<3)+s-'0';
            s=getchar();
        }
        return x*f;
    }

}

using opt::in;

#include<bitset>
#include<algorithm>

#define NN 10005

namespace mainstay {

    using std::bitset;

    typedef bitset<NN> bit;

    bit e1,e2,bas;

    u check(u x,const u &N) {
        if(x<=N/2) {
            bit _t1((bas>>(N-x))&(e1));
            _t1.set(0);
            bit _t2((bas>>(N-x))&(e2>>(N-2*x+1)));
            _t2.set(0);
            if(_t1!=_t2) return 1;
            else return 0;
        } else {
            bit _t1(e1>>(x-1));
            _t1.set(0);
            bit _t2((bas>>(x-1))&(e2>>(N-x)));
            _t2.set(0);
            if(_t1!=_t2) return 1;
            else return 0;
        }
    }

    inline void solve() {
        u T(in());
        while(T--) {
            u N(in()),flg(0);
            e1.reset(),e2.reset(),bas.reset();
            for(ri i(1); i<=N; ++i) bas.set(i);
            for(ri i(1); i<=N; ++i) {
                u _a(in());
                e1.set(_a),e2.set(N+1-_a);
                if(check(_a,N)) flg=1;
            }
            if(flg) printf("Y\n");
            else printf("N\n");
        }
    }

}

int main() {

    //freopen("x.txt","r",stdin);
    std::ios::sync_with_stdio(false);
    mainstay::solve();

}