結果

問題 No.1707 Simple Range Reverse Problem
ユーザー riano
提出日時 2021-10-15 23:03:13
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 16 ms / 2,000 ms
コード長 4,109 bytes
コンパイル時間 1,739 ms
コンパイル使用メモリ 175,692 KB
実行使用メモリ 18,944 KB
最終ジャッジ日時 2024-09-17 18:05:18
合計ジャッジ時間 2,625 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge2
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ファイルパターン 結果
sample AC * 1
other AC * 18
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ソースコード

diff #
プレゼンテーションモードにする

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define rep(i,n) for(int i=0;i<n;i++)
#define Pr pair<ll,ll>
#define Tp tuple<int,int,int>
#define all(v) v.begin(),v.end()
#define riano_ std::ios::sync_with_stdio(false);std::cin.tie(nullptr);typedef modint<mod> mint
using Graph = vector<vector<ll>>;
const ll mod = 998244353;
template<uint64_t mod>
struct modint{
    uint64_t val;
    constexpr modint(const int64_t val_=0) noexcept:val((val_%int64_t(mod)+int64_t(mod))%int64_t(mod)){}
    constexpr modint operator-() const noexcept{
        return modint(*this)=mod-val;
    }
    constexpr modint operator+(const modint rhs) const noexcept{
        return modint(*this)+=rhs;
    }
    constexpr modint operator-(const modint rhs) const noexcept{
        return modint(*this)-=rhs;
    }
    constexpr modint operator*(const modint rhs) const noexcept{
        return modint(*this)*=rhs;
    }
    constexpr modint operator/(const modint rhs) const noexcept{
        return modint(*this)/=rhs;
    }
    constexpr modint &operator+=(const modint rhs) noexcept{
        val+=rhs.val;
        val-=((val>=mod)?mod:0);
        return (*this);
    }
    constexpr modint &operator-=(const modint rhs) noexcept{
        val+=((val<rhs.val)?mod:0);
        val-=rhs.val;
        return (*this);
    }
    constexpr modint &operator*=(const modint rhs) noexcept{
        val=val*rhs.val%mod;
        return (*this);
    }
    constexpr modint &operator/=(modint rhs) noexcept{
        uint64_t ex=mod-2;
        modint now=1;
        while(ex){
            now*=((ex&1)?rhs:1);
            rhs*=rhs,ex>>=1;
        }
        return (*this)*=now;
    }
    constexpr bool operator==(const modint rhs) noexcept{
        return val==rhs.val;
    }
    constexpr bool operator!=(const modint rhs) noexcept{
        return val!=rhs.val;
    }
    friend constexpr ostream &operator<<(ostream& os,const modint x) noexcept{
        return os<<(x.val);
    }
    friend constexpr istream &operator>>(istream& is,modint& x) noexcept{
        uint64_t t;
        is>>t,x=t;
        return is;
    }
};
//累乗 aのb乗、正しmを法として求める
long long pw(long long a,long long b,long long m){
    if(b==0) return 1;
    else if(b%2==0){
        long long x = pw(a,b/2,m);
        return (x*x)%m;
    }
    else{
        long long x = pw(a,b-1,m);
        return (a*x)%m;
    }
}
//mod逆元
long long modinv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}
ll ranmod(ll l,ll r,ll d,ll N){
    ll res = ((pw(10,r-l+1,mod)-1*d%mod)*pw(10,N-r,mod)%mod)*modinv(9,mod)%mod;
    //cout << (pw(10,r-l+2,mod)-1*d%mod)*pw(10,N-r,mod)%mod << endl;
    return res;
}
// 最大公約数を求める
ll gcd(ll x,ll y){
    ll r=1;
    if(x<0) x *= -1;
    if(y<0) y *= -1;
    if(x<=y) swap(x,y);
    if(y==0) r=0;
    while(r>0){
        r=x%y;
        x=y;
        y=r;
    }
    return x;
}
//Cの計算 C[i][j]でiCjを得る
void combination(vector<vector <long long> > &v){
    for(int i = 0;i <v.size(); i++){
        v[i][0]=1;
        v[i][i]=1;
    }
    for(int k = 1;k <v.size();k++){
        for(int j = 1;j<k;j++){
          v[k][j]=(v[k-1][j-1]+v[k-1][j])%mod;
        }
    }
}
    
int main() {
    riano_; string ans;
    ll T; cin >> T;
    rep(ii,T){
        ans = "No";
        ll N; cin >> N;
        vector<ll> org;
        rep(i,N){
            org.push_back(i+1);
        }
        rep(i,N){
            org.push_back(i+1);
        }
        vector<vector<ll>> s(N+1,org);
        rep(i,N){
            auto itr1 = s[i].begin();
            auto itr2 = s[i].begin();
            rep(j,i) itr1++;
            rep(j,N+1+i) itr2++;
            reverse(itr1,itr2);
        }
        vector<ll> a;
        rep(i,2*N){
            ll b; cin >> b;
            a.push_back(b);
        }
        rep(i,N+1){
            if(a==s[i]) ans = "Yes";
        }
        cout << ans << "\n";
    }
    
}
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