結果

問題 No.2442 線形写像
ユーザー mikammikam
提出日時 2023-08-26 11:01:03
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 35 ms / 2,000 ms
コード長 3,991 bytes
コンパイル時間 7,551 ms
コンパイル使用メモリ 400,296 KB
実行使用メモリ 5,376 KB
最終ジャッジ日時 2024-06-07 07:37:34
合計ジャッジ時間 8,888 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge4
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 2 ms
5,248 KB
testcase_01 AC 2 ms
5,376 KB
testcase_02 AC 2 ms
5,376 KB
testcase_03 AC 2 ms
5,376 KB
testcase_04 AC 2 ms
5,376 KB
testcase_05 AC 2 ms
5,376 KB
testcase_06 AC 2 ms
5,376 KB
testcase_07 AC 2 ms
5,376 KB
testcase_08 AC 2 ms
5,376 KB
testcase_09 AC 2 ms
5,376 KB
testcase_10 AC 2 ms
5,376 KB
testcase_11 AC 2 ms
5,376 KB
testcase_12 AC 2 ms
5,376 KB
testcase_13 AC 2 ms
5,376 KB
testcase_14 AC 2 ms
5,376 KB
testcase_15 AC 14 ms
5,376 KB
testcase_16 AC 14 ms
5,376 KB
testcase_17 AC 27 ms
5,376 KB
testcase_18 AC 35 ms
5,376 KB
testcase_19 AC 35 ms
5,376 KB
testcase_20 AC 35 ms
5,376 KB
testcase_21 AC 27 ms
5,376 KB
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include <atcoder/all>
using namespace atcoder;
#include <atcoder/internal_math>
using namespace internal;
#include <bits/stdc++.h>
using namespace std;
#include <boost/multiprecision/cpp_int.hpp>
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep2(i,a,b) for (int i = (int)(a); i < (int)(b); i++)
#define all(v) v.begin(),v.end()
#define inc(x,l,r) ((l)<=(x)&&(x)<(r)) 
#define Unique(x) sort(all(x)), x.erase(unique(all(x)), x.end())
#define pcnt __builtin_popcountll
typedef long long ll;
#define int ll
using ld = long double;
using vi = vector<int>;
using vs = vector<string>;
using P = pair<int,int>;
using vp = vector<P>;
using Bint = boost::multiprecision::cpp_int;
template<typename T1,typename T2> bool chmax(T1 &a, const T2 b) {if (a < b) {a = b; return true;} else return false; }
template<typename T1,typename T2> bool chmin(T1 &a, const T2 b) {if (a > b) {a = b; return true;} else return false; }
template<typename T> using priority_queue_greater = priority_queue<T, vector<T>, greater<T>>;
template<typename T1,typename T2> ostream &operator<< (ostream &os, const pair<T1,T2> &p){os << p.first <<" "<<p.second;return os;}
ostream &operator<< (ostream &os, const modint1000000007 &m){os << m.val();return os;}
istream &operator>> (istream &is, modint1000000007 &m){ll in;is>>in;m=in;return is;}
ostream &operator<< (ostream &os, const modint998244353 &m){os << m.val();return os;}
istream &operator>> (istream &is, modint998244353 &m){ll in;is>>in;m=in;return is;}
template<typename T> istream &operator>>(istream& is,vector<T> &v){for(T &in:v)is>>in;return is;}
template<class... T> void input(T&... a){(cin>> ... >> a);}
#ifdef LOCAL
template<typename T> ostream &operator<<(ostream &os,const vector<T> &v){os<<"\x1b[32m";rep(i,v.size())os<<v[i]<<(i+1!=v.size()?" ":"");os<<"\x1b[0m";return os;}
template<class T> int print(T& a){cout << "\x1b[32m"<< a<< '\n' << "\x1b[0m";return 0;}
template<class T,class... Ts> int print(const T&a, const Ts&... b){cout << "\x1b[32m" << a;(cout<<...<<(cout<<' ',b));cout<<'\n' << "\x1b[0m";return 0;}
#else
template<typename T> ostream &operator<<(ostream &os,const vector<T> &v){rep(i,v.size())os<<v[i]<<(i+1!=v.size()?" ":"");return os;}
template<class T> int print(T& a){cout << a<< '\n';return 0;}
template<class T,class... Ts> int print(const T&a, const Ts&... b){cout << a;(cout<<...<<(cout<<' ',b));cout<<'\n';return 0;}
#endif
#define VI(v,n) vi v(n); input(v)
#define INT(...) int __VA_ARGS__; input(__VA_ARGS__)
#define STR(...) string __VA_ARGS__; input(__VA_ARGS__)
#define CHAR(...) char __VA_ARGS__; input(__VA_ARGS__)
int sign(ll x){return x>0?1:x<0?-1:0;}
ll ceil(ll x,ll y){assert(y!=0);if(sign(x)==sign(y))return (x+y-1)/y;return -((-x/y));}
ll floor(ll x,ll y){assert(y!=0);if(sign(x)==sign(y))return x/y;if(y<0)x*=-1,y*=-1;return x/y-(x%y<0);}
ll abs(ll x,ll y){return abs(x-y);}
ll bit(int n){return 1ll<<n;}
template<class T> bool ins(string s,T t){return s.find(t)!=string::npos;}
P operator+ (const P &p, const P &q){ return P{p.first+q.first,p.second+q.second};}
P operator- (const P &p, const P &q){ return P{p.first-q.first,p.second-q.second};}
int yesno(bool ok,string y="Yes",string n="No"){ cout<<(ok?y:n)<<endl;return 0;}
int YESNO(bool ok,string y="YES",string n="NO"){ cout<<(ok?y:n)<<endl;return 0;}
int di[]={-1,0,1,0,-1,-1,1,1};
int dj[]={0,1,0,-1,-1,1,-1,1};
const ll INF = 8e18;
//using mint = modint1000000007;
//using mint = modint998244353;
//mint stom(const string &s,int b=10){mint res = 0;for(auto c:s)res *= b,res += c-'0';return res;}

signed main() {
    cin.tie(0);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(20);

    INT(n);
    using ull = unsigned long long;
    vector<ull> a(1<<n);
    cin>>a;
    bool ok = 1;
    rep(i,1<<n){
        ull x = 0;
        rep(j,n)if(i>>j&1)x ^= a[1<<j];
        if(x!=a[i])ok = 0;
    }
    yesno(ok);
    return 0;
}
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