結果

問題 No.2907 Business Revealing Dora Tiles
コンテスト
ユーザー amentorimaru
提出日時 2024-05-30 02:21:06
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 6,289 bytes
コンパイル時間 11,624 ms
コンパイル使用メモリ 242,028 KB
実行使用メモリ 6,948 KB
最終ジャッジ日時 2024-09-23 10:43:54
合計ジャッジ時間 18,764 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2 WA * 1
other AC * 16 WA * 41
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 # 

#define ATCODER
#define _USE_MATH_DEFINES
#include<stdio.h>
#include<iostream>
#include<fstream>
#include<algorithm>
#include<vector>
#include<string>
#include <cassert>
#include <numeric>
#include <unordered_map>
#include <unordered_set>
#include <queue>
#include <math.h>
#include <climits>
#include <set>
#include <map>
#include <list>
#include <random>
#include <iterator>
#include <bitset>
#include <chrono>
#include <type_traits>
using namespace std;

using ll = long long;
using ld = long double;
using pll = pair<ll, ll>;
using pdd = pair<ld, ld>;

//template<claLR T> using pq = priority_queue<T, vector<T>, greater<T>>;

#define FOR(i, a, b) for(ll i=(a); i<(b);i++)
#define REP(i, n) for(ll i=0; i<(n);i++)
#define ROF(i, a, b) for(ll i=(b-1); i>=(a);i--)
#define PER(i, n) for(ll i=n-1; i>=0;i--)
#define REPREP(i,j,a,b) for(ll i=0;i<a;i++)for(ll j=0;j<b;j++)
#define VV(type) vector< vector<type> >
#define VV2(type,n,m,val) vector< vector<type> > val;val.resize(n);for(ll i;i<n;i++)val[i].resize(m)
#define vec(type) vector<type>
#define VEC(type,n,val) vector<type> val;val.resize(n)
#define VL vector<ll>
#define VVL vector< vector<ll> >
#define VP vector< pair<ll,ll> >
#define SZ size()
#define all(i) begin(i),end(i)
#define SORT(i) sort(all(i))
#define BITI(i) (1<<i)
#define BITSET(x,i) x | (ll(1)<<i)
#define BITCUT(x,i) x & ~(ll(1)<<i)
#define EXISTBIT(x,i) (((x>>i) & 1) != 0)
#define ALLBIT(n) (ll(1)<<n-1)
#define CHMAX(n,v) n=n<v?v:n
#define CHMIN(n,v) n=n>v?v:n
#define MP(a,b) make_pair(a,b)
#define DET2(x1,y1,x2,y2) (x1)*(y2)-(x2)*(y1)
#define DET3(x1,y1,z1,x2,y2,z2,x3,y3,z3) (x1)*(y2)*(z3)+(x2)*(y3)*(z1)+(x3)*(y1)*(z2)-(z1)*(y2)*(x3)-(z2)*(y3)*(x1)-(z3)*(y1)*(x2)
#define INC(a) for(auto& v:a)v++;
#define DEC(a) for(auto& v:a)v--;
#define SQU(x) (x)*(x)
#define L0 ll(0)
#ifdef ATCODER
#include <atcoder/all>
using namespace atcoder;
using mint = modint1000000007;
using mint2 = modint998244353;
#endif
template<typename T = ll>
vector<T> read(size_t n) {
  vector<T> ts(n);
  for (size_t i = 0; i < n; i++) cin >> ts[i];
  return ts;
}

template<typename TV, const ll N> void read_tuple_impl(TV&) {}
template<typename TV, const ll N, typename Head, typename... Tail>
void read_tuple_impl(TV& ts) {
  get<N>(ts).emplace_back(*(istream_iterator<Head>(cin)));
  read_tuple_impl<TV, N + 1, Tail...>(ts);
}
template<typename... Ts> decltype(auto) read_tuple(size_t n) {
  tuple<vector<Ts>...> ts;
  for (size_t i = 0; i < n; i++) read_tuple_impl<decltype(ts), 0, Ts...>(ts);
  return ts;
}

using val = ll;
//using func = pair<ll, ll>;

val op(val a, val b) {
  return min(a, b);
}

val e() { return 1e18; }


using val2 = mint2;
//using func = pair<ll, ll>;

val2 op2(val2 a, val2 b) {
  return a * b;
}

val2 e2() { return 1; }
//
//val mp(func f, val a)
//{
//  if (f.first < 0)
//    return a;
//  return f;
//}
//func comp(func f, func g) {
//  if (g.first < 0)
//    return f;
//  return g;
//}
//
//func id() {
//  return MP(-1, -1);
//}


ll di[4] = { 1,0,-1,0 };
ll dj[4] = { 0,1,0,-1 };
ll si[4] = { 0,3,3,0 };
ll sj[4] = { 0,0,3,3 };
//ll di[4] = { -1,-1,1,1 };
//ll dj[4] = { -1,1,-1,1 };
ll di8[8] = { 0,-1,-1,-1,0,1,1,1 };
ll dj8[8] = { -1,-1,0,1,1,1,0,-1 };


class NimberManager {
private:
  using u32 = unsigned int;
  using u64 = unsigned long long;

  std::vector<u32> precalc1;
  void fill_precalc1() {
    precalc1.assign(1 << 16, 0);
    precalc1[(1 << 8) ^ 1] = 1;
    for (int dd = 1; dd < 8; dd <<= 1) {
      int d = 1 << dd;
      int c = d >> 1;
      for (int a0 = 0; a0 < d; a0++) for (int a1 = 0; a1 < d; a1++) if (a0 | a1) {
        for (int b0 = 0; b0 < d; b0++) for (int b1 = 0; b1 < d; b1++) if (b0 | b1) {
          u64 buf = 0;
          buf ^= precalc1[(a1 << 8) ^ b1];
          buf ^= precalc1[(a1 << 8) ^ b0];
          buf ^= precalc1[(a0 << 8) ^ b1];
          buf <<= dd;
          buf ^= precalc1[(c << 8) ^ precalc1[(a1 << 8) ^ b1]];
          buf ^= precalc1[(a0 << 8) ^ b0];
          precalc1[(((a1 << dd) ^ a0) << 8) ^ ((b1 << dd) ^ b0)] = buf;
        }
      }
    }
  }

  u64 product_full(u64 a, u64 b, int d = 6) noexcept {
    if (!(a && b)) return 0;
    if (d == 3) { return precalc1[(a << 8) ^ b]; }
    d--;
    u64 lm = ((u64)1 << (1 << d)) - 1;
    u64 us = ((u64)1 << d);
    u64 buf = 0;
    u64 a1b1 = product_full(a >> us, b >> us, d);
    u64 a2b2 = product_full(a & lm, b & lm, d);
    u64 aabb = product_full((a & lm) ^ (a >> us), (b & lm) ^ (b >> us), d);
    buf ^= (aabb ^ a2b2);
    buf <<= us;
    buf ^= a2b2;
    buf ^= product_full((u64)1 << (us - 1), a1b1, d);
    return buf;
  }
public:
  NimberManager() { fill_precalc1(); }

  unsigned long long product(unsigned long long a, unsigned long long b) noexcept { return product_full(a, b); }
  unsigned long long pow(unsigned long long a, unsigned long long idx) noexcept {
    if (idx == 0) return 1;
    auto b = pow(product(a, a), idx / 2);
    if (idx & 1) b = product(b, a);
    return b;
  }
};

void solve() {
  NimberManager nm;
  unsigned long long invpow = -2;
  //REP(i, 16) {
  //  REP(j, 16) {
  //    printf("%2d ", nm.product(i, j));
  //  }
  //  cout << endl;
  //}

  //FOR(i,1, 16) {    
  //  cout << i << ":" << nm.pow(i, invpow) << endl;
  //}
  ll n, t;
  cin >> n >> t;
  vector h(t, vector<unsigned long long>());
  REP(i, t) {
    h[i] = read<unsigned long long>(n);
    DEC(h[i]);
  }
  mint2 fr = mint2(2).pow(64) - 1;
  vector<bool> use(t);
  mint2 ans = 1;
  REP(j, n) {
    ll idx = -1;
    unsigned long long inv = 0;
    REP(i, t) {
      if (use[i] || h[i][j] == 0)
        continue;
      idx = i;
      inv = nm.pow(h[i][j], invpow);
      break;
    }
    if (idx == -1) {
      ans *= fr;
      continue;
    }
    use[idx] = true;
    REP(i, t) {
      if (i == idx || h[i][j] == 0)
        continue;
      unsigned long long mul = nm.product(inv, h[i][j]);
      REP(jj, n) {
        h[i][jj] ^= nm.product(mul, h[idx][jj]);
      }
    }
  }

  // まだ正解ではない
  REP(i, t) {
    ll notzero = 0;
    REP(j, n) {
      notzero += h[i][j] != 0;
      //cout << h[i][j] << " ";
    }
    if (notzero == 1) {
      cout << 0;
      return;
    }
    //cout << endl;
  }
  cout << ans.val();
}

int main() {
  ll t = 1;
  //cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
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