結果

問題 No.803 Very Limited Xor Subset
ユーザー drken1215
提出日時 2019-03-17 21:41:43
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 3 ms / 2,000 ms
コード長 5,392 bytes
コンパイル時間 1,080 ms
コンパイル使用メモリ 115,052 KB
実行使用メモリ 5,376 KB
最終ジャッジ日時 2024-07-07 21:22:45
合計ジャッジ時間 2,209 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge1
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 * 4
other AC * 43
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 #
プレゼンテーションモードにする

#include <iostream>
#include <sstream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <cstring>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <bitset>
#include <numeric>
#include <utility>
#include <iomanip>
#include <algorithm>
#include <functional>
#include <unordered_map>
using namespace std;
#define REP(i, s) for (int i = 0; i < s; ++i)
#define ALL(v) (v.begin(), v.end())
#define COUT(x) cout << #x << " = " << (x) << " (L" << __LINE__ << ")" << endl
#define EACH(i, s) for (__typeof__((s).begin()) i = (s).begin(); i != (s).end(); ++i)
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
template<class T1, class T2> ostream& operator << (ostream &s, pair<T1,T2> P)
{ return s << '<' << P.first << ", " << P.second << '>'; }
template<class T> ostream& operator << (ostream &s, vector<T> P)
{ for (int i = 0; i < P.size(); ++i) { if (i > 0) { s << " "; } s << P[i]; } return s; }
template<class T> ostream& operator << (ostream &s, vector<vector<T> > P)
{ for (int i = 0; i < P.size(); ++i) { s << endl << P[i]; } return s << endl; }
template<class T> ostream& operator << (ostream &s, set<T> P)
{ EACH(it, P) { s << "<" << *it << "> "; } return s << endl; }
template<class T1, class T2> ostream& operator << (ostream &s, map<T1,T2> P)
{ EACH(it, P) { s << "<" << it->first << "->" << it->second << "> "; } return s << endl; }
const long long MOD = 1000000007;
const int MAX_ROW = 610; // to be set
const int MAX_COL = 610; // to be set
struct BitMatrix {
    int n, m;
    bitset<MAX_COL> val[MAX_ROW];
    BitMatrix(int n_ = 1, int m_ = 1) {n = n_; m = m_;}
    inline bitset<MAX_COL>& operator [] (int i) {return val[i];}
    inline friend ostream& operator << (ostream& s, BitMatrix M) {
        s << endl;
        for (int i = 0; i < M.n; ++i) {
            for (int j = 0; j < M.m; ++j) s << M.val[i][j];
            s << endl;
        }
        return s;
    }
};
inline BitMatrix operator * (BitMatrix A, BitMatrix B) {
    BitMatrix R(A.n, B.m);
    BitMatrix tB(B.m, B.n);
    for (int i = 0; i < tB.n; ++i) for (int j = 0; j < tB.m; ++j) tB[i][j] = B[j][i];
    for (int i = 0; i < R.n; ++i) for (int j = 0; j < R.m; ++j) R[i][j] = (A[i] & tB[j]).any();
    return R;
}
inline BitMatrix pow(BitMatrix A, long long n) {
    BitMatrix R(A.n, A.n);
    for (int i = 0; i < A.n; ++i) R[i][i] = 1;
    while (n > 0) {
        if (n & 1) R = R * A;
        A = A * A;
        n >>= 1;
    }
    return R;
}
int calc_rank(BitMatrix &A) {
    int r = 0;
    for (int i = 0; i < A.m; ++i) {
        int pivot = -1;
        for (int j = r; j < A.n; ++j) {
            if (A[j][i]) {
                pivot = j;
                break;
            }
        }
        if (pivot != -1) {
            swap(A[pivot], A[r]);
            for (int j = 0; j < A.n; ++j) if (j != r && A[j][i]) A[j] ^= A[r];
            ++r;
        }
    }
    return r;
}
vector<vector<int> > linear_equation(BitMatrix A, vector<int> b) {
    int rank = 0;
    for (int i = 0; i < A.n; ++i) { A[i][A.m] = b[i]; }
    
    vector<int> core, rem;
    for (int i = 0; i < A.m; ++i) {
        int pivot = -1;
        for (int j = rank; j < A.n; ++j) {
            if (A[j][i]) {
                pivot = j;
                break;
            }
        }
        if (pivot != -1) {
            core.push_back(i);
            swap(A[pivot], A[rank]);
            for (int j = 0; j < A.n; ++j) if (j != rank && A[j][i]) A[j] ^= A[rank];
            ++rank;
        }
        else rem.push_back(i);
    }
    
    vector<vector<int> > res;
    for (int i = rank; i < A.n; ++i)
        if (A[i][A.m]) return res;     // return -1;
    
    vector<int> sol(A.m, 0);
    for (int i = 0; i < core.size(); ++i) sol[core[i]] = A[i][A.m];
    res.push_back(sol);
    
    for (int i = 0; i < rem.size(); ++i) {
        vector<int> temp(A.m, 0);
        temp[rem[i]] = 1;
        for (int j = 0; j < core.size(); ++j) temp[core[j]] = A[j][rem[i]];
        res.push_back(temp);
    }
    
    return res;     // return A[0].size()-rank;
};
long long modpow(long long a, long long n, long long mod) {
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}
const int DIGIT = 35;
int main() {
    int N, M; long long X;
    while (cin >> N >> M >> X) {
        vector<long long> a(N);
        for (int i = 0; i < N; ++i) cin >> a[i];
        BitMatrix A(DIGIT + M, N);
        vector<int> b(DIGIT + M);
        for (int d = 0; d < DIGIT; ++d) {
            for (int i = 0; i < N; ++i) {
                if (a[i] & (1LL<<d)) A[d][i] = 1;
            }
            if (X & (1LL<<d)) b[d] = 1;
        }
        for (int d = 0; d < M; ++d) {
            int type, l, r; cin >> type >> l >> r;
            --l, --r;
            for (int i = l; i <= r; ++i) A[d + DIGIT][i] = 1;
            b[d + DIGIT] = type;
        }
        auto res = linear_equation(A, b);
        int jiyudo = (int)res.size() - 1;
        
        //COUT(A);
        //COUT(res);
        
        if (res.empty()) cout << 0 << endl;
        else cout << modpow(2LL, jiyudo, MOD) << endl;
    }
}
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