結果

問題 No.803 Very Limited Xor Subset
ユーザー veqcc
提出日時 2019-03-18 19:28:02
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 2,586 bytes
コンパイル時間 1,011 ms
コンパイル使用メモリ 110,920 KB
実行使用メモリ 6,948 KB
最終ジャッジ日時 2024-07-18 16:01:50
合計ジャッジ時間 2,350 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge3
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 <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <string>
#include <vector>
#include <random>
#include <bitset>
#include <queue>
#include <cmath>
#include <stack>
#include <set>
#include <map>
typedef long long ll;
using namespace std;
const ll MOD = 1000000007LL;
const int MAX_ROW = 600;
const int MAX_COL = 600;
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 << "\n";
        for (int i = 0; i < M.n; i++) {
            for (int j = 0; j < M.m; j++) s << M.val[i][j];
            s << "\n";
        }
        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;
}
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];
    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) {
            swap(A[pivot], A[rank]);
            for (int j = 0; j < A.n; j++) if (j != rank && A[j][i]) A[j] ^= A[rank];
            rank++;
        }
    }
    for (int i = rank; i < A.n; i++) if (A[i][A.m]) return -1;
    return rank;
}
ll modpow(ll a, ll n, ll mod) {
    ll res = 1;
    while (n) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}
const int digit = 35;
int main() {
    cin.sync_with_stdio(false);
    cin.tie(0);
    int n, m;
    ll x;
    cin >> n >> m >> x;
    vector <ll> 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] & (1 << d)) A[d][i] = 1;
        }
        if (x & (1 << 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;
    }
    int rank = linear_equation(A, b);
    if (rank == -1) cout << 0 << "\n";
    else cout << modpow(2LL, n - rank, MOD) << "\n";
    return 0;
}
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