結果

問題 No.803 Very Limited Xor Subset
ユーザー yuliicppy
提出日時 2019-11-13 17:37:16
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 3,714 bytes
コンパイル時間 882 ms
コンパイル使用メモリ 102,296 KB
実行使用メモリ 5,376 KB
最終ジャッジ日時 2024-09-21 21:09:53
合計ジャッジ時間 2,295 ms
ジャッジサーバーID
（参考情報） 		judge3 / 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 * 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 <string>
#include <vector>
#include <algorithm>
#include <stack>
#include <queue>
#include <cmath>
#include <tuple>
#include <cstdio>
#include <bitset>
#include <sstream>
#include <iterator>
#include <numeric>
#include <map>
#include <cstring>
#include <set>
#include <functional>
#include <iomanip>
using namespace std;
#define DEBUG_ //!!提出時にコメントアウト!!
#ifdef DEBUG_
    #define dump(x)  cerr << #x << " = " << (x) << endl;
#else
    #define dump(x)  ;
#endif
#define equals(a,b) (fabs((a)-(b)) < EPS)
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define SZ(x) ((int)(x).size())
#define pb push_back
#define eb emplace_back
//#define int long long
typedef long long LL;
typedef vector<int> VI;
typedef vector<LL> VL;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
template <typename T>
std::string printVector(const std::vector<T> &data)
{
    std::stringstream ss;
    std::ostream_iterator<T> out_it(ss, ", ");
    ss << "[";
    std::copy(data.begin(), data.end() - 1, out_it);
    ss << data.back() << "]";
    return ss.str();
}
template <typename T>
void print_array(const T &ary, int size){
    REP(i,size){
        cout << ary[i] << " ";
    }
    cout << endl;
}
const int MOD = 1e9+7;
const LL LINF = 1001002003004005006ll;
const int INF = 1001001001;
const double EPS = (1e-10);
const int MAX_ROW = 350;
const int MAX_COL = 350;
struct BitMatrix {
    int H, W;
    bitset<MAX_COL> val[MAX_ROW];
    BitMatrix(int m = 1, int n = 1) : H(m), W(n) {}
    inline bitset<MAX_COL>& operator [] (int i) {return val[i];}
};
int GaussJordan(BitMatrix &A, bool is_extended = false) {
    int rank = 0;
    for (int col = 0; col < A.W; ++col) {
        if (is_extended && col == A.W - 1) break;
        int pivot = -1;
        for (int row = rank; row < A.H; ++row) {
            if (A[row][col]) {
                pivot = row;
                break;
            }
        }
        if (pivot == -1) continue;
        swap(A[pivot], A[rank]);
        for (int row = 0; row < A.H; ++row) {
            if (row != rank && A[row][col]) A[row] ^= A[rank];
        }
        ++rank;
    }
    return rank;
}
int linear_equation(BitMatrix A, vector<int> b, vector<int> &res) {
    int m = A.H, n = A.W;
    BitMatrix M(m, n + 1);
    for (int i = 0; i < m; ++i) {
        for (int j = 0; j < n; ++j) M[i][j] = A[i][j];
        M[i][n] = b[i];
    }
    int rank = GaussJordan(M, true);
    // check if it has no solution
    for (int row = rank; row < m; ++row) if (M[row][n]) return -1;
    // answer
    res.assign(n, 0);
    for (int i = 0; i < rank; ++i) res[i] = M[i][n];
    return rank;
}
signed main(int argc, char const *argv[])
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << setprecision(12);
    int N,M,X; cin >> N >> M >> X;
    VI A(N);
    REP(i,N) cin >> A[i];
    BitMatrix B(32+M,N);
    VI res(32+M),res2;
    const int DIGIT = 32;
    REP(i,DIGIT){
        REP(j,N){
            if(A[j] & (1 << i)) B[i][j] = 1;
        }
        if(X & (1 << i)) res[i] = 1;
    }
    REP(i,M){
        int t,l,r; cin >> t >> l >> r;
        l--; r--;
        if(t == 0){
            for(int j = l; j <= r; j++){
                B[32+i][j] = 1;
            }
            res[32+i] = 0;
        }else{
            for(int j = l; j <= r; j++){
                B[32+i][j] = 1;
            }
            res[32+i] = 1;
        }
    }
    int rank = linear_equation(B,res,res2);
    if(rank == -1) cout << 0 << endl;
    else{
        LL ans = 1;
        REP(i,N-rank){
            ans = ans * 2 % MOD;
        }
        cout << ans << endl;
    }
}
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