結果

問題 No.803 Very Limited Xor Subset
ユーザー moricup
提出日時 2020-10-16 01:20:54
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 4 ms / 2,000 ms
コード長 4,196 bytes
コンパイル時間 1,849 ms
コンパイル使用メモリ 174,664 KB
実行使用メモリ 5,376 KB
最終ジャッジ日時 2024-07-20 20:13:40
合計ジャッジ時間 3,218 ms
ジャッジサーバーID
（参考情報） 		judge5 / 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<bits/stdc++.h>
using namespace std;
//typedef
typedef unsigned int UINT;
typedef unsigned long long ULL;
typedef long long LL;
typedef long double LD;
typedef pair<LL, LL> PLL;
typedef tuple<LL, LL, LL> TLL3;
typedef tuple<LL, LL, LL, LL> TLL4;
typedef set<LL, greater<LL> > setdownLL;
#define PQ priority_queue
typedef PQ<LL, vector<LL>, greater<LL> > pqupLL;
//container utill
#define ALL(v) (v).begin(),(v).end()
#define CR [](auto element1, auto element2){return element1>element2;}
#define LB lower_bound
#define UP upper_bound
#define PB push_back
#define MP make_pair
#define MT make_tuple
//constant
#define PI 3.141592653589793
const int MAX_ROW = 340; // to be set appropriately
const int MAX_COL = 310; // to be set appropriately
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];}
};
ostream& operator << (ostream& s, BitMatrix A) {
    s << endl; 
    for (int i = 0; i < A.H; ++i) {
        for (int j = 0; j < A.W; ++j) {
            s << A[i][j] << ", ";
        }
        s << endl;
    }
    return s;
}
inline BitMatrix operator * (BitMatrix A, BitMatrix B) {
    BitMatrix R(A.H, B.W);
    BitMatrix tB(B.W, B.H);
    for (int i = 0; i < tB.H; ++i) for (int j = 0; j < tB.W; ++j) tB[i][j] = B[j][i];
    for (int i = 0; i < R.H; ++i) for (int j = 0; j < R.W; ++j) R[i][j] = ((A[i] & tB[j]).count() & 1);
    return R;
}
inline BitMatrix pow(BitMatrix A, long long n) {
    BitMatrix R(A.H, A.H);
    for (int i = 0; i < A.H; ++i) R[i][i] = 1;
    while (n > 0) {
        if (n & 1) R = R * A;
        A = A * A;
        n >>= 1;
    }
    return R;
}
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;
}
const int MOD = 1000000007;
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;
}
int main() {
    //input
    LL N,M,X;
    cin >> N >> M >> X;
    LL A[N];
    LL i,j;
    for(i=0; i<N; i++){
        cin >> A[i];
    }
    long long ty[M], l[M], r[M];
    for(i=0; i<M; i++){
        cin >> ty[i] >> l[i] >> r[i];
        l[i]--, r[i]--;
    }
    //calc
    vector<int> b;
    for(i=0; i<30; i++){
        if(X%2==1){
            b.PB(1);
        }else{
            b.PB(0);
        }
        X/=2;
    }
    for(i=0; i<M; i++){
        b.PB(ty[i]);
    }
    BitMatrix AA(30+M,N);
    for(j=0; j<N; j++){
        for(i=0; i<30; i++){
            if(A[j]%2==1){
                AA[i][j]=1;
            }else{
                AA[i][j]=0;
            }
            A[j]/=2;
        }
    }
    for(i=0; i<M; i++){
        for(j=0; j<N; j++){
            if(l[i]<=j&&j<=r[i]){
                AA[i+30][j]=1;
            }else{
                AA[i+30][j]=0;
            }
        }
    }
    vector<int> c;
    LL rk=linear_equation(AA,b,c);
    LL ans;
    if(rk>=0){
        ans=1;
        for(i=0; i<N-rk; i++){
            ans*=2;
            ans%=MOD;
        }
    }else{
        ans=0;
    }
    //output
    cout << ans << endl;
    return 0;
}
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