#892905 (C++17) No.184 たのしい排他的論理和(HARD)

• 問題: No.184 たのしい排他的論理和(HARD)
• ユーザー: hiro71687k
• 提出日時: 2023-07-18 18:17:47
• 言語: C++17 (gcc 13.3.0 + boost 1.87.0)
• 結果: WA
• 実行時間: -
• コード長: 3,094 bytes
• コンパイル時間: 4,617 ms
• コンパイル使用メモリ: 253,080 KB
• 最終ジャッジ日時: 2025-02-15 15:37:37
• ジャッジサーバーID （参考情報）: judge4 / judge4

ファイルパターン	結果
sample	AC * 3
other	AC * 33 WA * 1

ソースコード

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#include <bits/stdc++.h>
#include <atcoder/all>
using namespace atcoder;
using namespace std;
using ll=long long;
using ld=long double;
ld pie=3.141592653589793;
ll mod=1000000007;
ld inf=10000999999999900;
const int MAX_ROW = 100002; // to be set appropriately
const int MAX_COL = 62; // 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 = 998244353;
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() { 
    int N, M; cin >> N;
    M=61;
    BitMatrix A(N, M);
    vector<ll>two(61,1);
    for (ll i = 1; i < two.size(); i++)
    {
        two[i]=two[i-1]*2;
    }
    for (int i = 0; i < N; ++i) {
        ll x;
        cin >> x;
        for (ll j = 0; j <=60; j++)
        {
            if (two[j]&x)
            {
                A[i][j]=1;
            }else{
                A[i][j]=0;
            }
        }
    }
    vector<int> res;
    int r = GaussJordan(A);  
    cout << two[r] << endl;
}
 
 
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