[ICPC2020济南A] Matrix Equation

[ICPC2020济南A] Matrix Equation - 线性基

Description

给一个 A 矩阵 一个 B 矩阵（矩阵元素为 0 或 1），求有多少个 C 矩阵 满足 A X C = B . C (叉乘 和 直积)。(n le 200)

Solution

C 的每一列是相互无关的，分开考虑，转化为解 (n) 个线性方程组的个数，实际上就是求系数矩阵的秩，用 bitset 线性基即可

#include <bits/stdc++.h>
using namespace std;

#define int long long

const int mod = 998244353;

int a[205][205], b[205][205];

int qpow(int p, int q)
{
    return (q & 1 ? p : 1) * (q ? qpow(p * p % mod, q / 2) : 1) % mod;
}

struct LinearBase
{
    // bitset<205> a[205];
    vector<bitset<205>> a;

    LinearBase()
    {
        a.resize(205);
    }
    int ans = 0;
    void insert(bitset<205> k)
    {
        for (int j = 204; j >= 0; j--)
        {
            if ((k >> j)[0])
            {
                if (a[j] == 0)
                {
                    a[j] = k;
                    ans++;
                    break;
                }
                else
                {
                    k ^= a[j];
                }
            }
        }
    }
};

signed main()
{
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            cin >> a[i][j];
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            cin >> b[i][j];

    int ans = 0;
    for (int c = 1; c <= n; c++)
    {
        LinearBase lb;
        for (int i = 1; i <= n; i++)
        {
            bitset<205> tmp;
            for (int j = 1; j <= n; j++)
            {
                tmp[j] = a[i][j];
                if (i == j && b[i][c])
                    tmp[j] = 1 - tmp[j];
            }
            lb.insert(tmp);
        }
        ans += n - lb.ans;
    }

    cout << qpow(2, ans) << endl;
}