【模板】矩阵求逆

\[PA=E, PE=P \]

按照定义，P 即为 A 的逆矩阵。矩阵乘法可以看作是对右边矩阵的一个线性变换，即：A 经过 P 的线性变换变成了 E，E 经过同样的线性变换变成了 P。因此，只需要在高斯消元 A 矩阵，将 A 变成单位矩阵的同时，维护一个单位矩阵，做与 A 完全相同的线性变换即可得到逆矩阵。

代码如下

#include <bits/stdc++.h>

using namespace std;

const int mod = 1e9 + 7;

typedef long long LL;

LL fpow(LL a, LL b) {
	LL ret = 1 % mod;
	for (; b; b >>= 1, a = a * a % mod) {
		if (b & 1) {
			ret = ret * a % mod;
		}
	}
	return ret;
}

struct matrix {
	vector<vector<LL>> mat;
	int n;
	matrix(int _n) {
		n = _n;
		mat.resize(n + 1, vector<LL>(n + 1, 0));
	}
	void identify() {
		for (int i = 1; i <= n; i++) {
			mat[i][i] = 1;
		}
	}
	vector<LL>& operator[](int x) {
		return mat[x];
	}
	void add(int i, int j, LL val) {
		for (int k = 1; k <= n; k++) {
			mat[i][k] = (mat[i][k] + mat[j][k] * val % mod + mod) % mod;
		}
	}
	void multiply(int i, LL val) {
		for (int j = 1; j <= n; j++) {
			mat[i][j] = (mat[i][j] * val % mod + mod) % mod;
		}
	}
	void print() {
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= n; j++) {
				cout << mat[i][j] << " ";
			}
			cout << endl;
		}
	}
};

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	int n;
	cin >> n;
	matrix a(n), b(n);
	b.identify();
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			cin >> a[i][j];
		}
	}
	auto gauss = [&]() -> bool {
		for (int i = 1; i <= n; i++) {
			if (a[i][i] == 0) {
				for (int j = i + 1; j <= n; j++) {
					if (a[j][i] != 0) {
						swap(a[i], a[j]);
						swap(b[i], b[j]);
						break;
					}
				}
			}
			if (a[i][i] == 0) {
				return 0;
			}
			LL inv = fpow(a[i][i], mod - 2);
			a.multiply(i, inv);
			b.multiply(i, inv);
			for (int j = i + 1; j <= n; j++) {
				b.add(j, i, -a[j][i]);
				a.add(j, i, -a[j][i]);
			}
		}
		for (int i = n - 1; i >= 1; i--) {
			for (int j = i + 1; j <= n; j++) {
				b.add(i, j, -a[i][j]);
				a.add(i, j, -a[i][j]);
			}
		}
		return 1;
	};
	if (gauss() == 1) {
		b.print();
	} else {
		cout << "No Solution" << endl;
	}
	return 0;
}