矩阵求逆

LuoguP4783

思路：

求A的逆矩阵，把A和单位矩阵I放在一个矩阵里

对A进行加减消元使A化成单位矩阵

此时原来单位矩阵转化成逆矩阵

原理大概就是 A(逆) * [A I] = [I A(逆)]

Code:

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const double eps = 1E-8;
const int p = 1e9 + 7;
const int N = 405;
int n, m;
ll a[N][N << 1];
ll pw(ll x, ll y) {
    ll re = 1;
    for (; y; y >>= 1) {
        if (y & 1) {
            re = re * x % p;
        }
        x = x * x % p;
    }
    return re;
}
void gauss() {//高斯消元 
    for (int i = 1; i <= n; i++) {
        int pre = i;
        for (int j = i + 1; j <= n; j++) {
            if (fabs(a[j][i] - a[pre][i]) >= eps) {
                pre = j;
            }
        }
        for (int k = 1; k <= m; k++) {//交换着两行,记得把右侧新添矩阵也交换 
            swap(a[i][k], a[pre][k]);
        }
        if (fabs(a[i][i]) <= eps) {
            printf("No Solution
");
            exit(0);
        }
        ll inv = pw(a[i][i], p - 2);//求逆元 
        for (int k = i; k <= m; k++) {
            a[i][k] = a[i][k] * inv % p;
        }
        for (int j = 1; j <= n; j++) {
            if (i != j) {
                for (int k = i + 1; k <= m; k++) {//在当前矩阵进行变换时,对所求矩阵也进行同样的变换 
                    a[j][k] = (a[j][k] - a[j][i] * a[i][k] % p + p) % p;
                }
                a[j][i] = 0;
            }
        }
    }
}
int main () {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            scanf("%lld", &a[i][j]);
        }
        a[i][n + i] = 1;//另一半初始化为元矩阵 
    }
    m = n << 1;
    gauss();
    for (int i = 1; i <= n; i++) {
        for (int j = n + 1; j <= m; j++) {
            printf("%lld%c", a[i][j], j == m ? '
' : ' ');
        }
    }
    return 0;
}