洛谷P4783 【模板】矩阵求逆(高斯消元)

题意

题目链接

Sol

首先在原矩阵的右侧放一个单位矩阵

对左侧的矩阵高斯消元

右侧的矩阵即为逆矩阵

// luogu-judger-enable-o2
#include<bits/stdc++.h> 
#define LL long long 
using namespace std;
const int MAXN = 2001, mod = 1e9 + 7;
const double eps = 1e-9;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, a[MAXN][MAXN];
int mul(int x, int y) {
    return 1ll * x * y % mod;
}
int fp(int a, int p) {
    int base = 1;
    while(p) {
        if(p & 1) base = mul(base, a);
        a = mul(a, a); p >>= 1;
    }
    return base;
}
int inv(int x) {
    return fp(x, mod - 2);
}
void add2(int &x, int y) {
    if(x + y < 0) x = x + y + mod;
    else x = (x + y >= mod) ? x + y - mod : x + y;
}
int MatrixInv() {
    for(int i = 1; i <= N; i++) 
        a[i][i + N] = 1;
    for(int i = 1; i <= N; i++) {
        int mx = i;
        for(int j = i + 1; j <= N; j++)
            if(a[j][i] > a[i][i]) mx = j;
        if(mx != i) swap(a[i], a[mx]);
        if(!a[i][i]) return -1; 
        int Inv = inv(a[i][i]);
        for(int j = i; j <= 2 * N; j++) a[i][j] = mul(a[i][j], Inv);
        for(int j = 1; j <= N; j++) {
            if(i != j) {
                int r = a[j][i];
                for(int k = i; k <= 2 * N; k++) 
                    add2(a[j][k], -mul(a[i][k], r));                
            }
        }
    }   
    return 0;
}
signed main() {
    //freopen("testdata.in", "r", stdin);
    N = read();
    for(int i = 1; i <= N; i++)
        for(int j = 1; j <= N; j++) 
            a[i][j] = read();
    if(MatrixInv() == -1) {puts("No Solution"); return 0;}
    for(int i = 1; i <= N; i++, puts(""))
        for(int j = N + 1; j <= 2 * N; j++)
            printf("%d ", a[i][j]);
    return 0;
}
/*
1
4 2 0 1 0
50 50
*/