Gauss elimination solves the inverse of matrix (Gauss)

The basic theory is in linear algebra ：
if A It's a reversible matrix ,(A,E) ~ (E,B), that B Namely A The inverse matrix .
Template title ：P4783 【 Templates 】 Matrix inversion

#include<cstdio>
#include<iostream>
#include<string>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<vector>
#include<map>
#include<queue>
#include<utility>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int maxn = 410;
const int inf = 0x3f3f3f3f3f;
const ll mod = 1e9+7;
const double eps = 1e-7;
int n;
ll a[maxn][maxn<<1];
ll exgcd(ll a,ll b,ll &x,ll &y)
{
    
    if(b==0){
    
        x = 1;
        y = 0;
        return a;
    }
    ll g = exgcd(b,a%b,y,x);
    y -= a/b * x;
    return g;
}
ll inverse(ll a,ll m)
{
    
    ll x,y;
    ll g = exgcd(a,m,x,y);
    return (x%m + m) % m;
}
bool gauss(){
    
    for(int i=1;i<=n;i++){
    
        int k=i;
        for(int j=i+1;j<=n;j++){
    
            if(a[j][i]>a[k][i]){
    
                k=j;
            }
        }
        if(!a[k][i]){
    
            printf("No Solution\n");
            return false;
        }
        if(i!=k)    swap(a[k],a[i]);
        ll kp = inverse(a[i][i],mod);
        for(k = 1;k<=n;k++){
    
            if(k!=i){
    
                ll p = a[k][i]*kp%mod;
                for(int j=i;j<=(n<<1);j++)
                    a[k][j] = ((a[k][j]-a[i][j]*p)%mod+mod)%mod;
            }
        }
        for(int j=1;j<=(n<<1);j++)
            a[i][j] = a[i][j]*kp%mod;
    }
    for(int i=1;i<=n;i++){
    
        for(int j=n+1;j<(n<<1);j++){
    
            printf("%lld ",a[i][j]);
        }
        printf("%lld\n",a[i][n<<1]);
    }
    return true;
}
int main(void)
{
    
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
    
        for(int j=1;j<=n;j++){
    
            scanf("%lld",&a[i][j]);
        }
        a[i][i+n] = 1;
    }
    gauss();

    return 0;
}