gym100923C. Por Costel and Bujor (高斯消元)

题意:简化一下 就是解N个 系数矩阵一样 等式右边列矩阵不一样的方程组

题解:系数矩阵一样 为什么我却毫无办法????

　　 其实只要把等式右边的矩阵都排在后面就好了啊

　　 就变成解一个N x 2N的方程组了 ...

#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-9;

int n;
double a[205][405];

void gauss()
{
    int now = 1, to;
    for(int i = 1; i <= n; i++)
    {
        to = now;
        for(int j = now; j <= n; j++) {
            if(fabs(a[j][i]) > fabs(a[to][i])) to = j;
        }
        //if(to > n) continue;
        if(fabs(a[to][i]) < eps) continue;
        if(to != now)
            for(int j = 1; j <= n + n; j++) swap(a[to][j], a[now][j]);

        double tmp = a[now][i];
        //for(int j = 1; j <= n + n; j++) a[now][j] /= tmp;
        for(int j = 1; j <= n + n; j++)
            if(j != now)
            {
                tmp = a[j][i];
                for(int k = 1; k <= n + n; k++) a[j][k] -= tmp * a[now][k];
            }
        now++;
    }
}
int main() {
    //freopen("bujor.in","r",stdin);
    //freopen("bujor.out","w",stdout);

    int T;
    scanf("%d", &T);
    while(T--) {
        scanf("%d", &n);
        for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++) scanf("%lf", &a[i][j]);

        for(int j = n + 1; j <= n * 2; j++)
        for(int i = 1; i <= n; i++) {
            if(i + n == j) a[i][j] = 1;
            else a[i][j] = 0;
        }
        gauss();

        for(int i = 1; i <= n; i++) {
            for(int j = n + 1; j <= 2 * n; j++) {
                double tmp = a[i][j] / a[i][i];
                if(fabs(tmp) < eps) tmp = 0;
                if(j != 2 * n) printf("%.9lf ", tmp);
                else printf("%.9lf\n", tmp);
            }
        }
    }
    return 0;
}