模拟，消元

#include 
     
      #define DB doubleconst int N = 110;const DB eps = 1e-7;DB A[N][N], Answer[N];int n;DB read() {DB a; scanf("%lf", &a); return a;}int main() {    std:: cin >> n;    for(int i = 1; i <= n; i ++)        for(int j = 1; j <= n + 1; j ++) A[i][j] = read();    for(int i = 1; i <= n; i ++) {        int r = i;        for(int j = i + 1; j <= n; j ++) if(abs(A[j][i]) > abs(A[r][i])) r = i;        if(abs(A[r][i]) <= eps) {std:: cout << "No Solution"; return 0;}        if(i != r) std:: swap(A[i], A[r]);        DB now = A[i][i];        for(int j = i; j <= n + 1; j ++) A[i][j] /= now;        for(int j = i + 1; j <= n; j ++) {            DB Now = A[j][i];            for(int k = i; k <= n + 1; k ++) A[j][k] -= Now * A[i][k];        }    }    Answer[n] = A[n][n + 1];    for(int i = n - 1; i >= 1; i --) {        Answer[i] = A[i][n + 1];        for(int j = n; j > n - (n - i); j --)            Answer[i] -= Answer[j] * A[i][j];    }    for(int i = 1; i <= n; i ++) printf("%.2lf\n", Answer[i]);        return 0;}/*31 -2 3 64 -5 6 127 -8 10 21*/