[Luogu 3389]【模板】高斯消元法

Description

给定一个线性方程组，对其求解

Input

第一行，一个正整数 n

第二至 n+1 行，每行 n+1 个整数，为a1,a2⋯an和 b，代表一组方程。1​​,a​2​​⋯a​n​​ 和 bbb，代表一组方程。

Output

共n行，每行一个数，第 i 行为 xi

如果不存在唯一解，在第一行输出"No Solution".

Sample Input

3
1 3 4 5
1 4 7 3
9 3 2 2

Sample Output

-0.97
5.18
-2.39

HINT

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题解

矩阵变换:
一、交换变换：$R_i<->R_j$，表示将$R_i$与$R_j$的所有元素对应交换
二、倍法变换：$R_i=R_i*k$，表示将$R_i$行的所有元素都乘上一个常数$k$
三、消去变换：$R_i=R_i+R_j*k$，表示将$R_i$行的所有元素对应的加上$R_j$行元素的$k$倍
------------------------------------------------------------------
实数解直接加减消元
整数解消元的时候用最小公倍数消去目标系数
------------------------------------------------------------------
① 无解 当方程中出现$(0, 0, …, 0, a)$的形式，且$a != 0$时，说明是无解的。
② 唯一解 形成了严格的上三角阵
③ 无穷解 不能形成严格的上三角形

 //It is made by Awson on 2017.10.10
 #include <set>
 #include <map>
 #include <cmath>
 #include <ctime>
 #include <cmath>
 #include <stack>
 #include <queue>
 #include <vector>
 #include <string>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #define LL long long
 #define Min(a, b) ((a) < (b) ? (a) : (b))
 #define Max(a, b) ((a) > (b) ? (a) : (b))
 #define sqr(x) ((x)*(x))
 using namespace std;
 const int N = ;

 int n;
 double a[N+][N+];

 void Gauss() {
   for (int line = ; line <= n; line++) {
     int Max_line = line;
     for (int i = line+; i <= n; i++)
       if (fabs(a[i][line]) > fabs(a[Max_line][line]))
     Max_line = i;
     if (Max_line != line) swap(a[line], a[Max_line]);
     if (a[line][line] == ) {
       printf("No Solution\n");
       return;
     }
     for (int i = line+; i <= n; i++) {
       double tmp = a[i][line]/a[line][line];
       for (int j = line; j <= n+; j++)
     a[i][j] -= a[line][j]*tmp;
     }
   }
   for (int i = n; i >= ; i--) {
     for (int j = i+; j <= n; j++)
       a[i][n+] -= a[j][n+]*a[i][j];
     a[i][n+] /= a[i][i];
   }
   for (int i = ; i <= n ;i++)
     printf("%.2lf\n", a[i][n+]);
 }
 void work() {
   scanf("%d", &n);
   for (int i = ; i <= n; i++)
     for (int j = ; j <= n+; j++)
       scanf("%lf", &a[i][j]);
   Gauss();
 }
 int main() {
   work();
   return ;
 }