Description

給定一個線性方程組，對其求解

Input

第一行，一個正整數 n

第二至 n+1 行，每行 n+1 個整數，為a1,a2?an和 b，代表一組方程。1??,a?2???a?n?? 和 bbb，代表一組方程。

Output

共n行，每行一個數，第 i 行為 xi

如果不存在唯一解，在第一行輸出"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

題解

矩陣變換:
一、交換變換：$R_i<->R_j$，表示將$R_i$與$R_j$的所有元素對應交換

 1 //It is made by Awson on 2017.10.10
 2 #include <set>
 3 #include <map>
 4 #include <cmath>
 5 #include <ctime>
 6 #include <cmath>
 7 #include <stack>
 8 #include <queue>
 9 #include <vector>
10 #include <string
 
>
11 #include <cstdio>
12 #include <cstdlib>
13 #include <cstring>
14 #include <iostream>
15 #include <algorithm>
16 #define LL long long
17 #define Min(a, b) ((a) < (b) ? (a) : (b))
18 #define Max(a, b) ((a) > (b) ? (a) : (b))
19 #define sqr(x) ((x)*(x))
20 using namespace std;
21 const int N = 100;
22 
23 int n;
24 double a[N+5][N+5];
25 
26 void Gauss() {
27   for (int line = 1; line <= n; line++) {
28     int Max_line = line;
29     for (int i = line+1; i <= n; i++)
30       if (fabs(a[i][line]) > fabs(a[Max_line][line]))
31     Max_line = i;
32     if (Max_line != line) swap(a[line], a[Max_line]);
33     if (a[line][line] == 0) {
34       printf("No Solution\n");
35       return;
36     }
37     for (int i = line+1; i <= n; i++) {
38       double tmp = a[i][line]/a[line][line];
39       for (int j = line; j <= n+1; j++)
40     a[i][j] -= a[line][j]*tmp;
41     }
42   }
43   for (int i = n; i >= 1; i--) {
44     for (int j = i+1; j <= n; j++)
45       a[i][n+1] -= a[j][n+1]*a[i][j];
46     a[i][n+1] /= a[i][i];
47   }
48   for (int i = 1; i <= n ;i++)
49     printf("%.2lf\n", a[i][n+1]);
50 }
51 void work() {
52   scanf("%d", &n);
53   for (int i = 1; i <= n; i++)
54     for (int j = 1; j <= n+1; j++)
55       scanf("%lf", &a[i][j]);
56   Gauss();
57 }
58 int main() {
59   work();
60   return 0;
61 }

[Luogu 3389]【模板】高斯消元法