【模板】高斯-约旦消元法

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int n;
 4 struct matrix{
 5     double mat[128][128];
 6     matrix() {memset(mat,0,sizeof(mat));}
 7     void get() {
 8         for(int i = 1;i <= n;++i) 
 9             for(int j = 1;j <= n+1;++j) 
10                 scanf("%lf",&mat[i][j]);
11         return;
12     }
13 } rix;
14 void gate() {
15     for(int i = 1;i <= n;++i) {
16         int maxn = i;
17         for(int j = i+1;j <= n;++j) 
18             if(fabs(rix.mat[i][j]) > fabs(rix.mat[maxn][i])) 
19                 maxn = j;
20         for(int j = 1;j <= n+1;++j) 
21             swap(rix.mat[i][j],rix.mat[maxn][j]);
22         if(!rix.mat[i][i]) {
23             printf("No Solution");
24             return;
25         }
26         for(int j = 1;j <= n;++j) 
27             if(i != j) {
28                 double temp = rix.mat[j][i]/rix.mat[i][i];
29                 for(int k = i+1;k <= n+1;++k) 
30                     rix.mat[j][k] -= rix.mat[i][k]*temp;
31             }
32     }
33     for(int i = 1;i <= n;++i) 
34         printf("%.2lf\n",rix.mat[i][n+1]/rix.mat[i][i]);
35     return;
36 }
37 int main() {
38     scanf("%d",&n);
39     rix.get();
40     gate();
41     return 0;
42 }

标签：mat,rix,int,maxn,元法,约旦,128,模板,matrix
来源： https://www.cnblogs.com/bikuhiku/p/Guass-Jodan.html