1 /**************************************************************
 2     Problem: 1013
 3     User: Tunix
 4     Language: C++
 5     Result: Accepted
 6     Time:0 ms
 7     Memory:1272 kb
 8 ****************************************************************/
 9  
10 //BZOJ 1013
11 #include<cmath>
12 #include<cstdio>
13 #include<cstring>
14 #include<cstdlib>
15 #include<iostream>
16 #include<algorithm>
17 #define rep(i,n) for(int i=0;i<n;++i)
18 #define F(i,j,n) for(int i=j;i<=n;++i)
19 #define D(i,j,n) for(int i=j;i>=n;--i)
20 using namespace std;
21 void read(int &v){
22     v=0; int sign=1; char ch=getchar();
23     while(ch<‘0‘||ch>‘9‘){ if (ch==‘-‘) sign=-1; ch=getchar();}
24     while(ch>=‘0‘&&ch<=‘9‘){ v=v*10+ch-‘0‘; ch=getchar();}
25     v*=sign;
26 }
27 /******************tamplate*********************/
28 const int N=11;
29 typedef double Matrix[N][N];
30 void gauss(Matrix a,int n){
31     int i,j,k,r;
32     for(i=0;i<n;++i){
33         //选一行r并与第i行交换
34         r=i;
35         for(j=i+1;j<n;++j)
36             if (fabs(a[j][i]) > fabs(a[r][i])) r=j;
37         if (r!=i) for(j=0;j<=n;++j) swap(a[r][j],a[i][j]);
38          
39         //与第i+1～n行进行消元
40         for(k=i+1;k<n;++k){
41             double f=a[k][i]/a[i][i];//为了让a[k][i]=0,第i行所乘的倍数
42             for(j=i;j<=n;++j) a[k][j]-=f*a[i][j];
43         }
44     }
45     //回代过程
46     for(i=n-1;i>=0;i--){
47         for(j=i+1;j<n;j++)
48             a[i][n]-=a[j][n]*a[i][j];
49         a[i][n]/=a[i][i];
50     }
51 }
52 Matrix f;
53 int n;
54 int main(){
55     scanf("%d",&n);
56     F(i,0,n){
57         rep(j,n){
58             scanf("%lf",&f[i][j]);
59             f[i][n]-=f[i][j]*f[i][j];
60             f[i][j]*=-2.0;
61             if (i) f[i-1][j]-=f[i][j];
62         }
63         if (i) f[i-1][n]-=f[i][n];
64     }
65     gauss(f,n);
66     printf("%.3f",f[0][n]);
67     for(int i=1;i<n;++i) printf(" %.3f",f[i][n]);
68     return 0;
69 }