高斯消元

 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 101311 #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 }