内容发布更新时间 : 2025/5/25 20:29:04星期一 下面是文章的全部内容请认真阅读。

#include #include #include using namespace std; int main() {

int n,i,j,k,m; cout<<\输入维数:\ cin>>n;

double **A=new double*[(n+1)]; for(i=1;i<=n;i++)

A[i]=new double[n+1]; double *b=new double[n+1]; double *x=new double[n+1]; double *y=new double[n+1];

cout<<\输入系数对称正定矩阵A[][]:\ for(i=1;i<=n;i++) for(j=1;j<=n;j++) cin>>A[i][j]; cout<<\输入向量b[]:\ for(i=1;i<=n;i++)

}

cin>>b[i]; cout<

for(k=1;k<=n;k++) {

double sum=0;

for(m=1;m<=k-1;m++) {

sum=sum+pow(A[k][m],2.0); }

sum=A[k][k]-sum; A[k][k]=sqrt(sum); for(i=k+1;i<=n;i++) {

double temp1=0;

for(m=1;m<=k-1;m++) {

temp1=temp1+A[i][m]*A[k][m]; }

temp1=A[i][k]-temp1; A[i][k]=temp1/A[k][k]; }

double temp2=0;

for(m=1;m<=k-1;m++) {

temp2=temp2+A[k][m]*y[m]; }

y[k]=(b[k]-temp2)/A[k][k]; }

x[8]=y[8]/A[8][8]; for(k=n-1;k>=1;k--) {

double temp3=0;

for(m=k+1;m<=n;m++) {

temp3=temp3+A[m][k]*x[m]; }

x[k]=(y[k]-temp3)/A[k][k]; }

cout<<\输出结果向量x[]:\

for(i=1;i<=n;i++) cout<

3、三对角形线性方程组

?4?100000000??x1??7???14?10000000??x??5?2???????0?14?1000000??x3???13???????x00?14?1000002???4????000?14?10000??x5??6? ???? = ??

?0000?14?1000??x6???12??00000?14?100??x??14?7???????000000?14?10??x8???4??0000000?14?1??x??5?9?????????00000000?14?????5???x10??x= ( 2, 1, -3, 0, 1, -2, 3, 0, 1, -1 )

追赶法

*T

#include #include #include using namespace std; int main() {

int n,i;

cout<<\输入系数矩阵的维数:\ cin>>n;

double *a=new double[n+1]; double *c=new double[n+1]; double *d=new double[n+1]; double *b=new double[n+1]; double *x=new double[n+1]; double *y=new double[n+1];

cout<<\输入系数矩阵A[]数据：\ for(i=1;i<=n;i++) cin>>a[i]; for(i=1;i<=n;i++) cin>>c[i]; for(i=1;i<=n;i++) cin>>d[i]; cout<<\输入b[] :\

}

for(i=1;i<=n;i++) cin>>b[i]; for(i=1;i<=n-1;i++) {

c[i]=c[i]/a[i];

a[i+1]=a[i+1]-d[i+1]*c[i]; }

cout<<\输出解向量a[]:\

for(i=1;i<=n;i++)cout<

for(i=1;i<=n;i++)cout<

y[i]=(b[i]-d[i]*y[i-1])/a[i]; }

cout<<\输出解向量y[]:\

for(i=1;i<=n;i++)cout<

for(i=n-1;i>=1;i--) {

x[i]=y[i]-c[i]*x[i+1]; }

cout<<\输出解向量x[]:\

for(i=1;i<=n;i++)cout<