#include?
#include??
#include??
#include??
using?namespace?std;

void?hhbg?（double?a[5][5]int?n）;
void?a_hhqr?（double?a[5][5]int?ndouble?*?udouble?*?vdouble?epsint?jt）?;
int?main（int?argc?char?*argv[]）
{
????QCoreApplication?b（argc?argv）;




????int?ijjt=60;//jt為迭代次數
????double?eps=0.000001;//迭代精度
????static?double?u[5]v[5];
????static?double?a[5][5]={{1.06.0-3.0-1.07.0}{8.0-15.018.05.04.0}
????????????????????????????????{-2.011.09.015.020.0}{-13.02.021.030.0-6.0}
?????????????????????????????????{17.022.0-5.03.06.0}};

???/*?static?double?a[5][5]={{110.94126.185678.26783-3}{1714.647124.87325.7797-5}
????????????????????????????????????{0-19.269935.00965.839117.1765}{00-61.3733-45.55446.18675}
?????????????????????????????????????{000-22.852625.8977}};*/

????hhbg（a5）;//把矩陣進行赫申伯格換

//輸出赫申伯格矩陣
????printf（“MAT?H?is:\n“）;
????for?（i=0;i<=4;i++）
????{

????????for（j=0;j<=4;j++）
????????????printf（“%f??“a[i][j]）;
????????printf（“\n“）;
????}
????printf（“\n“）;

????a_hhqr?（a5?uvepsjt）;//對矩陣進行QR變換求特征值

?//輸出矩陣特征值
????for（i=0;i<=4;i++）
????{
????????if（v[i]<0）

????????????printf（“??%f%fi\n?“u[i]v[i]）;
????????else?if（v[i]>0）
????????????printf（“??%f+%fi\n?“u[i]v[i]）;
????????else
????????????printf（“??%f\n?“u[i]）;
????}




????return?b.exec（）;
}


void?hhbg?（double?a[5][5]int?n）??????????//執行初等相似變換
{
????int?ijk;
????double?dt;
????for?（k=1;?k<=n-2;?k++）
????{
????????d=0.0;
????????for?（j=k;?j<=n-1;?j++）
????????{
????????????t=a[j][k-1];
????????????if?（fabs（t）>fabs（d））?{?d=t;?i=j;}
????????}
????????if?（fabs（d）+1.0!=1.0）
????????{
????????????if?（i!=k）
????????????{
????????????????for?（j=k-1;?j<=n-1;?j++）
????????????????{
????????????????????t=a[i][j];?a[i][j]=a[k][j];?a[k][j]=t;
????????????????}
????????????????for?（j=0;?j<=n-1;?j++）
????????????????{
????????????????????t=a[j][i];?a[j][i]=a[j][k];?a[j][k]=t;
????????????????}
????????????}
????????????for?（i=k+1;?i<=n-1;?i++）
????????????{
????????????????t=a[i][k-1]/d;?a[i][k-1]=0.0;
????????????????for?（j=k;?j<=n-1;?j++）
????????????????????a[i][j]=a[i][j]-t*a[k][j];
????????????????for?（j=0;?j<=n-1;?j++）
????????????????????a[j][k]=a[j][k]+t*a[j][i];
????????????}
????????}
????}
}
void??a_hhqr?（double?a[5][5]int?ndouble?*?udouble?*?vdouble?epsint?Max）?????????//用QR方法計算全部特征值
?{
?????int?mitijkl;
?????double?bcwgxypqrxsefzy;
?????it=0;?m=n;
?????while?（m!=0）
?????{
?????????l=m-1;
?????????while?（（l>0）&&（fabs（a[l][l-1]）>eps*
???????????????（fabs（a[l-1][l-1]）+fabs（a[l][l]））））?l=l-1;
?????????if?（l==m-1）
?????????{
?????????????u[m-1]=a[m-1][m-1];?v[m-1]=0.0;
?????????????m=m-1;?it=0;
?????????}
?????????else?if?（l==m-2）
?????????{
?????????????b=-（a[m-1][m-1]+a[m-2][m-2]）;
?????????????c=a[m-1][m-1]*a[m-2][m-2]-a[m-1][m-2]*a[m-2][m-1];