JSOI球形空间产生器 （高斯消元）

按照朴素的列方程，可以列出n+1个n元2次方程。

将相邻的两个方程相减就可以得到n个n元1次方程，进行高斯消元就可以了。

 var a,b:array[..,..] of extended;
     temp,ans:array[..] of extended;
     i,j,k,n:longint;
     cnt:extended;
 begin
     readln(n);
     for i:= to n+ do
         for j:= to n do
             read(b[i,j]);
     for i:= to n do
         for j:= to n do
         begin
             a[i,j]:=*(b[i+,j]-b[i,j]);
             a[i,n+]:=a[i,n+]+b[i+,j]*b[i+,j]-b[i,j]*b[i,j];
         end;
 //--------------------------------rpCardinal Orz--------------------------------------
     for i:= to n- do
     begin
         for j:=i+ to n do
             if a[j,i]>a[i,i] then
             begin
                 temp:=a[i];
                 a[i]:=a[j];
                 a[j]:=temp;
             end;
         for j:=i+ to n do
         begin
             cnt:=a[j,i]/a[i,i]; a[j,i]:=;
             for k:=i+ to n+ do
                 a[j,k]:=a[i,k]*cnt-a[j,k];
         end;
     end;
     ans[n]:=a[n,n+]/a[n,n];
     for i:=n- downto  do
     begin
         for j:=i+ to n do a[i,n+]:=a[i,n+]-ans[j]*a[i,j];
         ans[i]:=a[i,n+]/a[i,i];
     end;
     for i:= to n do write(ans[i]::,' ');
 end.