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[0..100,0..100] of extended;
    temp,ans:array[0..100] of extended;
    i,j,k,n:longint;
    cnt:extended;
begin
    readln(n);
    for i:=1 to n+1 do
        for j:=1 to n do
            read(b[i,j]);
    for i:=1 to n do
        for j:=1 to n do
        begin
            a[i,j]:=2*(b[i+1,j]-b[i,j]);
            a[i,n+1]:=a[i,n+1]+b[i+1,j]*b[i+1,j]-b[i,j]*b[i,j];
        end;
//--------------------------------rpCardinal Orz--------------------------------------
    for i:=1 to n-1 do
    begin
        for j:=i+1 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+1 to n do
        begin
            cnt:=a[j,i]/a[i,i]; a[j,i]:=0;
            for k:=i+1 to n+1 do
                a[j,k]:=a[i,k]*cnt-a[j,k];
        end;
    end;
    ans[n]:=a[n,n+1]/a[n,n];
    for i:=n-1 downto 1 do
    begin
        for j:=i+1 to n do a[i,n+1]:=a[i,n+1]-ans[j]*a[i,j];
        ans[i]:=a[i,n+1]/a[i,i];
    end;
    for i:=1 to n do write(ans[i]:0:3,' ');
end.