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. ..,..] of extended; temp,ans:..] of extended; i,j,k,n:longint; cnt:extended; begin readln(n); do to n do read(b[i,j]); to n do to n do begin a[i,j]:=*(b[i+,j]-b[i,j]); a[i,n+]:=a[i,n+]+b[i+…