生成矩阵.prg

来自「常用算法」· PRG 代码 · 共 20 行

clear
dime a(5,5)
for i=1 to 5
  for j=1 to 5
    do case 
      case i<j 
      a(i,j)=2
      case i=j
       a(i,j)=1
      otherwise
        a(i,j)=3
     endcase
   next j
   next i
  for i=1 to 5
    for j=1 to 5
      ?? a(i,j)
    next j
     ?
   next i