赞 brother44 幼苗
共回答了12个问题采纳率：91.7% 举报
我的理解,你是一个9*9的矩阵,那么:
dim s as integer
dim i as integer
dim j as integer
dim d(9,9) as integer '假设二维数组为d,且主对角线元素的值为1-9
s=0
for i=1 to 9
for j=1 to 9
if i=j then s = s + d(i,j)
next j
next i
print s
dim s as integer
dim i as integer
dim j as integer
dim d(9,9) as integer '假设二维数组为d,且主对角线元素的值为1-9
s=0
for i=1 to 9
for j=1 to 9
if i=j then s = s + d(i,j)
next j
next i
print s
1年前 追问
5
哦,这个这样: dim s as integer dim i as integer dim j as integer dim d(3,3) as integer dim n as integer '先构成矩阵的二维数组 n=0 for i=1 to 3 for j=1 to 3 n = n + 1 d(i,j) = n next j next i '输出矩阵 For i = 1 To 3 For j = 1 To 3 Print d(i, j), Next j Print Next i '计算主对角线和并输出 s = 0 For i = 1 To 3 For j = 1 To 3 If i = j Then s = s + d(i, j) Next j Next i Print s
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你能帮帮他们吗