Dim x(3, 5)
 For I = 1 To 3
    For j = 1 To 5
        x(I, j) = x(I - 1, j - 1) + I + j
    Next j
 Next I
 Print x(3, 4)              

为什么Print x(3,4)等于15   ？？？？

x(1,2)=x(0,1)+1+2=3
x(2,3)=x(1,2)+2+3=5
x(3,4)=x(2,3)+3+4=15

解释下每一步如何来的？（我刚学不太懂）

数组索引基本都从0开始～Dim后不赋值～VB6的预设都给0~基本上你的第0组初始值皆为0~
x(0,0)=0 .... x(0,5)=0

x(1,0)=0
i=1:j=1:x(1, 1) = x(1- 1, 1- 1) + 1+ 1=>x(0,0)+2
i=1:j=2:x(1, 2) = x(1- 1, 2- 1) + 1+ 2=>x(0,1)+3
i=1:j=3:x(1, 3) = x(1- 1, 3- 1) + 1+ 3=>x(0,2)+4
i=1:j=4:x(1, 4) = x(1- 1, 4- 1) + 1+ 4=>x(0,3)+5
i=1:j=5:x(1, 5) = x(1- 1, 5- 1) + 1+ 5=>x(0,4)+6

x(2, 0)=0
i=2:j=1:x(2, 1) = x(2- 1, 1- 1) + 2+ 1=>x(1,0)+3
i=2:j=2:x(2, 2) = x(2- 1, 2- 1) + 2+ 2=>x(1,1)+4
i=2:j=3:x(2, 3) = x(2- 1, 3- 1) + 2+ 3=>x(1,2)+5
i=2:j=4:x(2, 4) = x(2- 1, 4- 1) + 2+ 4=>x(1,3)+6
i=2:j=5:x(2, 5) = x(2- 1, 5- 1) + 2+ 5=>x(1,4)+7

x(3,0)=0
i=3:j=1:x(3, 1) = x(3- 1, 1- 1) + 3+ 1=>x(2,0)+4
i=3:j=2:x(3, 2) = x(3- 1, 2- 1) + 3+ 2=>x(2,1)+5
i=3:j=3:x(3, 3) = x(3- 1, 3- 1) + 3+ 3=>x(2,2)+6
i=3:j=4:x(3, 4) = x(3- 1, 4- 1) + 3+ 4=>x(2,3)+7 = 15 ?
i=3:j=4:x(3, 5) = x(3- 1, 4- 1) + 3+ 5=>x(2,4)+8

so 反推 x(2,3)+7=15 -> x(2,3)=8 -> x(1,2)=3 -> x(0,1)=0

故得证~

p.s 但没有 as 类型 就难说了～

好样的！

Option Base 1

加这行就是都从1开始了~