1樓：夏永烽
private sub command1_click()dim i as integer
dim s as integer
s = 0: i = 0
do while i <= 1000
i = i + 1:
if i mod 2 = 1 then
s = s + i
end if
loop
print "計算1至1000之間的奇數之和為："; send sub
沒有學過vb但是樓主的程式確實存在問題，這樣改回一下應該答差不多了，對於樓上說的溢位問題就要看整形數的表示範圍了。
2樓：聽不清啊
這個題目抄
不需要if語句的
。bais定義為integer會溢位的du。
private sub command1_click()dim i as integer
dim s as long
s = 0: i = 1
do while i < 1000
s = s + i: i = i + 2
loop
print "計算1至zhi1000之間的奇數之dao和為："; send sub
vb用函式過程1到9中奇數階乘的和
3樓：兄弟連教育北京總校
private sub command1_click()dim n,i,a,b
n = 100 '假設是100之內的數
for i = 1 to n - 1
if i mod 2 = 1 then
a = a + fc(i)
else
b = b + fc(i)
end if
next
print n & "之內奇數內階乘之
容和=" & a & vbnewline & n & "之內偶數階乘之和=" & b
end sub
function fc(n)
dim i
fc = 1
for i = 1 to n
fc = fc * i
next
end function
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