1樓：匿名使用者
1,private sub form_click()
dim a(1 to 50) as integer
dim sum as long
dim sumdel as long
for i = 1 to 50
a(i) = i ^ 2
sum = sum + a(i)
next
for i = 1 to 50
if a(i) mod 7 = 0 or a(i) mod 13 = 0 then
sumdel = sumdel + a(i)
end if
next
print "所求的和為:" & sum - sumdel
end sub
3，private sub form_click()
i = asc(inputbox("請輸入字元", "輸入框"))
if i >= 48 and i <= 57 then
msgbox "你輸入的是數字字元"
elseif (i >= 65 and i <= 90) or (i >= 97 and i <= 122) then
msgbox "你輸入的是字母字元"
else
msgbox "你輸入的是其它字元"
end if
end sub
4，'通用的排序過程
private sub rank(n as integer, a() as long)
for i = 1 to n - 1
imin = i
for j = i + 1 to n
if a(j) > a(imin) then imin = j
next j
if imin <> i then t = a(i): a(i) = a(imin): a(imin) = t
next i
end sub
private sub form_click()
dim s(1 to 3) as long
for i = 1 to 3
s(i) = val(inputbox("輸入第i個資料"))
next
call rank(3, s())
for i = 1 to 3
list1.additem s(i)
next
end sub
5,private sub form_click()
dim m as long
dim n as long
m = val(inputbox("請輸入m的值", "輸入框"))
n = val(inputbox("請輸入n的值", "輸入框"))
print "最大公約數為：" & max公約數(m, n)
end sub
function max公約數(a as long, b as long) '求出兩個數的最大公約數
dim x as long, y as long, k as long
x = iif(a >= b, a, b) 'x存入最大值
y = iif(a <= b, a, b) 'y 存入最小值
do '輾轉相除法
k = x mod y: if k = 0 then exit do
x = y: y = k
loop
max公約數 = y
end function
6,'通用宣告
option explicit
dim m as integer, n as integer
private sub form_load()
don = val(inputbox("n:"))
m = val(inputbox("m:"))
loop until n <= m
end sub
private function jc(byval x as integer) as double
dim i as integer
dim result as double
result = 1
for i = 1 to x
result = result * i
next i
jc = result
end function
private sub command1_click()
dim doufact as double
doufact = jc(m) / (jc(n) * jc(m - n))
text1.text = str(doufact)
end sub
2樓：匿名使用者
1.dim sum as long
dim i as integer
for i= 1 to 50
if i mod 7<>0 and i mod 13<>0 thensum=sum+i^2
end if
next i
3樓：匿名使用者
1,dim s, a, n as integers = 0
a = 1
do until a >= 50
s = s + a ^ 2
a = a + 1
for n = 1 to 7
if a = 7 * n or a = 13 * n thendo while a = 7 * n or a = 13 * na = a + 1
loop
end if
next n
loop
label1.text = s
2,上課了，下次再給你。。呵呵。。。不好意思
4樓：昆明錦鯉設計
第四題（非條件句解法）：仔細看看你的題目發現有點問題，是不是輸入三個數排序，按從大到小排序然後輸出。下面就是最簡便的方法（vb2005）：
dim x,y,z as double
x=val(textbox1.text)
y=val(textbox2.text)
z=val(textbox3.text)
dim a() as double =
array.sort(a)
msgbox("三個數從大到小排序是："&a(0)&"、"&a(1)&"、"&a(2))
*****===
條件句的話要判斷很多句啊。慢慢寫就出來了。。
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