1樓：岔路程式緣
第一題'建立一個text1,
dim r as single
dim s(8) as integer
private sub form_load()
form1.autoredraw = true
text1.text = 578.62
form1.width = 11000
text1.top = 1000
end sub
private sub text1_keypress(keyascii as integer)
select case keyascii
case 46, 48 to 57
case 13
if text1.text = "" then exit sub
r = format(val(text1.text), "0.00")
call zh(r)
print "一百元", "五十元", "十元", "五元", "一元", "五角", "一角", "五分", "一分"
for i = 0 to 8
print s(i),
next
case else
keyascii = ""
end select
end sub
function zh(byval i as single)
i = i * 100
s(0) = i \ 10000
i = i mod 10000
s(1) = i \ 5000
i = i mod 5000
s(2) = i \ 1000
i = i mod 1000
s(3) = i \ 500
i = i mod 500
s(4) = i \ 100
i = i mod 100
s(5) = i \ 50
i = i mod 50
s(6) = i \ 10
i = i mod 10
s(7) = i \ 5
s(8) = i mod 5
end function
第三題'建立七個command1-7
option explicit
dim str as string
dim s() as single, s1() as single, min as single, max as single
dim n as integer, i as integer, k as integer
dim shu as boolean
private sub command1_click()
clsstr = inputbox("陣列元素個數", "輸入", 10)
if str = "" then exit sub
if val(str) < 1 then exit sub
n = int(val(str)) - 1
redim s(n)
call sjsz
for i = 0 to n
print s(i);
if i mod 10 = 9 then print
next
command3.enabled = true
command4.enabled = true
command5.enabled = true
command6.enabled = true
end sub
private sub command2_click()
clsstr = inputbox("陣列元素個數", "輸入", 10)
if str = "" then exit sub
if val(str) < 1 then exit sub
n = int(val(str)) - 1
redim s(n)
for i = 0 to n
s(i) = srsz(i + 1)
print s(i);
if i mod 10 = 9 then print
next
command3.enabled = true
command4.enabled = true
command5.enabled = true
command6.enabled = true
end sub
private sub command3_click()
redim s1(n)
for i = 0 to n
s1(i) = s(i)
next
n = n + 1
redim s(n)
for i = 0 to n - 1
s(i) = s1(i)
next
s(n) = srsz(n + 1)
for i = 0 to n
print s(i);
if i mod 10 = 9 then print
next
end sub
private sub command4_click() '3)在陣列中第k個元素前插入一個元素;
str = inputbox("在第幾個元素前插入?(" & 1 & "-" & n + 1 & ")", "輸入", n)
if str = "" then exit sub
k = int(val(str))
if k < 1 or k > n + 1 then exit sub
redim s1(n)
for i = 0 to n
s1(i) = s(i)
next
n = n + 1
redim s(n)
s(k - 1) = srsz(k)
for i = 0 to n
if i < k - 1 then
s(i) = s1(i)
elseif i > k - 1 then
s(i) = s1(i - 1)
end if
next
for i = 0 to n
print s(i);
if i mod 10 = 9 then print
next
end sub
private sub command5_click() '4)刪除陣列中的第k個元素;
str = inputbox("刪除第幾個元素?(" & 1 & "-" & n + 1 & ")", "輸入", n)
if str = "" then exit sub
k = int(val(str))
if k < 1 or k > n + 1 then exit sub
redim s1(n)
for i = 0 to n
s1(i) = s(i)
next
n = n - 1
redim s(n)
for i = 0 to n
if i < k - 1 then
s(i) = s1(i)
elseif i >= k - 1 then
s(i) = s1(i + 1)
end if
next
for i = 0 to n
print s(i);
if i mod 10 = 9 then print
next
end sub
private sub command6_click()
str = inputbox("刪除哪個元素?", "輸入", n)
if str = "" then exit sub
min = val(str)
domax = cxys(min)
if max <> n + 2 then
msgbox "刪除第" & max + 1 & "個元素"
redim s1(n)
for i = 0 to n
s1(i) = s(i)
next
n = n - 1
redim s(n)
for i = 0 to n
if i < max then
s(i) = s1(i)
elseif i >= k then
s(i) = s1(i + 1)
end if
next
for i = 0 to n
print s(i);
if i mod 10 = 9 then print
next
end if
loop while max <> n + 2
end sub
private sub command7_click()
endend sub
private sub form_load()
form1.autoredraw = true
form1.width = 8000
form1.height = 6000
command1.caption = "產生隨機陣列"
command2.caption = "輸入陣列"
command3.caption = "後面新增"
command4.caption = "中間新增"
command5.caption = "按位置刪除"
command6.caption = "按值刪除"
command7.caption = "退出"
command3.enabled = false
command4.enabled = false
command5.enabled = false
command6.enabled = false
end sub
function srsz(byval l as integer) as single
dostr = inputbox("第" & l & "個元素", "輸入", l)
if str = "" then
shu = false
else
shu = true
srsz = val(str)
end if
loop while not shu
end function
private sub sjsz()
str = inputbox("陣列元素最小值數", "輸入", -100)
if str = "" then exit sub
min = int(val(str))
str = inputbox("陣列元素最大值數", "輸入", 100)
if str = "" then exit sub
max = int(val(str))
if max < min then exit sub
randomize
for i = 0 to n
s(i) = int(rnd * (max - min)) + min
next
end sub
function cxys(byval l as single) as integer '檢索要刪除的元素
for i = 0 to n
if s(i) = l then
cxys = i
exit for
else
cxys = n + 2
end if
next
end function
都執行過。
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