福建信息技術(shù)會考VB程序題匯總

1、信息技術(shù)(選修)輸入2個數(shù)，實現(xiàn)2個數(shù)的互換：-Dim a as single, b as single, t as single a bt a b輸入一個兩位整數(shù)，十位和個位數(shù)值互換 啟輸出：Dim x as integerDim ge as integer, shi as integerx二shi二x10ge二x mod 10二ge*10+shi輸入秒鑿，輸出時分秒：Dim a as integerDim h as integer, m as integer, s as integerh=a3600計算小時部分m二(a-3600*h) 60或m=(a mod 3600) 60計算分鐘部分s

2、二a mod 60“計算秒部分二h&“小時”&m&“分”&m&“秒”輸入一個年份，判斷是否是閏年(年份能 被400整除或者年份能被4整除但不被100整除)：Dim y as integery二If (y mod 4=0 and y mod 10000 ) or y mod 400=0then二”是閏年”Else二”非閏年”End if求S二1+2+3+. +NDim N as integer , s as integer N二S二0For i二1 To NS=S+iNext i程序題匯總求S二1+3+5+(2*N -1 )Dim N as integer

3、 , s as integer N二S二0For i二1 To (2*N -1) step 2S二S+iNext iPrint “S二”;S求S二1 - 2+3 - 4+NDim N as integer , s as integer N二S二0For i二1 To NS二S+I * (-1) ( i+1 )Next iPrint “S二”;S求S二1+1/2+1/3+.+1/NDim N as integer , s as integer N二S二0For i二1 To NS二S+ 1/ iNext iPrint “S二”;S 求S二1/2+2/3+3/4+N/CN+1) Dim N asi

4、nteger , s as integer N二S二0For i二1 To NS二S+ i/ (i+1)Next iPrint “S二”;S輸入三角形三邊，判斷能否構(gòu)成三角形，如可以則輸出面積，否則輸出“不能構(gòu)成三 角形，謖三邊是a, b, Cr q= (a+b+c)/2則 面積s=sqr(q*(q-a) (qb) (q-c):Dim a as single, b as single, c as single Dimq as single, s as singleabif (a+b) c and (b+c) a and (a+c) b then s二sqr (q*(q-a) (q-b) (q-

5、c)二“三角形面積是”& s else二“不能構(gòu)成三角形”end if統(tǒng)計100以內(nèi)能被3整除的數(shù)的個數(shù)：Dim n as integern=0For i=l to 100If i mod 3=0 thenPrint “S二”;Sq二(a+b+c)/2n-n+1$當作計數(shù)器，符合條件就+1end辻next iprint n輸入5個數(shù)輸出其中的最大值:Dim max as single,xa single Max二inputbox(“輸入一個數(shù)”)For i=l to 4x=inputbox(“輸入一個數(shù)”)If max=80 thenPrint“優(yōu)”Elseif score60 the

6、nPrint“不及格”ElsePrint“良”End ifEnd if求5*6*25的積Dim I As IntegerDim S As longS二1For I二5 to 25 Step 1S二S * INext IPrint求5*6*25的積：；S345678910 11求2 3 4 5 6 7 8910的值。Dim I As IntegerDim S As DoubleS = 0For I = 1 To 10 Step 1S = S + (I + 1) / INext IPrint2+3/2+4/3+5/4+6/5+ll/10二；S在Textl輸入整數(shù)a、Text2輸入整數(shù)b,判斷a和b

7、的差是否比50小，若比50小， 則在Text3文本框中輸出小于，否則在Text3文本框中輸出：大于或等于。Dim a As IntegerDim b As Integer b =If Abs(a - b) 5 And a + 5 b And b + 5 aThen1+求2 3 4 56 78 910的值。Dim i As IntegerDim S As singleS =0For i = 1 To 10 Step 1S =S+1 / iNext IPrintl+l/2+l/3+l/4+l/5+l/10二；S在Textl輸入整數(shù)a、Text2輸入整數(shù)b,判斷a和b的和能否被3整除，若能，則在Te

8、xt3文本框中輸出能，否則在Text3文 本框中輸出：不能Dim a As integerDim b As integera =b =If (a + b) Mod 3=0 Then=能ElseElse二不能=不能End IfEnd If 求5+10+15+50的和Dim S As integerS二0For i = 50 To 5 Step -5S =S+iNext I=s輸入一個整數(shù)，判斷它的奇偶性Dim x As integerx=If x mod 2=0 Then=偶數(shù)Else=奇數(shù)End If輸入一個數(shù)，判斷它其是否是整數(shù)。(Fix (x)取整)Dim x As singlex=If fix(x)=x Then=整數(shù)ElseEnd If輸入兩個正整數(shù)，求他們的公約數(shù)并輸出Dim x As Integer, y As Integerx二y二For i = 2 To xIf x Mod i = 0 And y Mod i = 0 ThenPrint iEnd IfNext i輸入兩個正整數(shù)，求他們的最大公約數(shù)并輸出Dim x As Integer, y As Integer, m