1、Priv ate Su b Form_C li ck ()Dim n%, i %n = 0F o r i = 1 T o 2 0 0I f i Mod 1 1 = 0 O r i M od 5 = 0 T h en n = n + iEn d IfN e x t iPr in t nEn d Sub4、输入一年份,判断它就是否为闰年,并显示有关信息。(判断闰年得条件就是:年份能被 4整除但不能被1 00整除,或者能被400整除)Pr iv ate Sub mand1_C lie k()Dim y%y = I np utB ox (”请输入年数)If y Mod 4 = 0 A n d y Mo
2、 d 10 0 00 r y Mod 40 0 = 0 The nMs g Box (y & 年就是闰年E ls eMsgBox ( y & ”年就是平年En d I fEnd S ub 5、已知x,y,z 3个变量中存放了 3个不同得数,比较它们得大小并进行调整,使得x y T he n t = x : x = y: y = t If x z The n t = x: x = z: z = tIf y z Then t = y: y = z:Pr int 排序后” x; ”; y; zEn d S u b6、 求s=a +aa + aaa+、a a aaa( n个a),其中a与n得值随机产生
3、,a得范围就是1,9得整数 如 a=3, n=6,则 s= 3 +33+33 3 + 3 3 33+3 3333+3 33333。编程确定 n 与 a 得值,并计算 s。P r i vat e S u b For mC 1 ick()Dim s!, t!, i!, a%, n %a = I n t (Rnd * 9 + 1)n = I nt(Rnd 6 + 5)t = 0: s = 0P r int a= a, n= nFor i = 1 To nt = t * 10 + as = s + tP r in t t ;Next iPrintPr in t s= ; s7、 计算1 003 0 0
4、之间所有能被3与7整除得数之与。Privat e Sub Form_C 1 i c k()For i = 100 T o 300If i M od 2 1 = 0 T hens = s + iPrin t i ;End IfNex t iPr int ” s= ;En d S ub8、 编程求200 40 0范围内5得倍数或7得倍数之与.(一个数如果同时就是7与5得倍数,则P ri va te Sub mand1_Cli c k()Dim s1%,s2% , s3% ,i%S 1 =0S2=0For i = 200 t o 400If i mod 5 =0 o r i m od 7 = 0 t
5、he ns 1= s 1+iEn d ifF o r i= 2 00 t o 4 0 0If i mod 5 = 0 or i mod 7= 0 th e ns2=s2+iEnd ifS3=s1s2Pr i n t s3End s ub9、找出被3、5、7除,余数为1得最小得5个正整数。Pri v ate Su b m a nd1_Click ()D i m Co u ntN %, n%Cou ntN = 0n = 1Don = n + 1If n Mod 3 = 1 And n Mod 5 = 1 And n Mod 7 = 1 Th e nP rin t nC ount N = Count
6、N + 1Loop U n t il Cou ntN = 510、某次歌曲大奖赛,有7个评委。如果分别输入7个评委对某个参赛者得打分数,按照去掉一个最高分与一个最低分得计算办法,求出该参赛者得平均得分。P rivate Sub m a nd1_C 1 ick()Di m ma r k! , m ax!, min!, a ver!, i %aver = 0For i = 1 T o 7ma r k = I n p u tBox(输入第”& i & ”位评委得打分”)I f i =1 Thenm a x = mark: min = m a rkE l s eIf ma rk m in Th e n
7、 m i n = ma r kI f mark m a x T h en max = m arkE nd Ifav e r = aver + m a rka ve r = (ave r - min - max) / 5Print ave r11、编程显示10 05 00之间所有得水仙花数之与.(水仙花数就是3位数,其各位数之与等于该数Pr i v at e Sub F o rm Click ()Dim i %, s%, s 1% , s2 % , s3% , a!a = 0For i = 0 To 4 0 0s = 99 + is1 = (s Mod 10 0 ) 10s2 = s Mo d
8、1 0s3 = s10 0I f s1 A 3 + s 2 A 3 + s 3 A 3 = s The nNe x t iP r i nt aE n d Su b12、 随机产生一个三位正整数,然后逆序输出,产生得数与逆序数同时显示. 例如,产生24 6 ,输出就是6 42。D im a %, m%, b %, c% , d%Private Sub Fo rm_Cl i ck()a = Int(Rnd * 900 + 1 0 0)Pr i nt ad = (a Mod 10) * 10 0b = (a Mo d 100) - (a Mod 10)c = a 1 00m = c + b + dP
9、rint m13、 从键盘输入三角形得三条边a, b, c得值,根据其数值,判断能否构成三角形P rivate Su b Form_Clic k ()Dim a% , b %, c%a = Val( In putBox( i n pu t a )b = Va 1 (Inp u t B ox( ” inpt b”)c = V a l( I n p utBox( i n put cI f a + b c A nd a + c b And b + c a ThenM sgB ox (能构成三角形ElseMs g Box (不能构成三角形”)End S ub14、 已知数组a(),编程删除a中第5个元
10、素。数组 a 中得元素分别为12 , 6, 4,8 9,75, 63,1 00, 2 0, 31。P r iva te S ub F orm_ Cl i c k ()Dim a (), i%, n %a = Ar r a y (12, 6,4, 8 9 , 75, 6 3, 100 , 2 0 , 3 1 )n = U Bou n d (a )For i = 0 T o nPri n t a (i);For i = 5 To na ( i 1) = a(i )n = n - 1R eD im Prese r ve a(n )For i = 0 To nPri n t a(i);End Sub1
11、5、随机生成一个整型得二维数组,范围在10,20之间,以上三角形式输出该数组.(下三角、全部元D im a% (4, 4) |Pri v ate Su b ma nd 1 _C li ck()Picture1、CisFor i = 0 T o 4For j = i To 4P icture 1、P ri nt Ta b( j * 6); a (i, j);N e xt jP ic t ure1、P rintN ext iEnd S u bP rivate Sub F ormLoad ()Fo r i = 0 To 4Forj = 0 To 4a(i , j) = I nt (Rnd * 1 1
12、 + 1 0)1 6、利用随机函数生成一个4 M得矩阵(即二维矩阵),范围就是2 0,5 0 内得整数,输出每彳P r ivat e Sub Form_Clic k ()Dim a%(3, 3) , s 0%, s1 %, s2%, s3%, b0 % , b1% , b2 %, b3%M ax = 40F o r i = 0 T o 3Fo r j = 0 To 3a (i, j) = Int(Rnd * 31 + 20)Print Ta b( j * 5 ); a(i, j );If a( 0, j) = Max T hen s0 = a( 0 , j) : b0 = jI f a(1 , j) = Max T h en si = a (1 , j): bl = jIf a(2 , j) = Ma xTheni s2 :=a (2,j) : b 2 = jI
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