vb编程求数列2 1 3 2 5 3
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PrivateSubForm_Click()Dima(1To40)AsLongFori=1To40Ifi=1Ori=2Thena(i)=1Elsea(i)=a(i-2)+a(i-1)EndIfPrin
Dimn,x,r,sAsSinglen=2s=0DoWhilen再问:能说一下算法的思路吗?再答:首先1不是素数,就用枚举将2到1000的每个数列举出来,再逐个检验,即从2到n/2一一举例,判断是否能
#include#includemain(void){intn,i;printf("请输入n的值\n");scanf("%d",&n);intsum1=1,sum2=2,sum;for(i=3;i
是没有思路,还是不会编程?至少有一位数字是5用以下的子函数boolIsInclude5(intnum){while(num!=0){if(num%10==5)returntrue;num=num/10
判断素数的原理是:素数是只能被1和本身整除的数.例如3只能被1和3整除,17只能被1和17整除等等,想9就不是素数(能被1,3,9整除).编程的算法是:穷举法,就是将需要判断的数除2、除3.一直除到这
Private Sub Command1_Click()Dim a As Long, b As Long, c
functionsum(nasinteger)dima(30)asintegerdimsasintegers=0a(1)=0a(2)=0a(3)=1fori=4to30a(i)=a(i-1)+a(i-
Dima1,a2,a3,a4,iAsIntegera1=3:a2=4:a3=5fori=4to26a4=a1+a2+a3a1=a2a2=a3a3=a4next输出a4即可
PrivateSubForm_Click()DimnAsIntegern=Val(InputBox("请输入N:"))Dima,bAsLonga=1:b=1Fori=1TonPrinta&""&b&"
OptionExplicitDimf(40)AsLongPrivateSubCommand1_Click()DimiAsByteDimsAsLongf(1)=1f(2)=1s=2Print"No1:"
简单就是不会再问:....再答:不是不会是好久没做了差不多都忘了呵呵
PrivateSubForm_Load()Fori=1To50n=1Forj=2Toin=n*jNexts=s+1/nNextMsgBoxsEndSu
非递归:staticvoidf(intn){longp1=1,p2=1,p=1;for(inti=1;i
PrivateSubCommand1_Click()DimaAsLongDimiAsLongDimsAsBooleans=Truea=InputBox("请输入一个自然数","输入")'-------
代码:Private Sub Command1_Click() Call max100End SubSub&nbs
假设级数表达式为f(i),随i值的变化而减小,则可在循环时利用级数f(i)和f(i-1)的差值来与10^-6比较,当两次计算的结果满足dpp=Abs(dc)b=b*-1n=n+1mv=rad^(2*(
Private Sub Command1_Click()Dim F(11), i As LongF(0) = 
Private Sub Form_Load()Dim I As IntegerForm1.AutoRedraw = TrueFor
我给你代码:#include <stdio.h>#include <stdlib.h>#define N 47int fibo
很简单的~不过你问错地方了