用递归法求数列第n项的值,数列定义如下:f(n)=
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#includeunsignedintFibonacci(intn);intmain(void){inti;for(i=1;i
添加一个文本框输入前N项的N值,再添加一个命令按钮即可PrivateFunctionF(NAsLong)AsLongIfN>2ThenF=F(N-1)+F(N-2)ElseF=1EndIfEndFun
dimf()asdoublen=inputbox("in","NO.")redimf(n)asdoublef(1)=1f(2)=1fori=3tonf(i)=f(i-1)+f(i-2)nextprin
#includelongintfn(int);voidmain(){printf("%d",fn(10));}longintfn(intm){longinttemp;if((1==m)|(2==m))
#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
#include/*非递归求:f(1)+f(2)+...+f(m)其中f(n)=n*(n+1)*/unsignedintsum_fn(unsignedintm){intn,sum=0;for(n=1;
case 1: return 1; case 2: return 1;case1后面加个冒号和return1就行.
#include#includevoidfun(intn){\x09inti;\x09int*a=(int*)malloc(n*sizeof(int));\x09a[0]=a[1]=1;\x09for
#include#defineCOL10//一行输出10个longscan(){//输入求fibonacci函数的第N项intn;printf("InputtheN=");scanf("%d",&n)
用VB写的~PrivateSubForm_Load()Dimfn&,f1&,f2&,n%f1=1:f2=1n=Val(InputBox("请输入n"))fn=f1+f2Printf1Printf2Pr
观察1,1,2,4,7,13,24,44,...该数列从第四项起,每一项是其前面三项的和.因此:an=a(n-1)+a(n-2)+a(n-3)(n属于N,n>=4)
#includeintFibonacci(intn){if(n==1||n==2)//递归结束的条件,求前两项return1;elsereturnFibonacci(n-1)+Fibonacci(n-
#includeusingnamespacestd;intmain(){intn,a=1,b=2;cout再问:^那这个是什么符号,这个没学过,有用temp做的么?再答:是位运算的异或符号;也可以用t
int F(int n){\x09if(n==0) //比较用==\x09\x09return 1; else&n
functionfibonacci(n:integer):integerbeginif(n=0)thenResult:=0;if(n=1)thenResult:=1;if(n>1)thenResult
publicclassFibonacci{\x09publicstaticvoidmain(Stringargs[]){intn,fn;//n为第n项,fn为第n项的值java.util.Scanne
1)a1=1,a2=1,a(n+2)=a(n+1)+an,a(n+2)+[(√5-1)/2]a(n+1)=[(√5+1)/2][a(n+1)+(√5-1)/2*an]==.=[(√5+1)/2]^n[
inta[3]={1,1,2};inttemp;for(inti=3;i再问:inta[3]={1,1,2};inttemp;是什么意思啊?求解再答:存前面3个数啊
#includefib(intn){if(n==0)return(0);elseif(n==1)return(1);elsereturn(fib(n-1)+fib(n-2));}main(){intn
#includeintFibonacci(intn){if(n==1||n==2)//递归结束的条件,求前两项return1;elsereturnFibonacci(n-1)+Fibonacci(n-