用递归求fibonacci第n项
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#includeunsignedintFibonacci(intn);intmain(void){inti;for(i=1;i
dimf()asdoublen=inputbox("in","NO.")redimf(n)asdoublef(1)=1f(2)=1fori=3tonf(i)=f(i-1)+f(i-2)nextprin
#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
c:intfib(intn){return(n
#include <stdio.h>int sumn(int n,int *flag){ (*flag)++; if(n==1)
PrivateSubForm_Click()DimnAsIntegern=Val(InputBox("请输入N:"))Dima,bAsLonga=1:b=1Fori=1TonPrinta&""&b&"
#includeusingnamespacestd;intmain(void){\x09intn,i,*fib;\x09cin>>n;\x09if(n==1||n==2)\x09\x09cout
#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)
public static int fibonacci(int n) { if (n <=
递归算法intfib(intn){//求fibonacci数列第n个数if(n==1||n==2)return1;elsereturnfib(n-1)+fib(n-2);}非递归intfib(intn
int F(int n){\x09if(n==0) //比较用==\x09\x09return 1; else&n
代码如下:OptionExplicitPrivateSubCommand1_Click()MsgBoxP(2,2)EndSubFunctionP(ByValnAsInteger,ByValxAsDou
楼上的程序会慢死人的.给一个非递归实现.functionFibonacci(byvalnasLong)asLongdiml1aslong,l2aslong,l3aslongl1=1l2=1ifn
#includelongfib(intn){inta;if(n==1)a=1;elseif(n==2)a=1;elsea=fib(n-1)+fib(n-2);returna;}voidmain(){\
#includeintf(inti){if(i==1||i==2)return1;elsereturnf(i-1)+f(i-2);}voidmain(){intn,j;scanf("%d",&n);f
#includemain(){intn,i,j,k;while(scanf("%d",&n)==1){if(n==1||n==2){printf("%d\n",1);cont
longfib(intn) { if(n==0)return0; if(n==1)return1; if(n>1)returnfib(n-1)+fib(n-2); }
publicclassFibonacci1{publicstaticlongfib(intn){longf1=1,f2=1;longm=0;if(n
//fibonacci数列:11235813213455...#includedoublefib_val[100]={0};doublefibonacci_1(intn)//递归,计算时间长,n最好不