求斐波那契数列的n项之和.其中n为手动输入的项的数目.

帮你写好了.unsigned int fib(unsigned int n) {\x09if (n == 1 

#includelongintfn(int);voidmain(){printf("%d",fn(10));}longintfn(intm){longinttemp;if((1==m)|(2==m))

用先设为等比数列再求解的方法,详细请见图片.

*求斐波那契数列1,1,2,3,5,8,13,21,34,…的前30项的和.该数列从第3项开始每项等于前两项之和.3524577SETTALkOFFCLEAS=2F1=1F2=1I=2DOWHILEI

PrivateFunctionF(nAsLong)AsLongIfn>2ThenF=F(n-1)+F(n-2)ElseF=1EndIfEndFunctionPrivateSubCommand1_Cli

main(){inti,n,s=1,f[]={0,1,1};printf("Pleaseinputthenumberofterms:");scanf("%d",&n);if(n==0){s=0;f[2

现成的

老式写法longint格式用%ld--l是L小写.现在写%d就可以了,longint,shortint,int都用%d程序用ASCII码写成：#includemain(){longs,n,k,i;sc

dima()aslong,nasintegern=inputbox("请输入n的值：")redima(1ton)callFibonaccia()subFibonacci(a()aslong)dimia

#includevoidmain(){intf1,f2,f,i,n;printf("请输入项数:");scanf("%d",&n);f1=f2=1;if(n

publicclassFibonacci{\x09publicstaticvoidmain(Stringargs[]){intn,fn;//n为第n项,fn为第n项的值java.util.Scanne

#includemain(){longf1,f2,f;inti,n;scanf("%d",&n);f1=f2=1;if(n

#includeintmain(){inti=0;floatsum=0;intn;intx[n],y[n];printf("请输出计算的项数:");scanf("%d",&n);x[0]=2;x[1]

#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

functionfibonacci(n:integer):integerbeginif(n=0)thenResult:=0;if(n=1)thenResult:=1;if(n>1)thenResult

#includefib(intn){if(n==0)return(0);elseif(n==1)return(1);elsereturn(fib(n-1)+fib(n-2));}main(){intn

设这个数列为an,公差为d,前n项和为SnSn=n(a1+an)/2S(2n+1)=(2n+1)[a1+a(2n+1)]/2=36+30=66(1)数列an的奇数项共有n+1项,也构成等差数列,其和为

[oldboy@daiqingyangsscripts]$catlist.sh#!/bin/bash#set-xif[$#-eq0]thenexitficount=1index=$1while[$co

按照正常的逻辑是只要求a[2][2]={1,1,1,0}这个矩阵的n次方就可以得到斐波那契数列的第n项（即a[0][1]）的值.但是你忽略了一点,就是你在求a[0][1],a[1][0],a[1][1

递归很简单：描述如下f（n）if(n==1||n==2)return1;returnf(n-1)+f(n-2);非递归用循环就可以做到：a=b=1;for(i=3;i