matelab用递归函数表示斐波纳契数列

我前几天刚刚答过这道题.代码：#includeusingnamespacestd;templatevoidPerm(Typelist[],intk,intm){if(k==m){for(inti=0;

#include#include//note:只能处理n是正整数的情况floatf(floatm,intn){assert(n>=0);if(n==0)return1.0;if(n==1)return

#include "stdafx.h"#include <iostream>using namespace std;int&nb

#include#defineNUM4intdsum(intn){return(n==01:n==11:dsum(n-1)*n);}intfsum(intn){inttotal=1;for(inti=

#includeusingnamespacestd;doublepnx(int,double);intmain(){doublen,x;coutx;cout再问：谢谢能加个好友吗给个q也行以后请多指教

//递归intfun(intn){if(n==1||n==2)return1;elsereturnfun(n-1)+fun(n-2);}//非递归intfun(){intans[41];ans[0]=

voidprt(intn){printf("%d",n%10);if(n>10)prt(n/10);}intmain(void){inta;printf("请输入整数：");scanf("%d"

#includevoidfib(intn,intf0,intf1){intf;//当前项inti=0;if(n=2)printf("%8d,%8d",f0,f1);//f0,f1for(i=2;i

#include"stdio.h"intmain(){inta,b,c,i;a=1;b=1;printf("%d%d",a,b);for(i=3;i

intFibona(intn){intm;if(n==1)return(1);elseif(n==2)return(1);else{m=Fibona(n-1)+Fibona(n-2);return(m

#includelongfib(intn){inta;if(n==1)a=1;elseif(n==2)a=1;elsea=fib(n-1)+fib(n-2);returna;}voidmain(){\

#includedoublepow(doublex,intn){if(1==n){returnx;}else{doubletemp=pow(x,n-1);return(x*temp);}}voidma

#includevoidfun(intn,int*s)///求斐波那契序列中第n位的值{intf1,f2;if(n==0||n==1)*s=1;else{fun(n-1,&f1);fun(n-2,&f

你这里的斑块其实就是连通域.MATLAb自带计算连通域个数的函数：bwlabel.% 返回x中连通域个数function n = f( x 

1.#include"stdio.h"//#defineRECURSION1#ifdefRECURSIONlongfact(intn){if(n

粘贴.递归就是一个函数内出现调用本身的现象,举个最简单的例子,求阶乘：当n=0或1时,n!=1；当n>1时,n!=n*(n-1)!通过这样的思想,程序写为：intfun(intn){if(n&l

#includeusingnamespacestd:intfuntion(intn){if(n==0){return0;}if(n==1){return0;}returnn&funtion(n-1);

你先了解这个函数的作用,结果就是n*(n/(2^1)*(n/(2^2))*(n/(2^3))*(n/(2^4))……*1n*(n/2)*(n/4)*(n/8)*……*1while(n>=0){if(n

intN(intx){if(x==0){return1;}else{returnx*N(x-1)}}intiRet=0;for(inti=1;i

longfun(n){longresult=1;for(inti=1;i