用递归调用的方法编写一个返回长整形的函数

#includeunsignedintFibonacci(intn);intmain(void){inti;for(i=1;i

#include"stdio.h"intprime(intn){if(n>1)returnn*prime(n-1);elsereturn1;}intmain(){into;for(o=1;o&l

#include<stdio.h>int f(int x ){\x09if ( x>0 )\x09\x09return&n

#includelongintf(intn){if(n==0)return0;elseif(n==1)return1;elsereturnf(n-1)+f(n-2);}intmain

楼上写的有问题,1、result=double(r);函数名写错了2、printf("%d",result);不应该是%d,而应该是%lf#include"stdio.h"doublef（double

#include#defineN20main(){intf(intn);inti;for(i=1;i

// C++int F(int n) {if (n == 0) return 1;else if

#includedoubleH(intn,doublex){if(x>1){if(n==0)return1.0;//H0(x)=1;if(n==1)return2.0*x;//H1(x)=2x;//直

#include#includeintmax(inta,intb){if(a>b)returna;if(a

#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;intexc(inta,intb)//这个时候参数不分大小{if(a>b)//确保第二个数是大数{intc;c=a;a=b;b=

/>#includeusingnamespacestd;longunsignedfun(intn){if(n>1)returnn*fun(n-1);return1;}voidmain(){intn;c

#includeintfact(int);main(){inti,sum=0;for(i=1;i

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

#include#includefloatmyfunction(intn,intx){if(0==n){return1;}elseif(1==n){returnx;}else{return((2*n-

functiongqj=erfen(p,a,b,e)ifabs(b-a)

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

#includeusingnamespacestd;longunsignedfun(intn){if(n>1)returnn*fun(n-1);return1;}voidmain(){intn;cou

publicclassFibonacci1{publicstaticlongfib(intn){longf1=1,f2=1;longm=0;if(n

用递归法计算n!用递归法计算n!可用下述公式表示：n!=1(n=0,1)n×(n-1)!(n>1)按公式可编程如下：longff(intn){longf;if(n