用递归法求n阶勒让德多项式的值.(函数实现

#include"stdio.h"intgys(intm,intn){if(n>1){if(m%n!=0){returngys(n,m%n);}elsereturnn;}}intmain(){intm

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

#includedoublefun(intn);intmain(void){\x09intn;\x09printf("Entern:");\x09scanf("%d",&n);\x09printf("

#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

递归的时候逻辑有点混乱,你看这样写是不是更好#includeintgcd(intm,intn){intg;g=m%n;if(0==g){returnn;}else{returngcd(n,g);}}i

#include#includevoidfun(intn){\x09inti;\x09int*a=(int*)malloc(n*sizeof(int));\x09a[0]=a[1]=1;\x09for

//很简单,应该是答案印错了//不过这样才是正确的递归方式doublelegendre(intn,doublex){if(n==0)return1;elseif(n==1)returnx;elsere

#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

#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

varn:integer;functionfac(n:integer):longint;beginifn

的很大的时候是否会超出double的有限范围?

代码如下:OptionExplicitPrivateSubCommand1_Click()MsgBoxP(2,2)EndSubFunctionP(ByValnAsInteger,ByValxAsDou

longfac(int);这一步应该为longfac(int,float);y=fac(n);这一步应该为：y=fac(n,x);elseif(n=0)f=1;这一步应该为：elseif(n==0)f

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

什么语言再问：计算机程序设计有关的再问：编程题目再答：对啊，用什么语言编再问：再问：标准再问：判断下对错再答：我没学过数据库再问：十六进制数FO5D.7A用八进制怎么表示再答：170135.36#in

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

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

#includeintgys(inta,intb){\x09intr;\x09r=a%b;\x09if(r==0)returnb;\x09elsereturngys(b,r);}voidmain(vo