编写递归程序计算Fibonacci数列从第几项的值大于10000.

publiclongpower(intm,intn){if(nreturnm;elsereturnpower(m,n--)*m;}

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

你应该定义成doublegetPower(doublex,inty)就可以了目前你的定义的情况下getPower(b,m)找不到最匹配的就是(double,int)只能找次匹配的,找到了(double

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

PrivateSubCommand1_Click()Dima,xAsSinglea=Val(Text1.Text)Ifa再问：x=x+a*0.2+3000*0.2怎么都乘0.2呀？

#includemain(){doublex,y;scanf("%lf",&x);if(x

longpower(intm,intn){doublep=1;if(n>0){p=m*power(m,(n-1));returnp;}}voidmain(){intm,n;longk;scanf("%

x=[3,4,5,6,7,8,9,10];y=[4,5,6,7,8,9,10,11];z=21.89-9.87*x+8.09*y-7.98*x.*y-1.87*x.^2-7.98*y.^2z=1.0e

doublef(doublex){doubles=.0;if(x

longadd(intn){intt=n-1;if(t>1){longresult=n*t;longsum=result+add(t);returnsum;}else{returnn;}}楼上的方法,

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

#includelongfac(intn){inti;longx=1;for(i=2;i再问：谢谢咯！可是我说的是递归法哦！再答：#includelongfac(intn){if(n==0)retur

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

#include <stdio.h>int main(){  float x, tax = 0; 

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

问题太多.1,第四行doubleresult,product,case;中case不能作变量名,保留字.2,第十行result=fact(i);函数参数太少,你下面定义的函数应该有2个参数3,case

;MOVAX,AANDAX,B;AX=aANDbMOVBX,AXORBX,B;BX=aXORbADDAX,BXADDAX,BX;AX=2*(aXORb)+aANDbADDAX,A;AX=a+2*(aX

#includevoidmain(){intn=1,N,i;scanf("%d",&N);if(N20)printf("shuruwuxiao!\n");else{for(i=1;i

#includedoublef(doublex,intn){if(n==1)returnx;else{doubled=1.0;inti;for(i=1;i

intgetpower(intx,inty){if(y==1)returnx;elsereturnx*getpower(x,y-1);}doublegetpower(doublex,inty){if(