5.编写计算m的n次方的递归函数(m为整数,n为正整数).
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#include"stdio.h"intgys(intm,intn){if(n>1){if(m%n!=0){returngys(n,m%n);}elsereturnn;}}intmain(){intm
publiclongpower(intm,intn){if(nreturnm;elsereturnpower(m,n--)*m;}
#includeunsignedintFibonacci(intn);intmain(void){inti;for(i=1;i
一次循环就解决的,不是很难,为什么不自己尝试一下呢?给你提示一下,却是不会再给你说吧!真正想学习,最好多练习一下好不好?累乘初始值为1,循环变量循环n次,每次累乘m,循环结束即为m的n次方的值
#includelongintfn(int);voidmain(){printf("%d",fn(10));}longintfn(intm){longinttemp;if((1==m)|(2==m))
PrivateFunctionF(nAsLong)AsLongIfn>2ThenF=F(n-1)+F(n-2)ElseF=1EndIfEndFunctionPrivateSubCommand1_Cli
#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
longpower(intm,intn){doublep=1;if(n>0){p=m*power(m,(n-1));returnp;}}voidmain(){intm,n;longk;scanf("%
intfunc(intx,intn){inty;if(n
main(){longi,t=1;doublex,e=1.0;scanf("%f",&x);for(i=1;i
#include#includeintjiecheng(inte,intn){//intx=e;if(n==1)returne;else{returne*jiecheng(e,--n);}}intma
#includeintFibonacci(intn){if(n==1||n==2)//递归结束的条件,求前两项return1;elsereturnFibonacci(n-1)+Fibonacci(n-
新建工程,添加command1控件.vb6.0+xp调试成功代码如下:DimnAsIntegerPrivateSubCommand1_Click()m=InputBox("请输入底数m")n=Inpu
#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
程序已经调试过了,你看下吧,可以的话给个好评#include"iostream"usingnamespacestd;floatpower(floata,intn){if(n==1)returna;el
C描述functionttt(n){ returnn>1?n*ttt(n-1):1;}使用方法:ttt(21);
#includedoublepow(doublex,intn){if(1==n){returnx;}else{doubletemp=pow(x,n-1);return(x*temp);}}voidma
#includemain(){intn,i,j,k;while(scanf("%d",&n)==1){if(n==1||n==2){printf("%d\n",1);cont
if(n==1)\x05\x05return1;你可以改成n==0或者改成returnx取其中一种就可以了
intgetpower(intx,inty){if(y==1)returnx;elsereturnx*getpower(x,y-1);}doublegetpower(doublex,inty){if(