c 使用递归法实现计算m的n次方

#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;}

#includeintg(intn,intk){if(k==1)returnn;returng(n,k-1)*n;}intf(intn,intk){if(n==1)return1;returnf(n-

#includeintfun1(intn){intsum=1;for(inti=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

#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("%

实现计算N!的算法是A递归B迭代C排序D查找求阶乘用递归functionsolution=factorial(n)ifn==1solution=1;elsesoluion=n*factorial(n-

这个东西关键是递归算法的确定,需要一点数学知识.#include//递归计算函数intcalc(intx){if(xif(x==1){return1;}elseif(x==2){return3;}el

然后是用递归逆转输出数字的.programReverseNum;varnum:integer;procedurefun(n:integer);beginifn1thenbeginwrite(nmod1

前面两个题目我并不完全按照你所说的格式来输出,这个自己改了.首先是折半查找的.programfind;varstr:string;ch:char;lo,hi:integer;m:integer;beg

intfunc(intx,intn){inty;if(n

#include<stdio.h>int gcd(int m,int n){while (n){int t=m%n;m=n;

#include#includeintjiecheng(inte,intn){//intx=e;if(n==1)returne;else{returne*jiecheng(e,--n);}}intma

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

longmi(intx,intn){if(n=1)returnn;elsereturn((long)x*mi(x,n-1);}

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

#include#includeintmain(intargc,char*argv[]){intm,n,c,i,j,max;scanf("%d%d",&m,&n);if(m=2){printf("Th

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

#include#includefun(intx);intmain(void){char*pszBuff=NULL;inta;inti=1;intsumResult=0;printf("inputai