C 求 1 2 3 -- n 的和:(用递归法实现)
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/20 12:57:54
#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
#include <stdio.h>int sumn(int n,int *flag){ (*flag)++; if(n==1)
#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/*非递归求:f(1)+f(2)+...+f(m)其中f(n)=n*(n+1)*/unsignedintsum_fn(unsignedintm){intn,sum=0;for(n=1;
设0为数列的第一项递推:intf1(intn){inti,item=-2;for(i=1;i
//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)
#include<stdio.h>void main(){ int i,k,sum=0;  
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<stdio.h>int calc(int n){\x09int i,sum;\x09i=1,sum=0;\x09while(i<=n)
#includelongfib(intn){inta;if(n==1)a=1;elseif(n==2)a=1;elsea=fib(n-1)+fib(n-2);returna;}voidmain(){\
f函数里已经被传进一个参数值了为什么还要再输入一个n值呢?另外,n前还要加&,应该是scanf("%d",&n);
#include#includefloatmyfunction(intn,intx){if(0==n){return1;}elseif(1==n){returnx;}else{return((2*n-
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