用c语言中递归法计算20项的雯波那契数列(1.1.2.3.5
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![用c语言中递归法计算20项的雯波那契数列(1.1.2.3.5](/uploads/image/f/6248029-13-9.jpg?t=%E7%94%A8c%E8%AF%AD%E8%A8%80%E4%B8%AD%E9%80%92%E5%BD%92%E6%B3%95%E8%AE%A1%E7%AE%9720%E9%A1%B9%E7%9A%84%E9%9B%AF%E6%B3%A2%E9%82%A3%E5%A5%91%E6%95%B0%E5%88%97%281.1.2.3.5)
#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
#includefun(intx,intn){if(n==0)return(1);elsereturn(x*fun(x,n-1));}main(){intx,n;scanf("%d,%d",&x,&n
#includeintfun1(intn){intsum=1;for(inti=1;i
#include"stdio.h"doublemysqrt(doublea,doublex0){doublex1;x1=(x0+a/x0)/2.0;if(fabs(x1-x0)>0.000000
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#defineN20main(){intf(intn);inti;for(i=1;i
/*以下内容是在VC6.0环境下编译运行成功的,现在一般学习C语言均使用该环境*/#includeintFibonacci(intn){intm;if(n==1||n==2)m=1;elsem=Fib
#defineLENsizeof(structtree)#defineNULL0#include#includestructtree{chardata;structtree*lchild,*rchil
#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;
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
#includedoubleH(intn,doublex){if(x>1){if(n==0)return1.0;//H0(x)=1;if(n==1)return2.0*x;//H1(x)=2x;//直
#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#includeintjiecheng(inte,intn){//intx=e;if(n==1)returne;else{returne*jiecheng(e,--n);}}intma
不用那么麻烦inta=1,b=2,i,k,n;floatsum=0.0;scanf("%d",&n);for(i=0;i再问:不是题目要用递归函数
#include<stdio.h>int sum(int n){ return n>0?n+sum(n-1):0;}i
intFibona(intn){intm;if(n==1)return(1);elseif(n==2)return(1);else{m=Fibona(n-1)+Fibona(n-2);return(m
#includelongfib(intn){inta;if(n==1)a=1;elseif(n==2)a=1;elsea=fib(n-1)+fib(n-2);returna;}voidmain(){\
longmi(intx,intn){if(n=1)returnn;elsereturn((long)x*mi(x,n-1);}
原理是一样的,可以参考下面的稍微改一下,自己动手可以学的更多,#include//实现数据的全排序voidswap(int*a,intx,inty)//数据交换{inttemp=a[x];a[x]=a