试用递归法编写一个返回长整型函数计算裴波纳

#includeintdigit(intn,intk){returnk>1digit(n/10,k-1):n%10;}intmain(){printf("%d",digit(12345,3));}

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

#includelongfun(longx){intn=10,m=0;while(x){intt=x%10;if(t%2==0){m+=t*n/10;n*=10;}x/=10;}returnm;}vo

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

#includelongintf(intn){if(n==0)return0;elseif(n==1)return1;elsereturnf(n-1)+f(n-2);}intmain

/*90897:7980923098:8903233201:1023378653:3568736451:1546383409:90438Pressanykeytocontinue*/#include&

// C++int F(int n) {if (n == 0) return 1;else if

亲,这是我改好的程序#includevoidfun(long*s,intn){inti,j=0;longt[100];for(i=0;i

语法算法都没错没有头文件#include希望对楼主有用

#include#includevoidmaxp(int**ia,intr,intc,int*x,int*y,int*max){*max=ia[0][0];inti,j;for(i=0;i

longintlongdouble

#includeintFibonacci(intn){if(n==1||n==2)//递归结束的条件,求前两项return1;elsereturnFibonacci(n-1)+Fibonacci(n-

#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

#include <stdio.h>char* dg(char* instr, char* outstr, char* 

Function isum(n As Integer) As Double     isum&nbs

#include#includemain(){\x09longintx;\x09inti=0;printf("input:");\x09scanf("%ld",&x);do{x=x/10;i++;\x

intSumNums(intnum){if(num

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

#includeintfibo(intn){if(nreturn1;elsereturnfibo(n-1)+fibo(n-2);}intmain(){intn;scanf("%d",&n);print

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