编写函数fun(int t)求斐波那契数列中大于t的最小的一个数,

intfun(intn){inta=n,b=0;while(a>0){b=b*10;b=b+a%10;a=a/10;}printf("%d",b);getch();return0;}或者把后三行删掉,

intfun(intn){intm=1,sum=0,i,j;for(i=1;i

#include"stdafx.h"#includeintsct(intm,intn){inttemp,a,b;if(m>y;g=sct(x,y);cout再问：是fun函数吗？测试用的主函数

PrivateSubfun()ifimod2=0thenforx=itoastep2s=s*xnextxelseforx=(i+1)toastep2thens=s*xnextxendif试试看行不行吧

#includedoublefun(doubleh){returnfloor(h*100+0.5)/100;}

intfun(intn){inta=n,b=0;while(a>=1){b=10*b+a%10;a=a/10;}returnb;}已经调试通过了哦!

#includeusingnamespacestd;doublefun(intn,doubleh){if(n==1)returnh;elseif(n再问：usingnamespacestd;这句

#include#includeintfun(intx){intr=1,d,c;while(1){d=x%10;r=r*d;x=(x-d)/10;if(x==0)break;}returnr;}mai

includeincludeddoublefun(inta,intb,intc){intp;p=(a+b+c)/2;returnsqrt(p*(p-a)*(p-b)*(p-c));}再问：ok再问：

上面的错了应该是intfun(){inti,sum=1;for(i=1;i

#include#includevoidswap(intc[],intlen){inti=0;inttmp;for(;i{tmp=c[i];c[i]=c[len];c[len]=tmp;}}intmu

#include <stdio.h>#include <stdlib.h>#include <math.h>double f

intfun(intlim,intaa[MAX]){intk=0,i,j;//k用于表示数组下标,i、j循环临时变量for(i=lim;i>1;i--)//每一个i数,从大到小尝试{for(j=2;j

请楼主参考采纳intfun(intt){inti;intcurrent=0;for(i=1;totalt)break;current+=i;}returncurrent;}

Functions(ByValaAsInteger,ByValbAsInteger)AsIntegerDimiAsIntegerFori=1To1000IfiModa=0AndiModb=0Thens

1.intf1=0,f2=1,f3;2.returnf3;继续写：f3=f1+f2;f2=f3;f1=f2;

用这肯定可以intfun(intm){for(inti=m-1;i>1;i--){intj;booleanisPrime=true;for(j=2;jif(i%j==0){isPrime=false;

#includeusingnamespacestd;longunsignedfun(intn){if(n>1)returnn*fun(n-1);return1;}voidmain(){intn;cou

#include#includefloatfun(inta,intb){floatc;c=sqrt(a)+sqrt(b);returnc;}intmain(){inta=12,b=20;floatc;

functionfun(d,h){if(d