编写一个函数,计算n!,在主函数调用此函数计算1! 2! 3! 4! 5!的值

#includeintjiecheng(int);voidmain(){floate=1;for(inti=1;i=1;i--)m=m*i;returnm;}

已经通过测试,请查收,#include <iostream>#include <cmath>using namespace std;vo

用递归函数#include#includeintinv(intx);main(){intsum=0,i;for(i=1;i

是用C语言么?我用递归调用帮你写个#includevoidmain(){intfac(intn);inta;floatb;prinf("inputanintegernumber:")scanf(%d,

intjc(intx){returnx==1?1:jc(x-1)*x;}再问：可以编一个完整的么？我直接运行试一下。。。新手，不好意思，，，，谢谢再答：intjc(intx){returnx==1?1

#includemain(){\x05intm,n;\x05ints=1;\x05intpower(inta,intb);\x05printf("inputmn(m^n):\n");\x05scanf

#includeintfun(intn){\x09ints=1,i;\x09for(i=1;i

#includemain(){intn;printf("Pleaseenteranumber:");scanf("%d",&n);if(n%2==0)printf("Thenumber%dis偶数\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

==||童鞋.你是要返回输入的n个数的最大值,还是一个数组的最大值啊.返回输入n个数最大值的函数：#includeintMax(intn);//最大值函数声明intmain(){intn;scanf(

如果是要求前N个偶数之和：#includeusingnamespacestd;longFunctionadd(intn){longsum=0;for(inti=0;i

importjava.util.Scanner;publicclassOushu{Oushu(){try{System.out.println("请输入n值");Scannerin=newScanne

#include<stdio.h>unsignedlongcombinNum(intm,intn){ inti; unsignedlongx=1; if(n&

#includevoidfun(intn){inti;for(i=2;i

首先你那个a[i]=a[i]*(n-j)/j公式有点问题吧,这样会输出全是1的啊...如果还是按照你这样可以改成如下代码：#include"stdafx.h"#include#include#incl

#include#defineN9voidmain(){inti,j,a[N][N]={0};for(i=1;i

#includemain(){intn,i,j,k;while(scanf("%d",&n)==1){if(n==1||n==2){printf("%d\n",1);cont

voids(folata,folatb,folatc){folatp;p=(a+b+c)/2;S=√[p(p-a)(p-b)(p-c)];returns;}

#includeusingnamespacestd;voidmain(){inta=0,b=0;cin>>a>>b;cout

#includevoidmain(){inta,b,sum=0;printf("请输入两个整数：");//将两个改成n个就好了scanf("%d%d",&a,&b);sum=a+b;printf("%