C#:编写函数计算任一输入的整数的各位数字之和.

#includelongfactorial(intm,intn){longsum=1,sum1=1;inti;if(m-n>n){for(i=m;i>m-n;i--)sum*=i;for

intfun(longx){ints=0,j;while(x){j=x%10;s=s+j;x=x/10;}returns;}main(){longn;scanf("%ld",&n);printf("%

#includeintfun(intn){ints=0;while(n){s+=n%10;n=n/10;}returns;}main(){intn;scanf("%d",&n);printf("%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(){doublex,y;scanf("%lf",&x);if(x

#include#includevoidmain(){doublex;doubley;coutx;if(-2

首先C我长时间没碰了,但你的程序的问题还是能看出来的,还是看看书吧.main函数就不对,应是main(){程序主体}.if函数没有返回值,你的程序中没必要用这么多花括号,当if或else后只有一句条件

#include"stdio.h"#include"conio.h"#include?main(){intx,i;printf("输入一个正整数:\n");scanf("%d",&x);for(i=1

#include#defineN4intmain(void){inti,j;inta[N][N];for(i=0;i

#includevoidasum(doublea[3][3],doubleb[3][3],doublec[3][3]){inti,j;for(i=0;i

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

#includepower(floata,floatb){floatsum1=a*b;returnsum1;}intmain(){floatx,y;floatsum;printf("inputthef

#includeintgetArea(intlen){\x09returnlen*len;}intmain(){\x09intn;\x09intarea;\x09\x09printf("请输入边长:\

elsey=3*x+1；后面的封号是汉字的

function [ s ] = solve_area( a,b,c )  p=(a+b+c)./2;&nbs

#includeintisRunyear(intyear){if(year%400==0||year%4==0&&year%100!=0)return1;return0;}voidmain(){int

#includeintmain(){intlength,width,perimeter;printf("长：");scanf("%d",&length);printf("宽：");scanf("%d"

#includeintcal(intm,intn){intret=0;ret=m%n;returnret;}intmain(intargc,char**argv){intm,n,max,min

自带函数mean.还需要自己写函数么?如果是你就把mean函数调出来看看,然后简化一下就可以了.

//使用海伦公式#include/*ForIO*/#include/*Forsqrt()*/intmain(void){doublea,b,c,p,s;printf("请输入a,b和c：");scan