编写一函数,计算任一输入的整数的各位数字之和.主函数包括输入输出和调用该函数.

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\

#include#defineSIZE20voidmain(){printf("请输入10个整数：\n");inta[SIZE];intn=10,i,j,k,b;for(i=0;i

intsum(intx){\x09intsum=0,n;\x09do{\x09\x09n=x%10;\x09\x09sum=sum+n;\x09\x09x=x/10;\x09}while(x>0);\

1：#includeintmain(){intnum,sum=0;while(scanf("%d",&num)==1&&num>0){sum+=num;\x09}printf("结果为%d",sum)

楼主你好,这是源程序,是用辗转相除法写的#includeintmax(intp_a,intp_b){\x09if(p_b==0)\x09\x09returnp_a;\x09else\x09\x09re

%首先将这些数储存在向量a中a=input('inputnumber');%输入数据时要加[].例如,[8216]n=length(a);b=zeros(1,100);fori=1:nforj=1:1

#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

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

#include#includeintmax(inta,intb,intc){c=c>(a=a>b?a:b)?c:a;returnc;}intmain(){inta,b,c;scanf("%d%d%d

publicstaticvoidtest(){Scannersc=newScanner(System.in);longnum=0,negative=0,positive=0,sum=0;Listnum

给你两种答案选择：（都是自己编的,而且经过VC测试的!）方法一：#includevoidmain(){inti,a,b=0;scanf("%d",&a);if(a==1)a++;for(i=0;i

intSumNums(intnum){if(num

Dimi1AsIntegerDimi2AsIntegerDimi3AsIntegeri1=Val(InputBox("请输入第一个整数"))i2=Val(InputBox("请输入第二个整数"))i3

intflip(intin)//不用数组{intout=0,flag=0;if(in>-10&&inreturnin;elseif(in{flag=1;in=-in;}do{out+=in%10;ou

PrivateSubCommand1_Click()a=InputBox("请输入数字","提示")b=InputBox("请输入数字","提示")Print"和："&Val(a)+Val(b)Pri

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.

#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("%

就用最简单的EXCEL吧,假设你的数在A列,那么你可以在B1单元格输入="你输入了"&COUNTIF(INDIRECT("A1:A"&MATCH(-1,A:A,0)),ROW())&"个"&ROW()