输入正数n,用while计算n之内的奇数和,和偶数和,并输出

packageTest;importjava.util.Scanner;publicclassTest{publicstaticvoidmain(String[]args){Scannersca

varn,i,j,k:longint;s:array[1..24,1..24]oflongint;begink:=0;readln(n);fori:=1tondoforj:=1toidobegink:

首先明确gbs（最小公倍数）=m*n/gys（最大公约数）；然后求最大公约数用欧几里得辗转相除法；代码很短的.#include"stdio.h"intgys(intm,intn){returnn==0

//c语言#includeintmain(){intn;inti=1;intsum=0;scanf("%d",&n);while(i

programlong;vari,n:integer;beginreadln(n);whilen0dobeginwrite(nmod10,'');n:=ndiv10;end;writeln;end.

定义unsignedintn,longlongintn1,计算过程用for循环每次*10,保存到n1,最后输出n1再答：算法思路大致就是这样了，具体代码应该不难写再问：再答：哦，原来是这个再答：那么把

importjava.awt.*;importjava.awt.event.*;classTestimplementsActionListener{TextFieldintext,outtext;Bu

方法1#includevoidmain(){floate,a;inti,n=0;/*涉及到小数运算要用浮点型变量*/e=1,a=1;/*a=1在后面有乘积运算*/for(i=1;1/a>1e-5;i+

加点悬赏好有动力啊哎太费脑子和打字了虽然我知道我这有本C语言书上有现成的但我真的不想去找和打字给你顶下应该有人给你回复的

while的：DimnAsInteger,aAsInteger,totAsDoubletot=1a=1n=InputBox("请输入N的值")Whilea再问：...调试不出来再答：把我的代码放在某个

正常输入n且n不等于0

chari;while(1){printf("inputyorn");scanf("%d",&x);if(i=='y'||i=='n'){break;}else{printf("Error,input

classProgram{staticvoidMain(string[]args){intM;stringS=Console.ReadLine();boolTorF=int.TryParse(S,ou

doublee;inti,k;i=1;k=1;e=1;while(1.0/i*100000

楼上的把循环条件中的n换成n/2,效率更好#includeusingnamespacestd;intmain(intargc,char*argv[]){intn,i,t,sum;cin>>n;i=2;

#includevoidmain(){intn,min,a[100],i,t,e;while(scanf("%d",&n)!=EOF&&n

#includeintfact(intn){intt=1;for(inti=1;i

importjava.util.Scanner;publicclassTest{publicstaticvoidmain(String[]args){\x05\x05Scannerinput=newS

用while的,其它的dowhile的基本相同PrivateSubCommand1_Click()DimnAsIntegern=InputBox("请输入")f=1i=2s=0Wh

datasegmentmsgdb'inputN:$'dataendscodesegmentassumecs:code,ds:datamainprocnearstart:movax,datamovds,