编写一个程序,找出2至99之间的全部同构数(也称自守数).

@echo offsetlocal enabledelayedexpansionfor /l %%i in (100 1 

30n+1316191121151181211241271301

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

#include"stdio.h"voidmain(){inta,b,i;intsum=0;printf("Inputstartyear:");scanf("%d",&a);printf("

intcount=0;for(inti=1;i

1.思路：用一个字符串数组储存输入的数据，并分别输出。程序：Vara:array[1..3]ofstring;i:integer;BeginFori:=1to3doReadln(a[i]);Fori:

#includemain(){inti,k;for(i=1000;i

SubMain()MsgBoxFunc(100,2,5)'按照要求写的例子EndSubRem参数(最大值,被整除的数的数组)FunctionFunc(n&,ParamArrayx())Dimi&,b&

java版publicclassSushu{\x05/**\x05*@paramargs\x05*/\x05publicstaticvoidmain(String[]args){\x05\x05//T

privatefunctionppd(x1,y1,x2,y2)aslongppd=sqr((x2-x1)^2+(y2-y1)^2)endfunction(3,4)和(5,2)的距离d=ppd(3,4,

programa4;vara,b,c:integer;beginreadln(a,b,c);if(((a>0)and(b>0)and(c>0))and((a+b>c)and(a+c>b)and(b+c

下面程序是根据原理写的,已经调试成功.#include#includeintmain(){inti,flag[101];for(i=2;i

#includevoidsort(intc[],intn){inti,j,buf;for(i=0;ifor(j=i+1;jif(c[i]>c[j]){buf=c[i];c[i]=c[j];c[j]=b

使用VS的Windows应用程序代码如下：usingSystem;usingSystem.Collections.Generic;usingSystem.ComponentModel;usingSys

varn,i,min,max,maxi,mini,s:integer;x:array[1..100]ofinteger;ans:real;beginreadln(n);fori:=1tondoread

sum=0;num=num1;do{num++;if(num|2)sum+=num;}while(num再问：似乎不对啊！再答：intnum1;intnum2;cin>>num1;cin>>num2;

pt = {2, 2};ContourPlot[ Sqrt[(x - pt[[1]])^2 + (y -&nb

/>vart;varl=prompt('请输入边长',3.5);t=3.1415926*(l/2.0)*(l/2.0);document.write("边长:"+l);document.write("

把奇数的一起补上了#include"stdio.h"intmain(){inta=0,b=0;inti=0;for(i=0;i

datasegmentvaldb100dup(0)codesegmentassumecs:codeds:datastart:movax,datamovds,axleasi,valmovbx,1movc