输入整数若是素数pascaL程序

/>简单枚举肯定会超时这道题有两种思路：(1)用筛法求出1..1e8范围内的素数,然后判断每个素数是否是回文数.(2)生成1..1e8范围内的回文数,然后判断它是否是素数.思路1的复杂度是O(n),思

varn:longint;beginreadln(n);ifnmod7=0thenwriteln('yes'){除以7取余,是否等于0,等于则输出yes,否则输出no}elsewriteln('no'

vari,k,n1,n2:integer;beginn1:=0;n2:=0;fori:=1to20dobeginreadln(k);ifk>0thenn1:=n1+1elseifk

programPrime;vari,j,k,n:Longint;l:array[1..60000]ofBoolean;begink:=0;FillChar(l,Sizeof(l),True);Read

varn,s:longint;beginread(n);s:=0;whilen0dobegins:=s+nmod10;n:=ndiv10;end;writeln(s);end.自创!

programPrime;varsieve:array[1..32767]ofByte;n,m,sqrt_m,sum,i,j:Word;beginReadln(n,m);sieve[1]:=0;for

#include<iostream>using namespace std;bool isSushu(int);//检查是否是素数void main

vara,n,i:longint;beginfori:=1to10dobeginread(a);ifa>0thenn:=n+1;end;write(n);end.

感觉像是ACM的题.N

programsushu(input,output);vari,n,t:integer;beginread(n);fori:=2ton-1doifnmodi=0thent:=1;ift=1thenwr

vari,j,n:longint;z:boolean;beginreadln(n);writeln(2);fori:=3tondobeginz:=true;forj:=2toi-1doifimodj=

核心代码readln(n);whilen0dobeginifn=1thenwriteln(0)elsebeginforj:=2tondoifprime(j)theninc(ans);writeln(a

很基础的,楼主要好好学习啊#include#includevoidmain(){inti;scanf("%d",&i);for(intj=1;j

PrivateSubForm_Load()ShowDima,iAsIntegera=Val(InputBox("请输入一个整数"))Fori=2ToInt(Sqr(a))IfaModi=0ThenPr

http://zhidao.baidu.com/question/59371246.html

vari,n:longint;b:boolean;beginreadln(n);b:=true;ifn

vari,j,n,m:longint;procedureprint(c:char;a:integer);vari:longint;beginfori:=1toadowrite(c);end;begin

vars,n,a:longint;begins:=0;whilea0dobeginn:=amod10;s:=s+n;a:=adiv10;end;writeln(s);end.

集合的上限是255个,你那样肯定不能运行

#includemain(){intnum,yushu;printf("Pleaseenterthenumber\n");scanf("%d",&num);yushu=num%2;if(yushu==