编写函数,判断一个数a能否同时被7和11整除

#includevoidmain(){intn;printf("input:\n");scanf("%d",&n);if(n%3==0&&n%5==0)printf("yes\n");elseprin

多的我就不说了,函数如下（你只需把下面的代码贴在intmain()之前就行了,然后在开头加一个头文件：#include）：boolisPrime(intx){if(x==1)return0;for(i

//也不说用什么语言,我用C语言来实现!#include#includeintprime(intm){inti,n;if(m==1)return0;n=(int)sqrt((double)m);for

if(number%3==0&&number%5==0){System.out.println("number"+能被三和五整除);}else{System.out.println("number"+

#includeintmain(intargc,char*argv[]){intnum;printf("输入一整数:");scanf("%d",&num);if(num%3==0&&num%5==0&

PrivateFunctionjo(ByValnAsInteger)n=InputBox("请输入一个数字")IfnMod20ThenMsgBoxn&"是奇数"ElseMsgBoxn&"是偶数"End

#includemain(){intn;printf("Pleaseenteranumber:");scanf("%d",&n);if(n%2==0)printf("Thenumber%dis偶数\n

DimnAsIntegerPrivateSubForm_Click()Ifn/2=0ThenPrint"n是偶数"ElsePrint"n是奇数"EndIfEndSubPrivateSubForm_Lo

#includemain(){intm,b,s,g;for(m=100;m

'判断一个数m是否是完数,是则返回True,否则返回FalseFunctionTestWanShu(ByValmasLong)AsBooleanDimSumAsLongDimiAsLongIfm

100x+10y+z=x³+y³+z³

if(x%3==0)y=3;if(x%5==0)y=5;if(x%7==0)y=7;if(x%7==0&&x%5==0&&x%3==0)y=15;if(x%7==0&&x%5==0)y=12;if(x

/*打印所有水仙花数*/#include"stdio.h"#includevoidmain(){inti,j,k,sum;for(i=1;i

3和5的最小公倍数为15所以只要一个数能够被15整除,就能同时被3和5整除

我用C语言编一下啊.其他语言类似!#includevoidsort(intn){if(n%2==0)printf("thenumberisdouble\n");elseprintf("thenumbe

oolfun(inti)//判断i{if(i%3==0)returntrue;returnfalse;}

inta,b;scanf("%d%d",&a,&b);while(a{intt=a;a=b;b=t;}if(a%b==0)printf("a能被b整除");elsepr

“在列表框显示1000以内的完数”为什么要放在主调程序里呢?在IsWs函数中已经把完数的因子算出来了,却只能返回一个true,岂不在主调程序里还要算一遍因子才能显示成6=1+2+3这种形式.你又没说要

#includevoidmain(){intn;printf("pleaseinputthedata:\n");scanf("%d",&n);if(n%3==0&&n%5==0)printf("the

intsum(intn){intx;If(x%3==0&&x%5==0)sum(n)==1;elsesum(n)==0;}这样的代码里边sum(n)成了递归了,当然死循环了#include"stdio