判断一个数能否同时被3和5整除ns流程图

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

#includevoidmain(){intn;scanf("%d",&n);if(n%3==0&&n%5==0&&n%7==0)printf("3\n");elseif(n%3==0&&n%5==0

被3整除的数,各个数字和为3倍数.被5整除的数,末位数字为0或5.==>同时被3和5整除的数,各个数字和为3倍数且末位数字为0或5.或者更简单些,能被15整除.算法：1、判断是否为3倍数————把n拆

PrivateSubcommand1_click()Dimint1AsIntegerint1=InputBox("输入一个整数")Ifint1Mod15=0ThenMsgBox"此数能同时被3和5整除

inta,b;if((a%b==0)||(b%a==0))printf("can")；elseprintf("can't");不处理边界问题哦

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&

#includeintmain(void){intn,a;scanf("%d",&n);a=0;if(n%3==0)++a;if(n%5==0)++a;if(n%7==0)++a;if(a==0)pr

这需要流程图啊.那真不会写1.intsum=0;chari;for(i=1;i

#include"stdio.h"main(){intx;printf("请输入一个数字:");scanf("%d",&x);if(x%3==0)printf("该数字能被3整除.");if(x%5=

#includeintmain(){inta;intb=0,c=0,d=0;scanf("%d",&a);if(a%5==0)c=1;if(a%7==0)d=1;elseif(c==1&&d==1)p

#include"stdafx.h"voidmain(){intnum;printf("输入一个数:");scanf("%d",&num);if(num%3==0){if(num%5==0){prin

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

以下是关键代码intnum;//这个是要你判断的数if(num%3==0&&num%5==0)//判断能否被整除printf("能被3和5整除");

#includevoidmain(){intnumber;intre[3];inti=0;scanf("%d",&number);if(number%3==0){re[i]=3;i++}if(numb

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

我觉得这才是楼主的意思吧,首先判断能否同时被5和6整除,如果不能再单独判断是否能被5或者6整除.importjava.util.Scanner;publicclassTest{publicstatic

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

直接将代码复制一下运行测试即可#include"stdio.h"#include"conio.h"main(){inti;printf("请输入你要判断的数(输入0结束程序):\n");while(1