输入一个数n判断能否被3和5整除流程图

#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

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&

我把SCANF里的中文去掉就正常了.

第一个问题:#include#includevoidQuickSort(int*pData,intleft,intright);voidmain(){inta[]={20,11,0,5,8,9,5,3

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

PrivateSubCommand1_Click()Dima,b,cAsIntegerDimmaxRandomizea=Int(900*Rnd+100)b=Int(900*Rnd+100)c=Int(

#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(){\x09longa;\x09printf("inputanumber:\n");\x09scanf("%d",&a);\x09if(a%3==0&&a%5==0&

voidmain(){intn;printf("请输入一个整数:");scanf("%d",&n);if(n%3==0&&n%5==0&&n%7==0)printf("\n此数能被3,5,7整除\n"

各位数相加和能被3整除,那么这个数就能被3整除,如258各位数相加等于15,15能被3整除,那么258也能被3整除.能够被5整除的只要各位数是0或5就行啦.15和10都能被5整除.另外,0除以任何数都

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

#includeintmain(){inta;intb=0,c=0,d=0;scanf("%d",&a);if(a%3==0)b=1;if(a%5==0)c=1;if(a%7==0)d=1;if(b*

intnumber;scanf("%d",&number);if(number%3==0&&number%5==0){printf("YES");}else{printf("NO");}