hdu 5585 Numbers

Problem Description

There is a number N.You should output "YES" if N is a multiple of 2, 3 or 5,otherwise output "NO".
 

 

 

Input

There are multiple test cases, no more than 1000 cases.
For each case,the line contains a integer N.(0<N<1030)

 

 

 

Output

For each test case,output the answer in a line.

 

 

 

Sample Input

2
3
5
7

 

 

Sample Output

YES
YES
YES
NO

 

 

 

Source

BestCoder Round #64 (div.2)

 

读入%s字符串

2和5看大数的最后一位，3就把所有的位加起来看看是不是3的倍数

 

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<bitset>
#include<map>
#include<vector>
#include<stdlib.h>
#include <stack>
using namespace std;
#define PI acos(-1.0)
#define max(a,b) (a) > (b) ? (a) : (b)
#define min(a,b) (a) < (b) ? (a) : (b)
#define ll long long
#define eps 1e-10
#define MOD 1000000007
#define N 1000000
#define inf 1e12
char s[36];
int main()
{


     while(scanf("%s",s)!=EOF){
         int len=strlen(s);
         //printf("len=%d\n",len);
         int sum=0;
         for(int i=0;i<len;i++){
            sum=sum+s[i]-'0';
         }
         //printf("sum=%d\n",sum);
         if(sum%3==0){
            printf("YES\n");
            continue;
         }
         int ans=s[len-1]-'0';
         //printf("ans=%d\n",ans);
         if((ans==0) || (ans==2) || (ans==4) || (ans==6) || (ans==8) || (ans==5)){
            printf("YES\n");
            continue;
         }
         printf("NO\n");
     }


    return 0;
}