HDOJ 5585 Numbers (水)

Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 52    Accepted Submission(s): 41

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

2357

 

Sample Output

YESYESYESNO

 

题意：输入一个数，判断是不是2，3，5的倍数
思路：转换为字符串形式，找出符合整除的条件：
2的倍数：末尾数字为2的倍数
3的倍数：所有位数加起来整除3
5的倍数：最后一位为5的倍数或者最后一位为0（因为n>0）


ac代码：

#include<stdio.h>#include<string.h>#include<math.h>#include<iostream>#include<algorithm>#define fab(a) (a)>0?(a):(-a)#define LL long long#define MAXN 3000#define INF 0xfffffff using namespace std;char s[MAXN];int main(){int i;    while(scanf("%s",s)!=EOF)    {    int len=strlen(s);    int num=s[len-1]-'0';    int sum=0;    for(i=0;i<len;i++)    sum+=s[i]-'0';    if(num%2==0||sum%3==0||num%5==0||num==0)    printf("YES\n");    else    printf("NO\n");}    return 0;}