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


[align=left]Problem Description[/align]
There is a number N.You should output "YES" if N is a multiple of 2, 3 or 5,otherwise output "NO".
 

[align=left]Input[/align]
There are multiple test cases, no more than 1000 cases.

For each case,the line contains a integer N.(0<N<1030)
 

[align=left]Output[/align]
For each test case,output the answer in a line.
 

[align=left]Sample Input[/align]

2
3
5
7

 

[align=left]Sample Output[/align]

YES
YES
YES
NO

 

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;
}