HDU 5585(数论)

最新推荐文章于 2018-12-15 20:11:00 发布

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此博客介绍了如何通过简单的算法判断一个给定的整数是否为2、3或5的倍数，并提供了实现该功能的代码示例。

Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

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

Sample Output

  
   YES
YES
YES
NO

Source

BestCoder Round #64 (div.2)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <map>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <algorithm>
#define LL long long
#define INF 0x3f3f3f3f
#define RR freopen("in.txt","r",stdin)
#define WW freopen("out.txt","w",stdout)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

using namespace std;

int main()
{
    char s[1010];
    while(cin>>s)
    {
        int sum = 0;
        int len = strlen(s);
        for(int i=0; i<len; i++)
            sum += (s[i] - '0');
        if(sum % 3 == 0 || (s[len-1] - '0')%2==0 || s[len-1] - '0'==5)
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
    return 0;
}