hdu 5585 水题 Numbers 2015.11.28 bestcoder 1001

题意：给一个长度最多为30的数，问是否能被2,3或者5整除

思路：按照字符串存储，判断最后一个数是不是2的倍数、5，以及判断各个位相加是否能被3整除

Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 130    Accepted Submission(s): 86

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

 

Source

BestCoder Round #64 (div.2)

<pre name="code" class="cpp">#include <iostream>#include <stdio.h>#include <math.h>#include <stdlib.h>#include <string>#include <string.h>#include <algorithm>#include <vector>#include <queue>#include <iomanip>#include <time.h>#include <set>#include <map>#include <stack>using namespace std;typedef long long LL;const int INF=0x7fffffff;const int MAX_N=10000;//long long n;char A[100];int main(){    while(scanf("%s",&A)!=EOF){        if(A[strlen(A)-1]=='0'||A[strlen(A)-1]=='5'){            printf("YES\n");            continue;        }        if((A[strlen(A)-1]-'0')%2==0){            printf("YES\n");            continue;        }        int ans=0;        for(int i=0;i<strlen(A);i++){            ans+=A[i]-'0';        }        if(ans%3==0){            printf("YES\n");            continue;        }        else printf("NO\n");    }    return 0;}