hdu 5585 水题 Numbers 2015.11.28 bestcoder 1001

最新推荐文章于 2021-02-27 16:50:52 发布

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本文介绍了一种快速判断大数（长度最多为30）是否能被2、3或5整除的方法。通过检查个位数及各位数字之和来实现高效判断。

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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

Sample Output

  
   YES
YES
YES
NO

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;
}