hdoj 5585 Numbers 【水题】

Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 53    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：给定一个大数，问它是否为2、3或5的倍数。

AC代码：

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <vector>#define INF 0x3f3f3f#define eps 1e-8#define MAXN (5000+10)#define MAXM (100000)#define Ri(a) scanf("%d", &a)#define Rl(a) scanf("%lld", &a)#define Rf(a) scanf("%lf", &a)#define Rs(a) scanf("%s", a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%.2lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define W(a) while(a--)#define CLR(a, b) memset(a, (b), sizeof(a))#define MOD 1000000007#define LL long long#define lson o<<1, l, mid#define rson o<<1|1, mid+1, r#define ll o<<1#define rr o<<1|1using namespace std;int main(){    char str[40];    while(scanf("%s", str) != EOF)    {        int len = strlen(str);        int num;        if(len == 1)        {            num = str[0] - '0';            if(num % 3 == 0 || num % 2 == 0 || num % 5 == 0)                printf("YES\n");            else                printf("NO\n");            continue;        }        bool flag = false;        if(str[len-1] == '2' || str[len-1] == '0' || str[len-1] == '5' || str[len-1] == '4' || str[len-1] == '6' || str[len-1] == '8')            flag = true;        num = 0;        for(int i = 0; i < len; i++)            num += str[i] - '0';        if(num % 3 == 0)            flag = true;        if(flag)            printf("YES\n");        else            printf("NO\n");    }    return 0;}