A water problem


Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 412    Accepted Submission(s): 218


Problem Description


Two planets named Haha and Xixi in the universe and they were created with the universe beginning.

There is  73 days in Xixi a year and  137 days in Haha a year. 

Now you know the days  N


 



Input


5 huge test cases).

For each test, we have a line with an only integer  N(0≤N), the length of  N is up to  10000000.


 



Output


For the i-th test case, output Case #i: , then output "YES" or "NO" for the answer.


 



Sample Input


10001 0 333


 



Sample Output


Case #1: YES Case #2: YES Case #3: NO


 



Author


UESTC


 




题意：判断输入的数是否可以同时整除73与137，也就是能不能整除73与137的最小公倍数。



因为输入的数的长度是10000000以内，我们只能采用字符串来存储它，从ans初始化为0，最低位开始ans=(ans+s[i]-'0')%b; 计算到最后，如果ans 等于0，可以整除，否则，不可以整除。






AC代码：


#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>

using namespace std;

char s[11000000];
int ans,b=73*137;
int main()
{
    for(int i=1; gets(s); i++)
    {
        ans=0;
        int l=strlen(s);
        for(int j=0; j<l; j++)
            ans=(ans*10+(s[j]-'0'))%b;
        if(!ans)printf("Case #%d: YES\n",i);
        else printf("Case #%d: NO\n",i);
    }
    return 0;
}