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A water problem

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 60    Accepted Submission(s): 37


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 after Big Bang, you need to answer whether it is the first day in a year about the two planets.
Input There are several test cases(about 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
Source

題目大意：

就是給你一個大數，讓你判斷是不是 73 和 137 的公倍數。

解題思路：

大數 其實這個題一分析一下，因為 73 和 137 是素數，所以就是對 73∗137 取模，然後跑一個循

環就行了，坑的是 cin 會超時，然後 改成 scanf 還超時，等了一會兒又交一遍竟然過了，AC

了！！！

My Code：

/**
2016 - 08 - 14 下午
Author: ITAK

Motto:

今日的我要超越昨日的我，明日的我要勝過今日的我，
以創作出更好的程式碼為目標，不斷地超越自己。
**/
 


#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int INF = 1e9+5;
const int MAXN = 1e7+5;
const int MOD = 10001;
const double eps = 1e-7;
const double PI = acos(-1);
using namespace std;
char str[MAXN];
int main()
{

    int cas = 1;
    while(scanf("%s",str)!=-1)
    {
        int len = strlen(str);
        int sum = 0;
        for(int i=0; i<len; i++)
        {
            sum = sum*10;
            sum = (sum+str[i]-'0');
            sum = sum%MOD;
        }
        if(sum == 0)
            printf("Case #%d: YES\n",cas++);
        else
            printf("Case #%d: NO\n",cas++);
    }
    return 0;
}