A water problem

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.

InputThere 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.OutputFor 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

--------------------------------------------------------------

思路分析：这道题看起来很简单，但是自己被里面的坑卡了好久。关键是这句：the length of N is up to 10000000.

N的值是很大的，即使longlong类型也存不下，所以就要自己写一个字符数组的除法。

#include <iostream>
#include<stdio.h>
#include<string>
/* run this program using the console pauser or add your own getch, system("pause") or input loop */
using namespace std;
bool div(string a,int b)//判断是否能被除数整除
{
	int sum=a[0]-'0';
	int size=a.size();
	int k=0;
	int yu=0;
	while(k<size)
	{
		//如果sum大于b，sum=sum%b*10+a[k+1]-'0' ;
		//如果sum小于b，sum=sum+10+ a[k+1]-'0';
		if(sum<b)
		{
			yu=sum%b;
			if(k+1<size)
			sum=sum*10+a[k+1]-'0';
			else
			break;
			
		}
		else
		{
			yu=sum%b;
		    if(k+1<size)
			sum=sum%b*10+a[k+1]-'0';
			else
			break;
		}
		k++;
	}

	if(yu==0)
	return true;
	else
	return false;
}
int main(int argc, char** argv) {
	string day;
	int i=1;
	while(cin>>day)
	{
		   if(div(day,73)&&div(day,137))
		   {
		   
			printf("Case #%d: YES\n",i);
			i++;
		}
		else{
	
	
				printf("Case #%d: NO\n",i);
			i++;
		}
		
	}
	return 0;
}