A water problem

100010333Sample Output
Case #1: YESCase #2: YES
Case #3: NO
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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;}