A water problem

2019-11-06 06:28:15

字體：大中小

供稿：網(wǎng)友

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 Nafter 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

100010333Sample OutputCase #1: YESCase #2: YESCase #3: NO
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思路分析：這道題看起來很簡單，但是自己被里面的坑卡了好久。關(guān)鍵是這句：the length of N is up to 10000000.
N的值是很大的，即使longlong類型也存不下，所以就要自己寫一個字符數(shù)組的除法。
#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)//判斷是否能被除數(shù)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;}