LightOJ 1214 - Large Division 【同余定理】

LightOJ 1214 - Large Division 

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit  Status

Description

Given two integers, a and b, you should check whether a is divisible by b or not. We know that an integer a is divisible by an integer b if and only if there exists an integer c such that a = b * c.

Input

Input starts with an integer T (≤ 525), denoting the number of test cases.

Each case starts with a line containing two integers a (-10²⁰⁰ ≤ a ≤ 10²⁰⁰) and b (|b| > 0, b fits into a 32 bit signed integer). Numbers will not contain leading zeroes.

Output

For each case, print the case number first. Then print 'divisible' if a is divisible by b. Otherwise print 'not divisible'.

Sample Input

6

101 101

0 67

-101 101

7678123668327637674887634 101

11010000000000000000 256

-202202202202000202202202 -101

Sample Output

Case 1: divisible

Case 2: divisible

Case 3: divisible

Case 4: not divisible

Case 5: divisible

Case 6: divisible

Ac-code:

#include<cstdio>
#include<cstring>
#include<cmath>
int main()
{
	int T,j,b,i,len,ant,flag;
	char s[205];
	long long ans;
	scanf("%d",&T);
	for(j=1;j<=T;j++)
	{
		flag=0;
		scanf("%s%d",&s,&b);
		if(s[0]=='-')
		{
			ans=s[1]-'0';
			ant=2;
		 } 
		 else
		 {
		 	ans=s[0]-'0';
		 	ant=1;
		 }
		 if(b<0)
		 	b=-b;
		len=strlen(s);
		ans=ans%b;
		for(i=ant;i<len;i++)
		{
			ans=(ans*10%b+(s[i]-'0')%b)%b;
		 } 
		if(ans==0)
			printf("Case %d: divisible\n",j);
		else 
			printf("Case %d: not divisible\n",j);
	}
	return 0;
}