LightOJ 1214 - Large Division【同余定理】

1214 - Large Division

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Time Limit: 1 second(s)

Memory Limit: 32 MB

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	Output for Sample Input
6 101 101 0 67 -101 101 7678123668327637674887634 101 11010000000000000000 256 -202202202202000202202202 -101	Case 1: divisible Case 2: divisible Case 3: divisible Case 4: not divisible Case 5: divisible Case 6: divisible

 

PROBLEM SETTER: JANE ALAM JAN

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char map[10000];
int main()
{
	int t;
	int uu=0;
	scanf("%d",&t);
	while(t--)
	{
		uu++;
		int nn;
		scanf("%s",map);
		int len=strlen(map);
		scanf("%d",&nn);
		long long ans=0;
		if(map[0]=='-')
		{
			for(int i=1;i<len;i++)
			{
				ans=ans*10+(map[i]-'0');
				ans=ans%nn;
			}
		}
		else
		{
			for(int i=0;i<len;i++)
			{
				ans=ans*10+(map[i]-'0');
				ans=ans%nn;
			}
		}
		printf("Case %d: ",uu);
		if(ans%nn==0)
		{
			printf("divisible\n");
		}
		else
		printf("not divisible\n");
	}
	return 0;
}