F - Large Division LightOJ - 1214-优快云博客

本文介绍了一种用于检查两个大整数之间是否具备整除性的高效算法。输入包括多个测试用例，每个测试用例包含两个整数a和b，算法判断a是否能被b整除，并输出相应的结果。该算法通过将大整数a逐步转换为较小的数值形式来避免溢出，适用于非常大的整数。

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iven 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

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

#include<cstdio>
#include<iostream>
#include<cstring>

using namespace std;

typedef long long ll;

int main()
{
	int T, kase = 1;;
	cin >> T;
	while (T --)
	{
		char n[250];
		ll digit;
		cin >> n >> digit;
		ll i = 0, ans = 0;
		int len = strlen(n);
		for (i = 0; i < len; i++)
		{
			if (n[i] == '-') continue;
			ans = (ans * 10 + n[i] - '0') % digit;
		}
		cout << "Case " << kase++ << ":" << " ";
		printf ("%s\n", ans ? "not divisible" : "divisible");
	}
	return 0;
}