Light OJ 1078:Integer Divisibility【数学】

最新推荐文章于 2020-07-31 09:19:45 发布

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最新推荐文章于 2020-07-31 09:19:45 发布

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分类专栏：＝＝＝＝C&＆C++＝＝＝＝文章标签：数学同余定理

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本文提供了一个算法解决方案，用于找出给定整数n和唯一数字m之间的最小乘数，该乘数由m组成且能被n整除。详细解释了如何通过循环和取模操作实现这一目标。

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1078 - Integer Divisibility

Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu

Submit Status

Description

If an integer is not divisible by 2 or 5, some multiple of that number in decimal notation is a sequence of only a digit. Now you are given the number and the only allowable digit, you should report the number of digits of such multiple.

For example you have to find a multiple of 3 which contains only 1's. Then the result is 3 because is 111 (3-digit) divisible by 3. Similarly if you are finding some multiple of 7 which contains only 3's then, the result is 6, because 333333 is divisible by 7.

Input

Input starts with an integer T (≤ 300), denoting the number of test cases.

Each case will contain two integers n (0 < n ≤ 10⁶ and n will not be divisible by 2 or 5) and the allowable digit(1 ≤ digit ≤ 9).

Output

For each case, print the case number and the number of digits of such multiple. If several solutions are there; report the minimum one.

Sample Input

3 1

7 3

9901 1

Sample Output

Case 1: 3

Case 2: 6

Case 3: 12

题意：给你两个整数，n和m,问有最少有几个m组成的数能被n整除。。。。。不用同余会超时。。。。

AC-code：

#include<cstdio>
int yu(long long n,int a)
{
	int i;
	long long ans;
	ans=a%n;
	if(a%n==0)
		return 1;
	for(i=2;;i++)
	{
		ans=(ans*10+a)%n;
		if(ans%n==0)
			return i;
	}
}
int main()
{
	int T,i,m;
	long long n;
	scanf("%d",&T);
	for(i=1;i<=T;i++)
	{
		scanf("%lld%d",&n,&m);
		printf("Case %d: %d\n",i,yu(n,m));
	}
	return 0;
}