light oj 1078 - Integer Divisibility

最新推荐文章于 2023-03-07 20:06:57 发布

Never_Say-Never

最新推荐文章于 2023-03-07 20:06:57 发布

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分类专栏：同余定理

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探讨了如何找出一个不被2或5整除的数的倍数，该倍数仅由指定数字组成，并确定其位数。通过迭代计算解决这一问题。

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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: 3 111能被3整除.

Case 2: 6 333333能被7整除

Case 3: 12 111111111111能被9901整除

此题用同余定理

(a*b)%c=(a%c*b%c)%c;

(a+b)%c=(a%c+b%c)%c;

代码如下：

#include<stdio.h>
#include<algorithm>
#include<stdlib.h>
#include<math.h>
using namespace std;
int main()
{
	int t,k=0;
	scanf("%d",&t);
	while(t--)
	{
	__int64 n,m,flag;
	scanf("%I64d%I64d",&n,&m);
	flag=m%n;
	int ans=1;
	while(flag)
	{
		flag=(flag*10+m)%n;//其实这跟我们平常手算除法的思想差不多一样
		ans++;
	}

	k++;
	printf("Case %d: ",k);	
	printf("%d\n",ans);
	}
}