LightOJ 1078 Integer Divisibility （同余定理）

题目：

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 （10的6次方）and n will not be divisible by 2&nbs***bsp;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

3 1

7 3

9901 1

Sample Output

Case 1: 3

Case 2: 6

Case 3: 12

代码：

#include <iostream>
using namespace std;
int main()
{
	long long t,n,i,ans,m,tem;
	cin>>t;
	ans=1;
	while(t--)
	{
		cin>>m>>n;
		cout<<"Case "<<ans++<<": ";
		if(n%m==0)
		{
			cout<<"1"<<endl;
			continue;
		}
		tem=n;
		for(i=1;n!=0;i++)
		{
			n=(n*10+tem)%m;
		}
		cout<<i<<endl;
	}
	return 0;
}