Integer Divisibility（同余）

Integer Divisibility

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

3 1

7 3

9901 1

Sample Output
Case 1: 3

Case 2: 6

Case 3: 12

因为该数据属于大数，所以不能直接除，用字符串存大数，然后每位转换成十进制的数，边转换边求余，直到余数为0.
ac代码：

#include<iostream>
#include<algorithm>

using namespace std;

int main(){
	int n,m,t;
	long long ans;
	cin>>t;
	int sum=0;
	while(t--){
	cin>>n>>m;
		ans=m;
		int s=0;
		while(ans%n!=0){
			ans=ans*10+m;
			//cout<<ans<<endl;
			ans%=n;
			s++;
		}
		cout<<"Case "<<++sum<<": "<<++s<<endl;
	}
}