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 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
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;
}