題目：

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

代碼：

#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; }

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