POJ 2551 Ones

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

3 1

7 3

9901 1

Sample Output

Case 1: 3

Case 2: 6

Case 3: 12

用同余定理做，直接暴力会超时。能整除就是余数为零。。

代码：

#include<stdio.h>int main(){int t,n,m;scanf("%d",&t);int k=1;while(t--){scanf("%d%d",&n,&m);int sum=0,i;for(i=1;;i++){sum=(sum*10+m)%n;if(sum==0)  break;}printf("Case %d: %lld\n",k++,i);}return 0;}