【Light oj 1078】

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

同于定理 ： 今天体会到了它的强大之处 ：

#include<stdio.h>  #include<string.h>  int main()  {      int n,m,t,test=1;       scanf("%d",&t);      while(t--)      {          int ans=1;        scanf("%d%d",&n,&m);          int a=m%n;            while(a)          {              a=(a*10%n+m%n)%n;              ans++;          }          printf("Case %d: %d\n",test++,ans);      }      return 0;  }