【 LightOJ 1078 Integer Divisibility + 同余定理 】

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

#include<cstdio>int main(){    int T,nl = 0,a,b,ans,cut;    scanf("%d",&T);    while(T--)    {        scanf("%d %d",&a,&b);        cut = 1;        ans = b % a;        while(ans)        {            ans = (ans * 10 + b) % a;            cut++;        }        printf("Case %d: %d\n",++nl,cut);    }    return 0;}