LightOJ

Integer Divisibility

Time Limit: 2000MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

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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

Source

Problem Setter: Jane Alam Jan

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#include <cstdio>
int main()
{
    int t;
    scanf("%d", &t);
    long long Q=1; 
    while(t--)
    {
        int div, dig;
        scanf("%d%d", &div, &dig);
        int sum=dig;
        long long times=1;
        if(dig%div==0)
            times=0;
        while(sum)
        {
            sum=(sum*10+dig)%div;
            times++;
        }  
        printf("Case %lld: %lld
", Q++, times);
    }
    return 0;
}