LightOJ 1078

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<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#define N 101000
int main()
{

int i, n, T, h, k, sum, count, t=1;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&k);
h=(int)log10(n)+1;
sum=k;
count=h;
for(i=1;i<h;i++)
{
sum=sum*10+k;
}
while(sum%n!=0)
{
sum=sum%n;
sum=sum*10+k;
count++;
}
printf("Case %d: %d\n",t++,count);

}
return 0;
}