1078 - Integer Divisibility

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

题意;求有多少个d可以使n被整除；

思路：同余定理 (a+b)%n=(a%n+b%n)%n。

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int dp[11000];int main(){int t,d,n,i,j,h=1,m,k;scanf("%d",&t);while(t--){scanf("%d%d",&n,&d);        k=d%n;//先求出模，不然会wa；m=1;while(k) {k=(k*10+d)%n;m++;}printf("Case %d: %d\n",h++,m);}return 0;}