lightoj 1078 - Integer Divisibility 【同余定理】

1078 - Integer Divisibility

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Time Limit: 2 second(s)Memory Limit: 32 MB

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

Output for Sample Input

3

3 1

7 3

9901 1

Case 1: 3

Case 2: 6

Case 3: 12

 

同余定理：（a+b)%n==(a%n+b%n)%n；

代码：

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