light oj 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    111能被3整除.

Case 2: 6     333333能被7整除

Case 3: 12   111111111111能被9901整除

此题用同余定理

(a*b)%c=(a%c*b%c)%c;

(a+b)%c=(a%c+b%c)%c;

代码如下：

#include<stdio.h>#include<algorithm>#include<stdlib.h>#include<math.h>using namespace std;int main(){int t,k=0;scanf("%d",&t);while(t--){__int64 n,m,flag;scanf("%I64d%I64d",&n,&m);flag=m%n;int ans=1;while(flag){flag=(flag*10+m)%n;//其实这跟我们平常手算除法的思想差不多一样ans++;}k++;printf("Case %d: ",k);printf("%d\n",ans);}}