【Lightoj】1078 - 多少个可以整除？（同余定理）

Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu

Submit Status Practice 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 andn will not be divisible by2 or5) 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

用了同余定理，这类题写多了就会了。

代码如下：

#include <stdio.h>
int main()
{
	int u;		//总次数 
	int y;		//余数 
	int c;				//次数 
	int a,b;
	scanf ("%d",&u);
	for (int i=1;i<=u;i++)
	{
		scanf ("%d %d",&a,&b);		//被除数	基本数字 
		printf ("Case %d: ",i);
		y=b%a;
		c=1;
		while (y)
		{
			y=(y*10+b)%a;
			//printf ("%d\n",y) ;
			c++;
		}
		printf ("%d\n",c);
	}
	return 0;
}