LIGHTOJ-1078 - Integer Divisibility

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

求几个一位数能被第一个数整除，模拟两数相除的过程！

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
	int i,j,k,a,b,m,n,num1,num2,cot=1;;
	scanf("%d",&n);
	
	while(n--)
	{
		num1=0;
		num2=0;
		scanf("%d%d",&a,&b);
		int x=a,y=b;
		while(x>y)
		{
			y=y*10+b;
			num2++;
		}
		
		while(y%x)
		{
			y=(y%x)*10+b;
			num1++;
		}
		printf("Case %d: %d\n",cot++,num1+num2+1);
	}
	return 0;
}