LightOJ 1078 - Integer Divisibility【同余】

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

解题思路

         题目的大意就是看看有多少个M可以让N整除。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
	int t;
	scanf("%d",&t);
	int yy=0;
	while(t--)
	{
		yy++;
		long long a,b;
		scanf("%lld%lld",&a,&b);
		long long map=b%a;
		long long ans=1;
		while(map)
		{
			map=(map*10+b)%a;
			ans++;
		}
		printf("Case %d: %lld\n",yy,ans);
	}
	return 0;
}