light oj 1078 - Integer Divisibility【同余定理】

最新推荐文章于 2018-08-25 08:42:28 发布

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最新推荐文章于 2018-08-25 08:42:28 发布

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本文链接：https://blog.youkuaiyun.com/liu6886/article/details/50164263

本文探讨了一个特定的整除问题，即寻找一个整数的倍数，该倍数仅由指定的一个数字组成，并且是最短长度。文章描述了输入输出格式，并提供了一个C++示例代码来解决这个问题。

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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 andn 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 1

7 3

9901 1

Case 1: 3

Case 2: 6

Case 3: 12

#include<cstdio>
#include<cstring>
int main(){
	int T,t;
	scanf("%d", &T);
	for(t = 1; t <= T; t++){
		int n,d,a;
		scanf("%d%d", &n, &d);
		a = d%n;
		int cnt = 1;
		while(a)
		{
			a = (a*10+d)%n;//放在while位置  超时。。 
			cnt++;
		}
		printf("Case %d: %d\n", t, cnt);
	}
	return 0;
}